Limits and Continuity

Limits and Continuity are two of the most crucial concepts. These ideas form the foundation for understanding more advanced topics like derivatives and integrals. To grasp the essence of calculus, one must first understand the definition of limits and continuity and how they interrelate.

1.0What Are Limits and Continuity? 

The concept of a Limit is fundamental to the study of calculus. In simple terms, the limit of a function at a particular point refers to the value that the function approaches as the input (or variable) approaches a specific value. It allows us to analyze how functions behave as they get close to a certain point, even if they never actually reach that point.

Continuity, on the other hand, is a property of a function that ensures it behaves in a predictable manner. A function is continuous if, roughly speaking, you can sketch its graph without lifting your pen. More formally, a function is considered continuous at a point if the limit of the function exists as it approaches that point is equal to the function’s value at that point.

2.0Definition of Limits and Continuity

Limit Definition: The limit of a function f(x) as x approaches a value c is the value that f(x) gets closer to as x gets closer to c. Mathematically, this is written as:

${\lim }_{x\to c}f(x)=L$

This means that as x gets arbitrarily close to c, f(x) approaches L.

Continuity Definition: A function f(x) is continuous at a point x = c if the following three conditions are met:

1. f(c) is defined.
2. ${\lim }_{x\to c}f(x)exists.$
3. ${\lim }_{x\to c}f(x)=f(c)$

If a function meets these criteria at every point in its domain, it is considered continuous on that domain.

3.0The Difference Between Limits and Continuity

While limits deal with the behavior of functions as they approach a specific point, continuity concerns whether the function's graph is unbroken at that point. Simply put, limits are about approaching a value, whereas continuity is about the function actually attaining that value smoothly. If a function is continuous at a point, then the limit at that point equals the function's value there.

4.0Types of Discontinuity

Discontinuities occur when a function is not continuous at a point. There are several types of discontinuities:

1. Non Removable Discontinuity

• Jump Discontinuity: The function has two different limit values as it approaches the point from the left and the right.
• Infinite Discontinuity: The function increases or decreases without bound as it approaches the point.

2. Removable Discontinuity: The function has a "hole" at a point where the limit exists, but the function is not defined or is defined differently.

5.0Solved Examples of Limits and Continuity

Example 1: Find the limit of ${\lim }_{x\to 1}\frac{x^2-1}{x-1}$ .

Solution:  

If we substitute x = 1 into the function gives an indeterminate form $\frac{0}{0}$.

$={\lim }_{x\to 1}\frac{(x-1)(x+1)}{x-1}$ (Splitting the numerator)

$={\lim }_{x\to 1}(x+1)$

= 1 + 1 = 2

So, ${\lim }_{x\to 1}\frac{x^2-1}{x-1}=2$

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

Solution:

${\lim }_{x\to \infty }\frac{x^2+x+1}{3x^2+2x-5},\left(\frac{\infty }{\infty }\text{ form }\right)$

Put $\mathrm{x}=\frac{1}{y}$

$={\lim }_{y\to 0}\frac{1+y+y^2}{3+2y-5y^2}=\frac{1}{3}$

Example 3: If ${\lim }_{x\to \infty }\left(\frac{x^3+1}{x^2+1}-(ax+b)\right)=2$ , then

(A) a = 1, b = 1 (B) a = 1, b = 2 (C) a = 1, b = –2 (D) none of these

Ans. (C)

Solution:

${\lim }_{x\to \infty }\left(\frac{x^3+1}{x^2+1}-(ax+b)\right)=2$

$\Rightarrow {\lim }_{x\to \infty }\frac{x^3(1-a)-b{x}^2-ax+(1-b)}{x^2+1}=2$

$\Rightarrow {\lim }_{x\to \infty }\frac{x(1-a)-b-\frac{a}{x}+\frac{(1-b)}{x^2}}{1+\frac{1}{x^2}}=2$

⇒  1 – a = 0, –b = 2 

⇒ a = 1, b = –2   

Example 4: Evaluate: ${\lim }_{x\to 0}\frac{x^3\cot x}{1-\cos x}$

Solution:

${\lim }_{x\to 0}\frac{x^3\cos x}{\sin x(1-\cos x)}$

$={\lim }_{x\to 0}\frac{x^3\cos x(1+\cos x)}{\sin x\cdot {\sin }^{2}x}$

$={\lim }_{x\to 0}\frac{x^3}{{\sin }^{3}x}\cdot \cos x(1+\cos x)=2$

Example 5: If $f(x)=\{\begin{array}{ll}\sin \frac{\pi x}{2},& x<1\\ [x]& x\ge 1\end{array}$ then find whether f(x) is continuous or not at x=1, where [.] denotes greatest integer function.

Solution:

$f(x)=\{\begin{array}{l}\sin \frac{\pi x}{2},x<1\\ [x],x\ge 1\end{array}$

For continuity at x=1, we determine, f(1), ${}^{{\lim }_{x\to 1^{-}}}f(x)\text{ and }{\lim }_{x\to 1^{+}}f(x).$

Now, f(1) = [1]=1 

$f(x){\lim }_{x\to 1^{-}}={\lim }_{x\to 1^{-}}\sin \frac{\pi x}{2}=\sin \frac{\pi }{2}=1\text{ and }{\lim }_{x\to 1^{+}}f(x)={\lim }_{x\to 1^{+}}[x]=1$

so $f(1)={\lim }_{x\to 1^{-}}f(x)={}^{{\lim }_{x\to 1^{+}}}f(x)$

∴ f(x) is continuous at x = 1

Example 6: Discuss the continuity of $f(x)=\{\begin{array}{ll}|x+1|& ,x<-2\\ 2x+3,& -2\le x<0\\ x^2+3,& 0\le x<3\\ x^3-15,& x\ge 3\end{array}$

Solution:

We write f(x) as $f(x)=\{\begin{array}{ll}-x-1,& x<-2\\ 2x+3,& -2\le x<0\\ x^2+3,& 0\le x<3\\ x^3-15,& x\ge 3\end{array}$

As we can see, f(x) is defined as a polynomial function in each of intervals (–∞, –2), [–2, 0), [0, 3) and [3, ∞). Therefore, the function is continuous in each of these four open intervals. Thus, we check the continuity at x = –2, 0, 3.

At the point x = –2

${\lim }_{x\to -2^{-}}\mathrm{f}(\mathrm{x})={\lim }_{x\to -2^{-}}(-\mathrm{x}-1)$

= +2 – 1

= 1

${\lim }_{x\to -2^{+}}f(x)={\lim }_{x\to -2^{+}}(2x+3)$

= 2·(–2) + 3 = –1

Therefore, ${\lim }_{x\to -2}$ f(x) does not exist and hence f(x) is discontinuous at x = –2.

At the point x = 0

${\lim }_{x\to 0^{-}}f(x)={\lim }_{x\to 0^{-}}(2x+3)=3$

${\lim }_{x\to 0^{+}}f(x)={\lim }_{x\to 0^{+}}\left(x^2+3\right)=3$

f(0) = 0² + 3 = 3

Therefore f(x) is continuous at x=0.

At the point x = 3

${\lim }_{x\to 3^{-}}f(x)={\lim }_{x\to 3^{-}}\left(x^2+3\right)$

=3² + 3 = 12

${\lim }_{x\to 3^{+}}f(x)={}^{{\lim }_{x\to 3^{+}}}\left(x^3-15\right)$

=3³ – 15 = 12

f(3) =3³ – 15 = 12

Therefore, f(x) is continuous at x = 3.

We find that f(x)  is continuous at all points in R except at x = –2 .

6.0Practice Questions on Limits and Continuity

1. ${\lim }_{x\to 1}\frac{x-1}{2x^2-7x+5}$ is equal to

(A) $\frac{1}{3}$ (B) $-\frac{1}{3}$ (C) $\frac{1}{4}$ (D) $-\frac{1}{4}$

2. ${\lim }_{x\to 2}\frac{x^3-6x^2+11x-6}{x^2-6x+8}$ is equal to

(A) $\frac{3}{2}$ (B) $\frac{1}{2}$ (C) 1 (D) Does not exist

3. ${\lim }_{x\to 3}\frac{x^3-7x^2+15x-9}{x^4-5x^3+27x-27}$ is equal to

(A) 2/9 (B) 1/9 (C) –2/9 (D) Does not exist

4. The function $f(x)=\frac{4-x^2}{4x-x^3}$ , is -

(A) discontinuous at only one point in its domain.

(B) discontinuous at two points in its domain.

(C) discontinuous at three points in its domain.

(D) continuous everywhere in its domain.

5. A function y=f(x) is defined as $f(x)=\{\begin{array}{lll}k\sin \frac{(x+3)\pi }{6}& \text{ for }& x\le 2\\ \frac{3-\sqrt{11-x}}{x-2}& \text{ for }& x>2\end{array}$ . If f(x) is continuous at x=2, then find the value of 6k.

7.0Sample Questions on Limits and Continuity

1. What is the Formal Definition of a Limit?

Ans: The limit of a function f(x) as x approaches a value c is the value L that f(x) gets closer to as x gets closer to c. Formally, this is expressed as:

${\lim }_{x\to c}f(x)=L$

This indicates that as x nears c, f(x) approaches L, regardless of whether f(c) is defined.

2. What is the Meaning of Continuity?

Ans: A function is continuous at a point x = c if it is unbroken at that point. For a function to be continuous at c, the following must hold:

• f(c) is defined.
• ${\lim }_{x\to c}f(x)exists.$
• ${\lim }_{x\to c}f(x)=f(c).$

This means the function has no jumps, holes, or vertical asymptotes at x = c .