Limit

1.0Introduction:

The concept of limit of a function is one of the fundamental ideas that distinguishes calculus from algebra and trigonometry. We use limits to describe the way a function $f$ varies. Some functions vary continuously; small changes in $x$ produce only small changes in $f (x)$ . Other functions can have values that jump or vary erratically. We also use limits to define tangent lines to graphs of functions. This geometric application leads at once to the important concept of derivative of a function

2.0Definition:

Let $f (x)$ be defined on an open interval about ' $a$ ' except possibly at ' $a$ ' itself. If $f (x)$ gets arbitrarily close to L (a finite number) for all $x$ sufficiently close to ' $a$ ' we say that $f (x)$ approaches the limit L as $x$ approaches ' $a$ ' and we write $lim_{x \to a} f (x) = L$ and say "the limit of $f (x)$ , as $x$ approaches a, equals $L$ ". This implies if we can make the value of $f (x)$ arbitrarily close to L (as close to L as we like) by taking $x$ to be sufficiently close to a (on either side of a) but not equal to a.

3.0Left Hand Limit and Right Hand Limit of A Function:

The value to which $f (x)$ approaches, as $x$ tends to ' $a$ ' from the left hand side ( $x \to a^{-}$ ) is called left hand limit of $f (x)$ at $x = a$ . Symbolically, LHL $= lim_{x \to a^{-}} f (x) = lim_{h \to 0} f (a - h)$ . The value to which $f (x)$ approaches, as $x$ tends to ' $a^{'}$ ' from the right hand side $(x \to a^{+})$ is called right hand limit of $f (x)$ at $x = a$ . Symbolically, RHL $= lim_{x \to a^{+}} f (x) = lim_{h \to 0} f (a + h)$ . Limit of a function $f (x)$ is said to exist as, $x \to a$ when $lim_{x \to a^{-}} f (x) = lim_{x \to a^{+}} f (x) =$ Finite quantity.

Example:

Graph of $y = f (x)$

Fig. 1 $lim_{x \to - 1^{+}} f (x) = lim_{h \to 0} f (- 1 + h) = f (- 1^{+}) = - 1$ $lim_{x \to 0^{-}} f (x) = lim_{h \to 0} f (0 - h) = f (0^{-}) = 0$ $lim_{x \to 1^{-}} f (x) = lim_{h \to 0} f (1 - h) = f (1^{-}) = - 1$ $lim_{x \to 1^{+}} f (x) = lim_{h \to 0} f (1 + h) = f (1^{+}) = 0$ $lim_{x \to 2^{-}} f (x) = lim_{h \to 0} f (2 - h) = f (2^{-}) = 1$ $lim_{x \to 0} f (x) = 0$ and $lim_{x \to 1} f (x)$ does not exist.

Important Note:

In $lim_{x \to a} f (x), x \neq = a x \to a$ necessarily implies. That is while evaluating limit at $x = a$ , we are not concerned with the value of the function at $x = a$ . In fact, the function may or may not be defined at $x = a$ . Also, it is necessary to note that if $f (x)$ is defined only on one side of ' $x = a$ ', one sided limit are good enough to establish the existence of limits, $&$ if $f (x)$ is defined on either side of ' $a$ ' both sided limits are to be considered. As in $lim_{x \to 1} cos^{- 1} x = 0$ , though $f (x)$ is not defined for $x > 1$ , even in its immediate vicinity.

Illustration 1:

Consider the adjacent graph of $y = f (x)$ Find the following:

(a)

lim_{x \to 0^{-}} f (x)

(b)

lim_{x \to 0^{+}} f (x)

(c)

lim_{x \to 1^{-}} f (x)

(d)

lim_{x \to 1^{+}} f (x)

(e)

lim_{x \to 2^{-}} f (x)

(f)

lim_{x \to 2^{+}} f (x)

(g)

lim_{x \to 3^{-}} f (x)

(h)

lim_{x \to 3^{+}} f (x)

(i)

lim_{x \to 4^{-}} f (x)

(j)

lim_{x \to 4^{+}} f (x)

(k)

lim_{x \to \infty} f (x) = 2

(l)

lim_{x \to 6^{-}} f (x) = - \infty

Solution:

(a) As $x \to 0^{-}$ : limit does not exist (the function is not defined to the left of $x = 0$ ) (b) As $x \to 0^{+} : f (x) \to - 1 \Rightarrow lim_{x \to 0^{+}} f (x) = - 1$ . (c) As $x \to 1^{-} : f (x) \to 1 \Rightarrow lim_{x \to 1^{-}} f (x) = 1$ . (d) As $x \to 1^{+} : f (x) \to 2 \Rightarrow lim_{x \to 1^{+}} f (x) = 2$ . (e) As $x \to 2^{-} : f (x) \to 3 \Rightarrow lim_{x \to 2^{-}} f (x) = 3$ . (f) As $x \to 2^{+} : f (x) \to 3 \Rightarrow lim_{x \to 2^{-}} f (x) = 3$ . (g) As $x \to 3^{-} : f (x) \to 2 \Rightarrow lim_{x \to 3^{-}} f (x) = 2$ . (h) As $x \to 3^{+} : f (x) \to 3 \Rightarrow lim_{x \to 3^{+}} f (x) = 3$ . (i) As $x \to 4^{-} : f (x) \to 4 \Rightarrow lim_{x \to 4^{-}} f (x) = 4$ . (j) As $x \to 4^{+} : f (x) \to 4 \Rightarrow lim_{x \to 4^{+}} f (x) = 4$ . (k) As $x \to \infty : f (x) \to 2 \Rightarrow lim_{x \to \infty} f (x) = 2$ . (l) As $x \to 6^{-}, f (x) \to - \infty \Rightarrow lim_{x \to 6^{-}} f (x) = - \infty$ limit does not exist because it is not finite.

Illustration 2:

Let $[t$ ] denote the greatest integer $\leq t$ . If for some $λ \in R - {0, 1}, lim_{x \to 0} \frac{1 - x + ∣ x ∣}{λ - x + [ x ]} = L$ , then $L$ is equal to : (A) 1 (B) 2 (C) $\frac{1}{2}$ (D) 0

Ans. (B)

Solution:

LHL: $lim_{x \to 0^{-}} \frac{1 - x - x}{λ - x - 1} = \frac{1}{λ - 1}$ RHL : $lim_{x \to 0^{+}} \frac{1 - x + x}{λ - x + 1} = \frac{1}{λ}$ For existence of limit LHL = RHL $\Rightarrow \frac{1}{∣ λ - 1∣} = \frac{1}{∣ λ ∣} \Rightarrow λ = \frac{1}{2}$ $∴ L = \frac{1}{∣ λ ∣} = 2$

Illustration 3:

Let $a$ be an integer such that $lim_{x \to 7} \frac{18 - [ 1 - x ]}{[ x - 3 a ]}$ exists, where $[t]$ is greatest integer $\leq t$ . Then $a$ is equal to : (A) -6 (B) -2 (C) 2 (D) 6

Ans. (A)

Solution:

$lim_{x \to 7} \frac{18 - [ 1 - x ]}{[ x ] - 3 a}$ L.H.L. $lim_{x \to 7 -} \frac{18 - [ 1 - x ]}{[ x ] - 3 a}$ $= \frac{18 - ( - 6 )}{6 - 3 a}$ $= \frac{24}{6 - 3 a}$ R.H.L. $lim_{x \to 7 +} \frac{18 - [ 1 - x ]}{[ x ] - 3 a}$ $= \frac{18 - ( - 7 )}{7 - 3 a}$ $= \frac{25}{7 - 3 a}$ Now L.H.L. = R.H.L. $\frac{24}{6 - 3 a} = \frac{25}{7 - 3 a}$ $\Rightarrow 168 - 72 a = 150 - 75 a$ $\Rightarrow 18 = - 3 a$ $\Rightarrow a = - 6$

Illustration 4:

Let $f (x) = {x - 1, 2 x, x is even x is odd, x \in N$ . If for some $a \in N, f (ff (a))) = 21$ , then $lim_{x \to a^{-}} {\frac{∣ x ∣ ^{3}}{a} - [\frac{x}{a}]}$ where $[t]$ denotes the greatest integer less than or equal to $t$ , is equal to : (A) 121 (B) 144 (C) 169 (D) 225

Ans. (B)

Solution:

$f (x) = {x - 1, 2 x, x is even x is odd,$ $f (f (f (a))) = 21$ C-1: If $a =$ even

f (a) = a - 1 = odd f (f (a)) = 2 (a - 1) = even f (f (f (a))) = 2 a - 3 = 21 \Rightarrow a = 12

C-2: If $a =$ odd $f (a) = 2 a =$ even $f (f (a)) = 2 a - 1 =$ odd $f (f (f (a))) = 4 a - 2 = 21$ (Not possible) Hence $a = 12$ Now

lim_{x \to 1 2^{-}} (\frac{∣ x ∣ ^{3}}{2} - [\frac{x}{12}]) lim_{x \to 1 2^{-}} \frac{∣ x ∣ ^{3}}{12} - lim_{x \to 1 2^{-}} [\frac{x}{12}] = 144 - 0 = 144.

Illustration 5:

Let ${x}$ denote the fractional part of $x$ and $f (x) = \frac{c o s ^{- 1} 1 - { x } ^{2} s i n ^{- 1} ( 1 - { x })}{{ x } - { x } ^{3}}, x \neq = 0$ . If $L$ and $R$ respectively denotes the left hand limit and the right hand limit of $f (x)$ at $x = 0$ , then $\frac{32}{π ^{2}} (L^{2} + R^{2})$ is equal to .

Ans. (18)

Solution:

Finding right hand limit

lim_{x \to 0^{+}} f (x) = lim_{h \to \infty} f (0 + h) = lim_{h \to 0} f (h) = lim_{h \to 0} \frac{c o s ^{- 1} ( 1 - h ^{2} ) s i n ^{- 1} ( 1 - h )}{h ( 1 - h ^{2} )} = lim_{h \to 0} \frac{c o s ^{- 1} ( 1 - h ^{2} )}{h} (\frac{s i n ^{- 1} 1}{1})

Let $cos^{- 1} (1 - h^{2}) = θ \Rightarrow cos θ = 1 - h^{2}$ $= \frac{π}{2} lim_{θ \to 0} \frac{θ}{1 - c o s θ}$ $= \frac{π}{2} lim_{θ \to 0} \frac{1}{\frac{1 - c o s θ}{θ ^{2}}}$ $= \frac{π}{2} \frac{1}{1/2}$ $R = \frac{π}{2}$ Now finding left hand limit

L = lim_{x \to 0^{-}} f (x) = lim_{h \to 0} f (- h)

$= lim_{h \to 0} \frac{c o s ^{- 1} ( 1 - { - h } ^{2} ) s i n ^{- 1} ( 1 - { - h })}{{ - h } - { - h } ^{3}}$ $= lim_{h \to 0} \frac{c o s ^{- 1} ( 1 - ( - h + 1 ) ^{2} ) s i n ^{- 1} ( 1 - ( - h + 1 ))}{( - h + 1 ) - ( - h + 1 ) ^{3}}$ $= lim_{h \to 0} \frac{c o s ^{- 1} ( - h ^{2} + 2 h ) s i n ^{- 1} h}{( 1 - h ) ( 1 - ( 1 - h ) ^{2} )}$ $= lim_{h \to 0} (\frac{π}{2}) \frac{s i n ^{- 1} h}{( 1 - ( 1 - h ) ^{2} )}$ $\frac{π}{2} lim_{x \to 0} (\frac{s i n ^{- 1} h}{- h ^{2} + 2 h})$ $= \frac{π}{2} lim_{h \to 0} (\frac{s i n ^{- 1} h}{h}) (\frac{1}{- h + 2})$ $L = \frac{π}{4}$ $\frac{32}{π ^{2}} (L^{2} + R^{2}) = \frac{32}{π ^{2}} (\frac{π ^{2}}{2} + \frac{π ^{2}}{16})$ $= 16 + 2$ $= 18$

4.0Fundamental Theorems on Limits:

Let $lim_{x \to a} f (x) = l & lim_{x \to a} g (x) = m$ . If $l & m$ exist finitely then : (a) Sum rule: $lim_{x \to a} {f (x) + g (x)} = l + m$ (b) Difference rule: $lim_{x \to a} {f (x) - g (x)} = l - m$ (c) Product rule: $lim_{x \to a} f (x) \cdot g (x) = l \cdot m$ (d) Quotient rule: $lim_{x \to a} \frac{f ( x )}{g ( x )} = \frac{l}{m}$ , provided $m \neq = 0$ (e) Constant multiple rule : $lim_{x \to a} k f (x) = k lim_{x \to a} f (x)$ ; where $k$ is constant. (f) Power rule: If $m$ and $n$ are integers then $lim_{x \to a} [f (x)]^{m / n} = l^{m / n}$ provided $l^{m / n}$ is a real number. (g) $lim_{x \to a} f [g (x)] = f (lim_{x \to a} g (x)) = f (m)$ ; provided $f (x)$ is continuous at $x = m$ .

For example : $lim_{x \to a} ln (g (x)) = ln [lim_{x \to a} g (x)]$ $= ℓ n (m)$ ; provided $ℓ n x$ is continuous at $x = m, m = lim_{x \to a} g (x)$ . 5. Indeterminate Forms: $\frac{0}{0}, \frac{\infty}{\infty}, \infty - \infty, 0 \times \infty, 1^{\infty}, 0^{0}, \infty^{0}$ Initially we will deal with first five forms only and the other two forms will come up after we have gone through differentiation. Note : (i) Here 0,1 are not exact, in fact both are approaching to their corresponding values. (ii) We cannot plot $\infty$ on the paper. Infinity ( $\infty$ ) is a symbol & not a number It does not obey the laws of elementary algebra, (a) $\infty + \infty \to \infty$ (b) $\infty \times \infty \to \infty$ (c) $\infty^{\infty} \to \infty$ (d) $0^{\infty} \to 0$ 6. General Methods to be used to Evaluate Limits:

(a) Factorization:

Important factors:

(i) $x^{n} - a^{n} = (x - a) (x^{n - 1} + a x^{n - 2} +$ $+ a^{n - 1}), n \in N$ (ii) $x^{n} + a^{n} = (x + a) (x^{n - 1} - a x^{n - 2} +$ $+ a^{n - 1}), n$ is an odd natural number. Note: $lim_{x \to a} \frac{x ^{n} - a ^{n}}{x - a} = n a^{n - 1}$

Illustration 6:

Evaluate: $lim_{x \to 2} [\frac{1}{x - 2} - \frac{2 ( 2 x - 3 )}{x ^{3} - 3 x ^{2} + 2 x}]$

Solution:

We have

lim_{x \to 2} [\frac{1}{x - 2} - \frac{2 ( 2 x - 3 )}{x ^{3} - 3 x ^{2} + 2 x}] = lim_{x \to 2} [\frac{1}{x - 2} - \frac{2 ( 2 x - 3 )}{x ( x - 1 ) ( x - 2 )}] = lim_{x \to 2} [\frac{x ( x - 1 ) - 2 ( 2 x - 3 )}{x ( x - 1 ) ( x - 2 )}] = lim_{x \to 2} [\frac{x ^{2} - 5 x + 6}{x ( x - 1 ) ( x - 2 )}] = lim_{x \to 2} [\frac{( x - 2 ) ( x - 3 )}{x ( x - 1 ) ( x - 2 )}] = lim_{x \to 2} [\frac{x - 3}{x ( x - 1 )}] = - \frac{1}{2}

5.0(b) Rationalization or Double Rationalization:

Illustration 7:

Evaluate: $lim_{x \to 1} \frac{4 - 15 x + 1}{2 - 3 x + 1}$

Solution:

$lim_{x \to 1} \frac{4 - 15 x + 1}{2 - 3 x + 1} = lim_{x \to 1} \frac{( 4 - 15 x + 1 ) ( 2 + 3 x + 1 ) ( 4 + 15 x + 1 )}{( 2 - 3 x + 1 ) ( 4 + 15 x + 1 ) ( 2 + 3 x + 1 )}$ $= lim_{x \to 1} \frac{( 15 - 15 x )}{( 3 - 3 x )} \times \frac{2 + 3 x + 1}{4 + 15 x + 1} = \frac{5}{2}$

Illustration 8:

Evaluate: $lim_{x \to 1} (\frac{x ^{2} + 8 - 10 - x ^{2}}{x ^{2} + 3 - 5 - x ^{2}})$

Solution:

This is of the form $\frac{3 - 3}{2 - 2} = \frac{0}{0}$ if we put $x = 1$ To eliminate the $\frac{0}{0}$ factor, multiply by the conjugate of numerator and the conjugate of the denominator $∴$ Limit $= lim_{x \to 1} (x^{2} + 8 - 10 - x^{2}) \frac{( x ^{2} + 8 + 10 - x ^{2} )}{( x ^{2} + 8 + 10 - x ^{2} )} \times \frac{( x ^{2} + 3 + 5 - x ^{2} )}{( x ^{2} + 3 + 5 - x ^{2} ) ( x ^{2} + 3 - 5 - x ^{2} )}$ $= lim_{x \to 1} \frac{x ^{2} + 3 + 5 - x ^{2}}{x ^{2} + 8 + 10 - x ^{2}} \times \frac{( x ^{2} + 8 ) - ( 10 - x ^{2} )}{( x ^{2} + 3 ) - ( 5 - x ^{2} )} = lim_{x \to 1} (\frac{x ^{2} + 3 + 5 - x ^{2}}{x ^{2} + 8 + 10 - x ^{2}}) \times 1 = \frac{2 + 2}{3 + 3} = \frac{2}{3}$

Illustration 9:

If $a = lim_{x \to 0} \frac{1 + 1 + x ^{4} - 2}{x ^{4}}$ and $b = lim_{x \to 0} \frac{s i n ^{2} x}{2 - 1 + c o s x}$ , then the value of $a b^{3}$ is : (A) 36 (B) 32 (C) 25 (D) 30

Ans. (B)

Solution:

a = lim_{x \to 0} \frac{1 + 1 + x ^{4} - 2}{x ^{4}} = lim_{x \to 0} \frac{1 + x ^{4} - 1}{x ^{4} ( 1 + 1 + x ^{4} + 2 )} = lim_{x \to 0} \frac{x ^{4}}{x ^{4} ( 1 + 1 + x ^{4} + 2 ) ( 1 + x ^{4} + 1 )}

Applying limit $a = \frac{1}{4 2}$ $b = lim_{x \to 0} \frac{s i n ^{2} x}{2 - 1 + c o s x}$ $= lim_{x \to 0} \frac{( 1 - c o s ^{2} x ) ( 2 + 1 + c o s x )}{2 - ( 1 + c o s x )}$ $b = lim_{x \to 0} (1 + cos x) (2 + 1 + cos x)$ Applying limits $b = 2 (2 + 2) = 42$ Now, $a b^{3} = \frac{1}{4 2} \times (42)^{3} = 32$ (c) Limit when $x \to \infty$ : (i) Divide by greatest power of $x$ in numerator and denominator. (ii) Put $x = 1/ y$ and apply $y \to 0$

Illustration 10:

Evaluate: $lim_{x \to \infty} \frac{x ^{2} + x + 1}{3 x ^{2} + 2 x - 5}$

Solution:

$lim_{x \to \infty} \frac{x ^{2} + x + 1}{3 x ^{2} + 2 x - 5}, (\frac{\infty}{\infty}$ form $)$ Put $x = \frac{1}{y}$ Limit $= lim_{y \to 0} \frac{1 + y + y ^{2}}{3 + 2 y - 5 y ^{2}} = \frac{1}{3}$

Illustration 11:

If $lim_{x \to \infty} (\frac{x ^{3} + 1}{x ^{2} + 1} - (a x + b)) = 2$ , then (A) $a = 1, b = 1$ (B) $a = 1, b = 2$ (C) $a = 1, b = - 2$ (D) none of these

Ans. (C)

Solution:

lim_{x \to \infty} (\frac{x ^{3} + 1}{x ^{2} + 1} - (a x + b)) = 2 \Rightarrow lim_{x \to \infty} \frac{x ^{3} ( 1 - a ) - b x ^{2} - a x + ( 1 - b )}{x ^{2} + 1} = 2 \Rightarrow lim_{x \to \infty} \frac{x ( 1 - a ) - b - \frac{a}{x} + \frac{( 1 - b )}{x ^{2}}}{1 + \frac{1}{x ^{2}}} = 2 \Rightarrow 1 - a = 0, - b = 2 \Rightarrow a = 1, b = - 2

Illustration 12:

If $lim_{n \to \infty} (n^{2} - n - 1 + n α + β) = 0$ then $8 (α + β)$ is equal to : (A) 4 (B) -8 (C) -4 (D) 8

Ans. (C)

Solution:

$lim_{n \to \infty} n (1 - \frac{n + 1}{n ^{2}})^{\frac{1}{2}} + α n + β = 0$ $lim_{n \to \infty} {{1 - \frac{1}{2} (\frac{n + 1}{n ^{2}}) + \frac{( \frac{1}{2} ) ( - \frac{1}{2} )}{2 !} (\frac{n + 1}{n ^{2}})^{2} + \dots .} + α n + β = 0$ $lim_{n \to \infty} n - \frac{1}{2} + \frac{1}{n} + \dots . + n α + β = 0$ $α = - 1, β = \frac{1}{2}$ $8 (α + β) = - 4$ (d) Squeeze Play Theorem (Sandwich Theorem):

Statement: If $f (x) \leq g (x) \leq h (x); \forall x$ in the neighbourhood at $x = a$ and $lim_{x \to a} f (x) = ℓ = lim_{x \to a} h (x) then, lim_{x \to a} g (x) = ℓ$

Ex. $1 lim_{x \to 0} x^{2} sin \frac{1}{x} = 0$ ,

∵ sin (\frac{1}{x})

lies between

- 1&1

\Rightarrow - x^{2} \leq x^{2} sin \frac{1}{x} \leq x^{2} \Rightarrow lim_{x \to 0} x^{2} sin \frac{1}{x} = 0 as lim_{x \to 0} (- x^{2}) = lim_{x \to 0} x^{2} = 0

Ex. $2 lim_{x \to 0} x sin \frac{1}{x} = 0$

∵ sin (\frac{1}{x}) lies between - 11 \Rightarrow - x \leq x sin \frac{1}{x} \leq x \Rightarrow lim_{x \to 0} x sin \frac{1}{x} = 0 as lim_{x \to 0} (- x) = lim_{x \to 0} x = 0

Illustration 13:

Evaluate: $lim_{n \to \infty} \frac{[ x ] + [ 2 x ] + [ 3 x ] + \dots .. [ n x ]}{n ^{2}}$ (Where [.] denotes the greatest integer function.)

Solution:

We know that $x - 1 < [x] \leq x$ $\Rightarrow x + 2 x + \dots . n x - n < \sum_{r = 1}^{n} [r x] \leq x + 2 x + \dots\dots\dots + n x$ $\Rightarrow \frac{x n}{2} (n + 1) - n < \sum_{r = 1}^{n} [r x] \leq \frac{x . n ( n + 1 )}{2}$ $\Rightarrow \frac{x}{2} (1 + \frac{1}{n}) - \frac{1}{n} < \frac{1}{n ^{2}} \sum_{r = 1}^{n} [r x] \leq \frac{x}{2} (1 + \frac{1}{n})$ Now, $lim_{n \to \infty} \frac{x}{2} (1 + \frac{1}{n}) = \frac{x}{2}$ and $lim_{n \to \infty} \frac{x}{2} (1 + \frac{1}{n}) - \frac{1}{n} = \frac{x}{2}$ Thus, $lim_{n \to \infty} \frac{[ x ] + [ 2 x ] + \dots\dots . + [ n x ]}{n ^{2}} = \frac{x}{2}$

Illustration 14:

$lim_{n \to \infty} (1 + \frac{1 + \frac{1}{2} + \dots\dots\dots + \frac{1}{n}}{n ^{2}})^{n}$ is equal to : (A) $\frac{1}{2}$ (B) 0 (C) $\frac{1}{e}$ (D) 1

Ans. (D)

Solution:

Given limit is of $1^{\infty}$ form So, $l = exp (lim_{n \to \infty} \frac{1 + \frac{1}{2} + \frac{1}{3} + \dots\dots\dots + \frac{1}{n}}{n})$ Now, $0 \leq 1 + \frac{1}{2} + \frac{1}{3} + \dots . + \frac{1}{n} \leq 1 + \frac{1}{2} + \frac{1}{3} + \dots . + \frac{1}{n}$ So, $l = exp (0)$ (from sandwich theorem) = 1

Illustration 15:

Let $f :\to R \to (0, \infty)$ be strictly increasing function such that $lim_{x \to \infty} \frac{f ( 7 x )}{f ( x )} = 1$ . Then, the value of $lim_{x \to \infty} [\frac{f ( 5 x )}{f ( x )} - 1]$ is equal to (A) 4 (B) 0 (C) $7/5$ (D) 1

Ans. (B)

Solution:

$f : R \to (0, \infty)$ $lim_{x \to \infty} \frac{f ( 7 x )}{f ( x )} = 1$ $∵ f$ is increasing $∴ f (x) < f (5 x) < f (7 x)$ $∵ \frac{f ( x )}{f ( x )} < \frac{f ( 5 x )}{f ( x )} < \frac{f ( 7 x )}{f ( x )}$ $1 < lim_{x \to \infty} \frac{f ( 5 x )}{f ( x )} < 1$ $∴ [\frac{f ( 5 x )}{f ( x )} - 1]$ $\Rightarrow 1 - 1 = 0$

6.0Limit of Trigonometric Functions:

$lim_{x \to 0} \frac{s i n x}{x} = 1 = lim_{x \to 0} \frac{t a n x}{x} = lim_{x \to 0} \frac{t a n ^{- 1} x}{x} = lim_{x \to 0} \frac{s i n ^{- 1} x}{x}$ [where $x$ is measured in radians] If $lim_{x \to a} f (x) = 0$ , then $lim_{x \to a} \frac{s i n f ( x )}{f ( x )} = 1$ , e.g., $lim_{x \to 1} \frac{s i n ( l n x )}{( l n x )} = 1$

Illustration 16:

Evaluate: $lim_{x \to 0} \frac{x ^{3} c o t x}{1 - c o s x}$

Solution:

$lim_{x \to 0} \frac{x ^{3} c o s x}{s i n x ( 1 - c o s x )} = lim_{x \to 0} \frac{x ^{3} c o s x ( 1 + c o s x )}{s i n x \cdot s i n ^{2} x} = lim_{x \to 0} \frac{x ^{3}}{s i n ^{3} x} \cdot cos x (1 + cos x) = 2$

Illustration 17:

Evaluate: $lim_{x \to 0} \frac{( 2 + x ) s i n ( 2 + x ) - 2 s i n 2}{x}$

Solution:

lim_{x \to 0} \frac{2 ( s i n ( 2 + x ) - s i n 2 ) + x s i n ( 2 + x )}{x} = lim_{x \to 0} (\frac{2 \cdot 2 \cdot c o s ( 2 + \frac{x}{2} ) s i n \frac{x}{2}}{x} + sin (2 + x)) = lim_{x \to 0} \frac{2 c o s ( 2 + \frac{x}{2} ) s i n \frac{x}{2}}{\frac{x}{2}} + lim_{x \to 0} sin (2 + x) = 2 cos 2 + sin 2

Illustration 18:

Evaluate: $lim_{n \to \infty} \frac{s i n \frac{a}{n}}{t a n \frac{b}{n + 1}}$

Solution:

As $n \to \infty, \frac{1}{n} \to 0$ and $\frac{a}{n}$ also tends to zero $sin \frac{a}{n}$ should be written as $\frac{s i n \frac{a}{n}}{\frac{a}{n}}$ so that it looks like $lim_{θ \to 0} \frac{s i n θ}{θ}$ The given limit $= lim_{n \to \infty} (\frac{s i n \frac{a}{n}}{\frac{a}{n}}) (\frac{\frac{b}{n + 1}}{t a n \frac{b}{n + 1}}) \cdot \frac{a ( n + 1 )}{n \cdot b}$ $= lim_{n \to \infty} (\frac{s i n \frac{a}{n}}{\frac{a}{n}}) (\frac{\frac{b}{n + 1}}{t a n \frac{b}{n + 1}}) \cdot \frac{a}{b} (1 + \frac{1}{n}) = 1 \times 1 \times \frac{a}{b} \times 1 = \frac{a}{b}$

7.0Illustration 19:

$lim_{x \to 0} \frac{x c o t ( 4 x )}{s i n ^{2} x c o t ^{2} ( 2 x )}$ is equal to :- (A) 2 (B) 0 (C) 4 (D) 1

Ans. (D)

Solution:

$lim_{x \to 0} \frac{x t a n ^{2} 2 x}{t a n 4 x s i n ^{2} x} = lim_{x \to 0} \frac{x ( \frac{t a n ^{2} 2 x}{4 x ^{2}} ) 4 x ^{2}}{( \frac{t a n 4 x}{4 x} ) 4 x ( \frac{s i n ^{2} x}{x ^{2}} ) x ^{2}} = 1$

Illustration 20:

The value of $lim_{x \to 1} \frac{( x ^{2} - 1 ) s i n ^{2} ( π x )}{x ^{4} - 2 x ^{3} + 2 x - 1}$ is equal to: (A) $\frac{π ^{2}}{6}$ (B) $\frac{π ^{2}}{3}$ (C) $\frac{π ^{2}}{2}$ (D) $π^{2}$

Ans. (D)

Solution:

$lim_{x \to 1} \frac{( x ^{2} - 1 ) s i n ^{2} π x}{( x ^{2} - 1 ) ( x - 1 ) ^{2}} = lim_{x \to 1} (\frac{s i n (( 1 - x ) π ))}{π ( 1 - x )})^{2} π^{2} = π^{2}$

Illustration 21:

The value of the limit $lim_{θ \to 0} \frac{t a n ( π c o s ^{2} θ )}{s i n ( 2 π s i n ^{2} θ )}$ is equal to: (A) $lim_{θ \to 0} \frac{t a n ( π c o s ^{2} θ )}{s i n ( 2 π s i n ^{2} θ )}$ (B) $- \frac{1}{4}$ (C) 0 (D) $\frac{1}{4}$

Ans. (A)

Solution:

$lim_{θ \to 0} \frac{t a n ( π ( 1 - s i n ^{2} θ ) )}{s i n ( 2 π s i n ^{2} θ )}$ $= lim_{θ \to 0} \frac{- t a n ( π s i n ^{2} θ )}{s i n ( 2 π s i n ^{2} θ )}$ $= lim_{θ \to 0} - (\frac{t a n ( π s i n ^{2} θ )}{π s i n ^{2} θ}) (\frac{2 π s i n ^{2} θ}{s i n ( 2 π s i n ^{2} θ )}) \times \frac{1}{2}$ $= \frac{- 1}{2}$

Illustration 22:

If $lim_{x \to 0} {\frac{1}{x ^{8}} (1 - cos \frac{x ^{2}}{2} - cos \frac{x ^{2}}{4} + cos \frac{x ^{2}}{2} cos \frac{x ^{2}}{4})} = 2^{- k}$ , then the value of $k$ is Ans. (8)

Solution:

$lim_{x \to 0} {\frac{1}{x ^{8}} (1 - cos \frac{x ^{2}}{2} - cos \frac{x ^{2}}{4} + cos \frac{x ^{2}}{2} cos \frac{x ^{2}}{4})} = 2^{- k}$ $\Rightarrow lim_{x \to 0} \frac{( 1 - c o s \frac{x ^{2}}{2} )}{4 ( \frac{x ^{2}}{2} ) ^{2}} \frac{( 1 - c o s \frac{x ^{2}}{4} )}{16 ( \frac{x ^{2}}{4} ) ^{2}} = \frac{1}{8} \times \frac{1}{32} = 2^{- k}$ $\Rightarrow 2^{- 8} = 2^{- k} \Rightarrow k = 8$ .

Illustration 23:

If $α$ is the positive root of the equation, $p (x) = x^{2} - x - 2 = 0$ , then $lim_{x \to α^{+}} \frac{1 - c o s ( p ( x ))}{x + α - 4}$ is equal to (A) $\frac{3}{2}$ (B) $\frac{3}{2}$ (C) $\frac{1}{2}$ (D) $\frac{1}{2}$

Ans. (A)

Solution:

$x^{2} - x - 2 = 0$ roots are $2& - 1$ $\Rightarrow lim_{x \to 2^{+}} \frac{1 - c o s ( x ^{2} - x - 2 )}{( x - 2 )}$ $= lim_{x \to 2^{+}} \frac{2 s i n ^{2} \frac{( x ^{2} - x - 2 )}{2}}{( x - 2 )}$ $= lim_{x \to 2^{+}} \frac{2 s i n ( \frac{( x - 2 ) ( x + 1 )}{2} )}{( x - 2 )} = \frac{3}{2}$

Illustration 24:

If $α > β > 0$ are the roots of the equation $a x^{2} + b x + 1 = 0$ , and $lim_{x \to \frac{1}{α}} (\frac{1 - c o s ( x ^{2} + b x + a )}{2 ( 1 - αx ) ^{2}})^{\frac{1}{2}} = \frac{1}{k} (\frac{1}{β} - \frac{1}{α})$ , then $k$ is equal to (A) $2 β$ (B) $2 α$ (C) $α$ (D) $β$

Ans. (B)

Solution:

$α, β$ are roots of $a x^{2} + b x + 1 = 0$ $\frac{1}{α}, \frac{1}{β}$ are roots of $x^{2} + b x + a = 0$ , (by transformation) $x^{2} + b x + a = (x - \frac{1}{α}) (x - \frac{1}{β})$ $lim_{x \to \frac{1}{α}} [\frac{1 - c o s ( x - \frac{1}{α} ) ( x - \frac{1}{β} )}{2 ( 1 - αx ) ^{2}}]^{\frac{1}{2}} = L$ $($ By using $lim_{θ \to 0} \frac{1 - c o s θ}{θ ^{2}} = \frac{1}{2}$ ) $\Rightarrow [\frac{( \frac{1}{α} - \frac{1}{β} ) ^{2}}{4 α ^{2}}]^{\frac{1}{2}} = L$ $\Rightarrow \frac{\frac{1}{β} - \frac{1}{α}}{2 α} = L$ Comparing $k = 2 α$

8.0Limit of Exponential Functions:

(a) $lim_{x \to 0} \frac{a ^{x} - 1}{x} = ℓ$ n $a (a > 0)$ In particular $lim_{x \to 0} \frac{e ^{x} - 1}{x} = 1$ .

In general, if $lim_{x \to a} f (x) = 0$ , then $lim_{x \to a} \frac{a ^{f (x)} - 1}{f ( x )} = ℓ$ n $a, a > 0$

Illustration 25:

Evaluate: $lim_{x \to 0} \frac{e ^{t a n x} - e ^{x}}{t a n x - x}$

Solution:

$lim_{x \to 0} \frac{e ^{t a n x} - e ^{x}}{t a n x - x} = lim_{x \to 0} \frac{e ^{x} \times e ^{(t a n x - x)} - e ^{x}}{t a n x - x}$ $= lim_{x \to 0} \frac{e ^{x} ( e ^{t a n x - x} - 1 )}{t a n x - x} = lim_{x \to 0 y \to 0} \frac{e ^{x} ( e ^{y} - 1 )}{y}$ where $y = tan x - x$ and $lim_{y \to 0} \frac{e ^{y} - 1}{y} = 1$ $= e^{0} \times 1 [$ as $x \to 0, tan x - x \to 0]$ $= 1 \times 1 = 1$ (b) $($ i) $lim_{x \to 0} (1 + x)^{1/ x} = e = lim_{x \to \infty} (1 + \frac{1}{x})^{x}$ (Note: The base and exponent depend on the same variable.) In general, if $lim_{x \to a} f (x) = 0$ , then $lim_{x \to a} (1 + f (x))^{1/ f (x)} = e$ (ii) $lim_{x \to 0} \frac{l n ( 1 + x )}{x} = 1$ (iii) If $lim_{x \to a} f (x) = 1$ and $lim_{x \to a} ϕ (x) = \infty$ , then; $lim_{x \to a} [f (x)]^{ϕ (x)} = e^{k}$ where $k = lim_{x \to a} ϕ (x) [f (x) - 1]$

Illustration 26:

$lim_{x \to 0} \frac{x ( e ^{(1 + x^{2} + x^{4} - 1) / x} - 1 )}{1 + x ^{2} + x ^{4} - 1}$ (A) does not exist. (B) is equal to $e$ . (C) is equal to 0 . (D) is equal to 1 .

Ans. (D)

Solution:

$lim_{x \to 0} \frac{x ( e ^{(1 + x^{2} + x^{4} - 1) / x} - 1 )}{1 + x ^{2} + x ^{4} - 1}$ $∵ lim_{x \to 0} \frac{1 + x ^{2} + x ^{4} - 1}{x}$ (from) $lim_{x \to 0} \frac{( 1 + x ^{2} + x ^{4} ) - 1}{x ( 1 + x ^{2} + x ^{4} + 1}$ $lim_{x \to 0} \frac{x ( 1 + x ^{2} )}{( 1 + x ^{2} + x ^{4} + 1 )} = 0$ So $lim_{x \to 0} \frac{x e ^{(\frac{1 + x ^{2} + x ^{4} - 1}{x})} - 1}{1 + x ^{2} + x ^{4} - 1} (\frac{0}{0}$ from $)$ $lim_{x \to 0} \frac{e ^{\frac{1 + x ^{2} + x ^{4} - 1}{x}} - 1}{( \frac{1 + x ^{2} + x ^{4} - 1}{x} )} = 1$

Illustration 27:

Evaluate: $lim_{x \to 1} (lo g_{3} 3 x)^{l o g_{x} 3}$

Solution:

$lim_{x \to 1} (lo g_{3} 3 x)^{l o g_{x} 3} = lim_{x \to 1} (lo g_{3} 3 + lo g_{3} x)^{l o g_{x} 3}$ $= lim_{x \to 1} (1 + lo g_{3} x)^{1/ l o g_{3} x} = e ∵ lo g_{b} a = \frac{1}{l o g _{a} b}$

Illustration 28:

Evaluate: $lim_{x \to 0} \frac{x l n ( 1 + 2 t a n x )}{1 - c o s x}$

Solution:

$lim_{x \to 0} \frac{x l n ( 1 + 2 t a n x )}{1 - c o s x} = lim_{x \to 0} \frac{x l n ( 1 + 2 t a n x )}{\frac{1 - c o s x}{x ^{2}} \cdot x ^{2}} \cdot \frac{2 t a n x}{2 t a n x} = 4$

Illustration 29:

$lim_{x \to 0} \frac{l o g _{e} ( 1 - x + x ^{2} ) + l o g _{e} ( 1 + x + x ^{2} )}{s e c x - c o s x}$ is equal to : (A) 1 (B) -1 (C) e (D) 0

Ans. (A)

Solution:

$lim_{x \to 0} \frac{( l n ( 1 + x ^{2} + x ^{4} ) ) c o s x}{1 - c o s ^{2} x}$ $lim_{x \to 0} \frac{( \frac{l n ( 1 + x ^{2} + x ^{4} )}{x ^{2} + x ^{4}} ) x ^{2} ( 1 + x ^{2} ) c o s x}{( \frac{s i n ^{2} x}{x ^{2}} ) x ^{2}} = 1$

Illustration 30:

Evaluate: $lim_{x \to \infty} (\frac{2 x ^{2} - 1}{2 x ^{2} + 3})^{4 x^{2} + 2}$

Solution:

Since it is in the form of $1^{\infty}$ $lim_{x \to \infty} (\frac{2 x ^{2} - 1}{2 x ^{2} + 3})^{4 x^{2} + 2} = e^{l i m_{x \to \infty}} (\frac{2 x ^{2} - 1 - 2 x ^{2} - 3}{2 x ^{2} + 3}) (4 x^{2} + 2) = e^{- 8}$

Illustration 31:

Evaluate: $lim_{x \to \infty} (\frac{7 x ^{2} + 1}{5 x ^{2} - 1})^{\frac{x ^{5}}{1 - x ^{3}}}$

Solution:

Here $f (x) = \frac{7 x ^{2} + 1}{5 x ^{2} - 1}, ϕ (x) = \frac{x ^{5}}{1 - x ^{3}} = \frac{x ^{2} \cdot x ^{3}}{1 - x ^{3}} = \frac{x ^{2}}{\frac{1}{x ^{3}} - 1}$ $∴ lim_{x \to \infty} f (x) = \frac{7}{5} & lim_{x \to \infty} ϕ (x) \to - \infty$ $\Rightarrow lim_{x \to \infty} (f (x))^{ϕ (x)} = (\frac{7}{5})^{- \infty} = 0$

Illustration 32:

$lim_{x \to 0} (\frac{1 + t a n x}{1 + s i n x})^{cosec x}$ is equal to (A) $e$ (B) $\frac{1}{e}$ (C) 1 (D) None of these

Ans. (C)

Solution:

Given limit $= lim_{x \to 0} [(1 + tan x)^{cosec x} \times 1/ (1 + sin x)^{cosec x}]$ $= lim_{x \to 0} [{(1 + tan x)^{c o t x}}^{s e c x} \times {1/ (1 + sin x)^{cosec x}}] = e^{s e c 0} \cdot \frac{1}{e} = e \cdot \frac{1}{e} = 1$

Illustration 33:

If $lim_{x \to \infty} (1 + \frac{a}{x} + \frac{b}{x ^{2}})^{2 x} = e^{2}$ then the values of $a$ and $b$ are (A) $a = 1, b = 2$ (B) $a = 1, b \in R$ (C) $a \in R, b = 2$ (D) $a \in R, b \in R$

Ans. (B)

Solution:

Since, $lim_{x \to \infty} (1 + \frac{a}{x} + \frac{b}{x ^{2}}) = 1$ $∴ lim_{x \to \infty} [(1 + \frac{a x + b}{x ^{2}})^{\frac{x ^{2}}{a x + b}}]^{\frac{2 ( a x + b )}{x}} = e^{2} \Rightarrow lim_{x \to \infty} \frac{2 ( a x + b )}{x} = e^{2} \Rightarrow lim_{x \to \infty} \frac{2 ( a x + b )}{x} = 2 \Rightarrow 2 a = 2 \Rightarrow a = 1$ Thus $a = 1$ and $b \in R$ .

Illustration 34:

$Lim_{x \to 0} \frac{e - ( 1 + 2 x ) ^{\frac{1}{2 x}}}{x}$ is equal to : (A) $e$ (B) $\frac{- 2}{e}$ (C) 0 (D) $e - e^{2}$

Ans. (A)

Solution:

$Lim_{x \to 0} \frac{e - e ^{\frac{1}{2 x} l n (1 + 2 x)}}{x} = Lim_{x \to 0} (- e) \frac{( e ^{\frac{l n ( 1 + 2 x )}{2 x} - 1} - 1 )}{x}$ $= Lim_{x \to 0} (- e) \frac{l n ( 1 + 2 x ) - 2 x}{2 x ^{2}}$ $= (- e) \times (- 1) \frac{4}{2 \times 2} = e$

Illustration 35:

Given $h (x) = lim_{n \to \infty} \frac{x ^{2 n} f ( x ) + g ( x )}{x ^{2 n} + 1}, f (2) = 5$ and $g (\frac{1}{2}) = - 3$ , then value of $h (2) - 2 h (\frac{1}{2})$ is (Given $f (x)$ and $g (x)$ are bounded functions)

Ans. (11)

Solution:

$h (x) = lim_{n \to \infty} \frac{x ^{2 n} f ( x ) + g ( x )}{x ^{2 n} + 1} = ⎩ ⎨ ⎧ g (x) \frac{f ( x ) + g ( x )}{2} f (x) ∣ x ∣ < 1 ∣ x ∣ = 1 ∣ x ∣ > 1$ $∴ h (2) - 2 h (\frac{1}{2}) = f (2) - 2 g (\frac{1}{2}) = 5 - 2 (- 3) = 11$

9.0Limit using Series Expansion:

Expansion of function like binomial expansion, exponential & logarithmic expansion, expansion of $sin x, cos x, tan x should$ be remembered by heart which are given below : (a) $a^{x} = 1 + \frac{x ℓ na}{1 !} + \frac{x ^{2} ℓ n ^{2} a}{2 !} + \frac{x ^{3} ℓ n ^{3} a}{3 !} + \dots, x \in R, a > 0, a \neq = 1$ (b) $e^{x} = 1 + \frac{x}{1 !} + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !} + \dots, x \in R$ (c) $ln (1 + x) = x - \frac{x ^{2}}{2} + \frac{x ^{3}}{3} - \frac{x ^{4}}{4} + \dots$ for $- 1 < x \leq 1$ (d) $sin x = x - \frac{x ^{3}}{3 !} + \frac{x ^{5}}{5 !} - \frac{x ^{7}}{7 !} + \dots, x \in R$ (e) $cos x = 1 - \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} - \frac{x ^{6}}{6 !} + \dots, x \in R$ (f) $tan x = x + \frac{x ^{3}}{3} + \frac{2 x ^{5}}{15} + \dots, - \frac{π}{2} < x < \frac{π}{2}$ (g) $tan^{- 1} x = x - \frac{x ^{3}}{3} + \frac{x ^{5}}{5} - \frac{x ^{7}}{7} + \dots, x \in (- 1, 1)$ (h) $sin^{- 1} x = x + \frac{1 ^{2}}{3 !} x^{3} + \frac{1 ^{2} \cdot 3 ^{2}}{5 !} x^{5} + \frac{1 ^{2} \cdot 3 ^{2} \cdot 5 ^{2}}{7 !} x^{7} + \dots, x \in (- 1, 1)$ (i) $sec^{- 1} x = 1 + \frac{x ^{2}}{2 !} + \frac{5 x ^{4}}{4 !} + \frac{61 x ^{6}}{6 !} + \dots, x \in (- \infty, - 1) \cup (1, \infty)$ (j) $(1 + x)^{n} = 1 + n x + \frac{n ( n - 1 )}{2 !} x^{2} + \dots, n \in R, x \in (- 1, 1)$

Illustration 36:

$lim_{x \to 0} \frac{e ^{x} - e ^{- x} - 2 x}{x - s i n x}$ Solution: $lim_{x \to 0} \frac{e ^{x} - e ^{- x} - 2 x}{x - s i n x} \Rightarrow lim_{x \to 0} \frac{1 + x + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !} + \dots\dots - ( 1 - x + \frac{x ^{2}}{2 !} - \frac{x ^{3}}{3 !} + \dots .. ) - 2 x}{x - ( x - \frac{x ^{3}}{3 !} + \frac{x ^{5}}{5 !} \dots .. )}$ $\Rightarrow lim_{x \to 0} \frac{2 \cdot \frac{x ^{3}}{6} + 2 \cdot \frac{x ^{5}}{5 !} + \dots ...}{\frac{x ^{3}}{6} + \frac{x ^{5}}{5 !} \dots ..} \Rightarrow lim_{x \to 0} \frac{x ^{3} ( \frac{1}{3} + \frac{1}{60} x ^{2} + \dots .. )}{x ^{3} ( \frac{1}{6} + \frac{1}{120} x ^{2} + \dots .. )} = \frac{1/3}{1/6} = 2$

Illustration 37:

If $lim_{x \to 0} \frac{s i n ^{- 1} x - t a n ^{- 1} x}{3 x ^{3}}$ is equal to $L$ , then the value of $(6 L + 1)$ is (A) $\frac{1}{6}$ (B) $\frac{1}{2}$ (C) 6 (D) 2

Ans. (D)

Solution:

$lim_{x \to 0} \frac{( x + \frac{x ^{3}}{3 !} \dots ) - ( x - \frac{x ^{3}}{3} \dots )}{3 x ^{3}} = \frac{1}{6}$ So $6 L + 1 = 2$

Illustration 38:

If $lim_{x \to 0} \frac{a x ^{2} e ^{x} - b l o g _{e} ( 1 + x ) + c x e ^{- x}}{x ^{2} s i n x} = 1$ , then $16 (a^{2} + b^{2} + c^{2})$ is equal to . Ans. (81)

Solution:

lim_{x \to 0} \frac{+ c x ( 1 - x + \frac{x ^{2}}{x !} - \frac{x ^{3}}{3 !} + \dots\dots .. )}{x ^{3} \cdot \frac{s i n x}{x}}

$= lim_{x \to \infty} \frac{( c - b ) x + ( \frac{b}{2} - c + a ) x ^{2} + ( a - \frac{b}{3} + \frac{c}{2} ) x ^{3} + \dots\dots .}{x ^{3}} = 1$ $c - b = 0, \frac{b}{2} - c + a = 0$ $a - \frac{b}{3} + \frac{c}{2} = 1 a = \frac{3}{4} b = c = \frac{3}{2}$ $a^{2} + b^{2} + c^{2} = \frac{9}{16} + \frac{9}{4} + \frac{9}{4}$ $16 (a^{2} + b^{2} + c^{2}) = 81$

Illustration 39:

If $lim_{x \to 0} \frac{3 + α s i n x + β c o s x + l o g _{e} ( 1 - x )}{3 t a n ^{2} x} = \frac{1}{3}$ , then $2 α - β$ is equal to: (A) 2 (B) 7 (C) 5 (D) 1

Ans. (C)

Solution:

$lim_{x \to 0} \frac{3 + α s i n x + β c o s x + l o g _{e} ( 1 - x )}{3 t a n ^{2} x} = \frac{1}{3}$ $\Rightarrow lim_{x \to 0} \frac{3 + α [ x - \frac{x ^{3}}{3 !} + \dots ] + β [ 1 - \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} \dots ] + ( - x - \frac{x ^{2}}{2} - \frac{x ^{3}}{3} \dots )}{3 t a n ^{2} x} = \frac{1}{3}$ $\Rightarrow lim_{x \to 0} \frac{( 3 + β ) + ( α - 1 ) x + ( - \frac{1}{2} - \frac{β}{2} ) x ^{2} + \dots}{3 x ^{2}} \times \frac{x ^{2}}{t a n ^{2} x} = \frac{1}{3}$ $\Rightarrow β + 3 = 0, α - 1 = 0$ and $\frac{- \frac{1}{2} - \frac{β}{2}}{3} = \frac{1}{3}$ $\Rightarrow β = - 3, α = 1$ $\Rightarrow 2 α - β = 2 + 3 = 5$

Illustration 40:

Evaluate $lim_{x \to 0} sin \frac{π}{x}$ .

Solution:

Again, the function $f (x) = sin (π / x)$ is undefined at 0 . Evaluating the function for some small values of $x$ , we get $f (1) = sin π = 0, f (\frac{1}{2}) = sin 2 π = 0$ , $f (0.1) = sin 10 π = 0, f (0.01) = sin 100 π = 0$ . Based on this information, we might be tempted to guess that $lim_{x \to 0} sin \frac{π}{x} = 0$ but this time our guess is wrong. Note that although $f (1/ n) = sin nπ = 0$ for any integer $n$ , it is also true that $f (x) = 1$ for infinitely many values of $x$ that approach 0 . [In fact, $sin (π / x) = 1$ when $\frac{π}{x} = \frac{π}{2} + 2 nπ$ and solving for $x$ , we get $x = 2/ (4 n + 1)]$ . The graph of $f$ is given in following figure

The dashed line indicates that the values of $sin (π / x)$ oscillate between 1 and -1 infinitely often as $x$ approaches 0 . Since the values of $f (x)$ do not approach a fixed number as $x$ approaches 0 , $\Rightarrow lim_{x \to 0} sin \frac{π}{x}$ does not exist.

Illustration 41:

$lim_{x \to 0} {\frac{s i n x - x + \frac{x ^{3}}{6}}{x ^{5}}} =$ (A) $1/120$ (B) $- 1/120$ (C) $1/20$ (D) None of these

Ans. (A)

Solution:

Expand $sin x$ and then solve. $sin x = x - \frac{x ^{3}}{3 !} + \frac{x ^{5}}{5 !} - \frac{x ^{7}}{7 !} + \dots .. lim_{x \to 0} \frac{x - \frac{x ^{3}}{6} + \frac{x ^{5}}{120} - \dots .. x + \frac{x ^{3}}{6}}{x ^{5}} = \frac{1}{120}$

Illustration 42:

Let $f : (- \infty, \infty) - {0} \to R$ be a differentiable function such that $f^{'} (1) = lim_{a \to \infty} a^{2} f (\frac{1}{a})$ . Then $lim_{a \to \infty} \frac{a ( a + 1 )}{2} tan^{- 1} (\frac{1}{a}) + a^{2} - 2 lo g_{e} a$ is equal to (A) $\frac{3}{2} + \frac{π}{4}$ (B) $\frac{3}{8} + \frac{π}{4}$ (C) $\frac{5}{2} + \frac{π}{8}$ (D) $\frac{3}{4} + \frac{π}{8}$

Ans. (C)

Solution:

$f : (- \infty, \infty) - {0} \to R$ $f^{'} (1) = lim_{a \to \infty} a^{2} f (\frac{1}{a})$ $lim_{a \to \infty} \frac{a ( a + 1 )}{2} tan^{- 1} (\frac{1}{a}) + a^{2} - 2 ln (a)$ $lim_{a \to \infty} a^{2} (\frac{( 1 + \frac{1}{a} )}{2} tan^{- 1} (\frac{1}{a}) + 1 - \frac{2}{a ^{2}} ln (a))$ $f (x) = \frac{1}{2} (1 + x) tan^{- 1} (x) + 1 - 2 x^{2} ln (x)$ $f^{'} (x) = \frac{1}{2} (\frac{1 + x}{1 + x ^{2}} + tan^{- 1} (x) + 4 x ln (x)) + 2 x$ $f^{'} (1) = \frac{1}{2} (1 + \frac{π}{4}) + 2$ $f^{'} (1) = \frac{5}{2} + \frac{π}{8}$

Illustration 43:

The value of $lim_{x \to 0} 2 (\frac{1 - c o s x c o s 2 x 3 c o s 3 x \dots .. 10 c o s 10 x}{x ^{2}})$ is - Ans. (55)

Solution:

$lim_{x \to 0} 2 (\frac{1 - ( 1 - \frac{x ^{2}}{2 !} ) ( 1 - \frac{4 x ^{2}}{2 !} ) ( 1 - \frac{9 x ^{2}}{2 !} ) \dots .. ( 1 - \frac{100 x ^{2}}{2 !} )}{x ^{2}})$ By expansion $lim_{x \to 0} \frac{2 ( 1 - ( 1 - \frac{x ^{2}}{2} ) ) ( 1 - \frac{1}{2} \cdot \frac{4 x ^{2}}{2} ) ( 1 - \frac{1}{3} \cdot \frac{9 x ^{2}}{2} ) \dots\dots ( 1 - \frac{1}{10} \cdot \frac{100 x ^{2}}{2} )}{x ^{2}}$ $lim_{x \to 0} 2 (\frac{1 - ( 1 - \frac{x ^{2}}{2} ) ( 1 - \frac{2 x ^{2}}{2} ) ( 1 - \frac{3 x ^{2}}{2} ) \dots ( 1 - \frac{10 x ^{2}}{2} )}{x ^{2}})$ $lim_{x \to 0} \frac{2 ( 1 - 1 + x ^{2} ( \frac{1}{2} + \frac{2}{2} + \frac{3}{2} + \dots .. + \frac{10}{2} ) )}{x ^{2}}$ $2 (\frac{1}{2} + \frac{2}{2} + \frac{3}{2} + \dots .. + \frac{10}{2})$ $1 + 2 + \dots\dots + 10 = \frac{10 \times 11}{2} = 55$

Illustration 44:

If $α, β$ are the distinct roots of $x^{2} + b x + c = 0$ , then $lim_{x \to β} \frac{e ^{2 (x^{2} + b x + c)} - 1 - 2 ( x ^{2} + b x + c )}{( x - β ) ^{2}}$ is equal to: (A) $b^{2} + 4 c$ (B) $2 (b^{2} + 4 c)$ (C) $2 (b^{2} - 4 c)$ (D) $b^{2} - 4 c$

Ans. (C)

Solution:

$lim_{x \to β} \frac{e ^{2 (x^{2} + b x + c)} - 1 - 2 ( x ^{2} + b x + c )}{( x - β ) ^{2}}$ $\Rightarrow lim_{x \to β} \frac{1 ( 1 + \frac{2 ( x ^{2} + b x + c )}{1 !} + \frac{2 ^{2} ( x ^{2} + b x + c ) ^{2}}{2 !} + \dots ) - 1 - 2 ( x ^{2} + b x + c )}{( x - β ) ^{2}} \Rightarrow lim_{x \to β} \frac{2 ( x ^{2} + b x + 1 ) ^{2}}{( x - β ) ^{2}}$ $\Rightarrow lim_{x \to β} \frac{2 ( x - α ) ^{2} ( x - β ) ^{2}}{( x - β ) ^{2}}$ $\Rightarrow 2 (β - α)^{2} = 2 (b^{2} - 4 c)$

10.0Illustration 45:

$lim_{x \to 2} (\sum_{n = 1}^{9} \frac{x}{n ( n + 1 ) x ^{2} + 2 ( 2 n + 1 ) x + 4})$ is equal to : (A) $\frac{9}{44}$ (B) $\frac{5}{24}$ (C) $\frac{1}{5}$ (D) $\frac{7}{36}$

Ans. (A)

Solution:

$S = lim_{x \to 2} \sum_{n = 1}^{9} \frac{x}{n ( n + 1 ) x ^{2} + 2 ( 2 n + 1 ) x + 4}$ $S = \sum_{n = 1}^{9} \frac{2}{4 ( n ^{2} + 3 n + 2 )} = \frac{1}{2} \sum_{n = 1}^{9} (\frac{1}{n + 1} - \frac{1}{n + 2})$ $S = \frac{1}{2} (\frac{1}{2} - \frac{1}{11}) = \frac{9}{44}$

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