Limits and Derivatives

Limits and Derivatives are fundamental concepts in calculus that lay the groundwork for more advanced topics. Understanding them is crucial for solving problems related to rates of change, optimization, and motion, among others.

1.0What are Limits and Derivatives in Mathematics?

Limits help us understand the behavior of functions as they approach a certain point. Formally, the limit of a function f(x) as x approaches a value a is the value that f(x) gets closer to as x gets closer to a. It’s expressed as:

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

Where L is the limit value. 

Following fig. show the graphs of three functions. Note that in part(c), f(a) is not defined and part (b), f(a) ≠ L. But in each case, it happens at a. It is true that ${\lim }_{x\to a}f(x)=L$ _.

Derivatives, on the other hand, measure how a function changes as its input changes. It's the rate at which the function's value changes with respect to change in the variable. The derivative of a function f(x) at a point x = a is given by:

$f^{\prime }(a)={\lim }_{h\to 0}\frac{f(a+h)-f(a)}{h}$

This formula is essentially the limit of the average rate of change as the interval shrinks to zero.

2.0Left hand Limits and Right Hand limit of a Function

${\lim }_{x\to a^{-}}f(x)=L$ ${\lim }_{x\to a^{+}}f(x)=L$

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

If ${\lim }_{x\to a^{-}}f(x)={\lim }_{x\to a^{+}}f(x)=L(\text{ finite })$

Geometrical Meaning: 

Tangent to the curve y = f(x) at a points (a, f(a)) is a limiting case of slope of secant through A.

${\lim }_{B\to A}{m}_{AB}={\lim }_{h\to 0}\frac{f(a+h)-f(a)}{h}$

${\lim }_{C\to A}{m}_{Ac}={\lim }_{h\to 0}\frac{f(a-h)-f(a)}{-h}$

Derivative exists if 

${\lim }_{h\to 0}\frac{f(a-h)-f(a)}{-h}={\lim }_{h\to 0}\frac{f(a+h)-f(a)}{h}$

3.0Limits and Derivatives Formulas

1. Algebra of Limits:

• ${\lim }_{x\to a}c=c$, where c is a constant.
• ${\lim }_{x\to a}x=a$
• ${\lim }_{x\to a}[f(x)\pm g(x)]={\lim }_{x\to a}f(x)\pm {\lim }_{x\to a}g(x)$
• ${\lim }_{x\to a}[f(x)\cdot g(x)]={\lim }_{x\to a}f(x)\cdot {\lim }_{x\to a}g(x)$
• ${\lim }_{x\to a}\frac{f(x)}{g(x)}=\frac{{\lim }_{x\to a}f(x)}{{\lim }_{x\to a}g(x)}$
• ${\lim }_{x\to a}[(f)(x)]={\lim }_{x\to a}f(x)$

2. Limits of Some Basic Trigonometric Functions

• ${\lim }_{x\to 0}\frac{\sin x}{x}=1$
• ${\lim }_{x\to 0}\frac{\tan x}{x}=1$
• ${\lim }_{x\to 0}\frac{{\sin }^{-1}x}{x}=1$
• ${\lim }_{x\to 0}\frac{{\tan }^{-1}x}{x}=1$(Where x is measured in radians)
• If ${\lim }_{x\to 0}f(x)=0than{\lim }_{x\to 0}\frac{\sin f(x)}{f(x)}=1$e.g. ${\lim }_{x\to 1}\frac{\sin (\ln x)}{(\ln x)}=1$
• ${\lim }_{x\to 0}\frac{1-\cos x}{x}=0$

3. Limits of Some Exponential and Logarithmic Functions

• ${\lim }_{x\to 0}{e}^x=1$
• ${\lim }_{x\to 0}\frac{e^x-1}{x}=1$
• ${\lim }_{\mathrm{x}\to \infty }{\left(1+\frac{1}{\mathrm{x}}\right)}^{\mathrm{x}}=\mathrm{e}$
• ${\lim }_{x\to \infty }{\left(1+\frac{a}{x}\right)}^x=e^a$
• ${\lim }_{x\to 0}(1+x{)}^{1/x}=e$
• ${\lim }_{x\to 0}\frac{a^x-1}{x}={\log }_{e}a$
• ${\lim }_{x\to 0}\frac{\log (1+x)}{x}=1$

4. xⁿ Formula

• ${\lim }_{x\to a}\frac{x^n-a^n}{x-a}=n(a{)}^{n-1}$

5. Basic Derivative Formulas:

• $\frac{d}{dx}[c]=0$, where c is a constant.
• $\frac{d}{dx}\left[x^n\right]=n{x}^{n-1}$
• $\frac{d}{dx}[\sin x]=\cos x$
• $\frac{d}{dx}[\cos x]=-\sin x$

6. Algebra of Derivative of Functions:

$\frac{d}{dx}[f(x)\pm g(x)]=\frac{d}{dx}f(x)\pm \frac{d}{dx}g(x)$

$\frac{d}{dx}[f(x)\times g(x)]=f(x)\cdot \frac{d}{dx}g(x)+g(x)\cdot \frac{d}{dx}f(x)$

$\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]=\frac{g(x)\cdot \frac{d}{dx}f(x)-f(x)\cdot \frac{d}{dx}g(x)}{(g(x){)}^2}$

4.0Relationship Between Derivatives and Limits

The concept of derivatives is fundamentally built on the concept of limits. A derivative is essentially a limit that describes the rate of change of a function at a specific point. Without limits, the definition of a derivative wouldn't exist. When we calculate a derivative using the definition, we're actually finding the limit of the function's average rate of change as the interval over which the change is measured approaches zero.

5.0Solved Examples of Limits and Derivatives

Example 1: Evaluate the limit: ${\lim }_{x\to 3}\frac{x^2-9}{x-3}$

Solution:

Given: ${\lim }_{x\to 3}\frac{x^2-9}{x-3}$

$={\lim }_{x\to 3}\frac{x^2-9}{x-3}$

Simplify the expression

$={\lim }_{x\to 3}\frac{(x-3)(x+3)}{x-3}$

Cancelling the common factor (x – 3)

$={\lim }_{x\to 3}x+3$

Now, substitute x = 3 into the simplified expression

= 3 + 3 = 6

So, ${\lim }_{x\to 3}\frac{x^2-9}{x-3}=6$

Example 2: Find the derivative of f(x) = x² at x = 2.

Solution:

Using the definition of a derivative:

$f^{\prime }(x)={\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}$

Substituting f(x) = x²:

$f^{\prime }(2)={\lim }_{h\to 0}\frac{(2+h{)}^2-4}{h}$

$f^{\prime }(2)={\lim }_{h\to 0}\frac{4+4h+h^2-4}{h}$

$f^{\prime }(2)={\lim }_{h\to 0}(4+h)$

$f^{\prime }(2)=4$

So, f'(2) = 4.

Example 3: Evaluate the limit: ${\lim }_{x\to 2}\frac{x^2-4}{x-2}$

Solution:

Given: ${\lim }_{x\to 2}\frac{x^2-4}{x-2}$

$={\lim }_{x\to 2}\frac{x^2-4}{x-2}$

Simplify the expression

$={\lim }_{x\to 2}\frac{(x-2)(x+2)}{x-2}$

Cancelling the common factor (x – 2)

$={\lim }_{x\to 2}x+2$

Now, substitute x = 2 into the simplified expression

= 2 + 2 = 4

So, ${\lim }_{x\to 2}\frac{x^2-4}{x-2}=4$

Example 4: Find the derivative of f(x) = 3x³ – 5x² + 2x – 7.

Solution:

Using the power rule $\frac{d}{dx}\left[x^n\right]=n{x}^{n-1}$, differentiate each term:

$f^{\prime }(x)=3\cdot 3x^2-5\cdot 2x^1+2\cdot 1x^0-0$

$f^{\prime }(x)=9x^2-10x+2$

Example 5: Find the derivative of f(x) = x² + 3x using the definition of a derivative.

Solution: 

The derivative is given by:

$f^{\prime }(x)={\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}$

Substitute f(x) = x² + 3x:

Expand and simplify:

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

$f^{\prime }(x)={\lim }_{h\to 0}\frac{2xh+h^2+3h}{h}$

Factor out h:

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

$f^{\prime }(x)=2x+3$

So, the derivative is f'(x) = 2x + 3.

Example 6: Evaluate the limit ${\lim }_{x\to \infty }\frac{5x^3-2x^2+4}{3x^3+x-7}$

Solution: 

Given: ${\lim }_{x\to \infty }\frac{5x^3-2x^2+4}{3x^3+x-7}$

Divide each term by x³ (the highest power of x in the denominator)

${\lim }_{x\to \infty }\frac{5-\frac{2}{x}+\frac{4}{x^3}}{3+\frac{1}{x^2}-\frac{7}{x^3}}$

Putting the limit x = infinite

${\lim }_{x\to \infty }\frac{5-\frac{2}{x}+\frac{4}{x^3}}{3+\frac{1}{x^2}-\frac{7}{x^3}}=\frac{5}{3}$

So, ${\lim }_{x\to \infty }\frac{5x^3-2x^2+4}{3x^3+x-7}=\frac{5}{3}$

Example 7: Evaluate ${\lim }_{x\to 0}\left\{\frac{\sqrt{1+x}-\sqrt{1-x}}{x}\right\}$

Solution: 

We have 

${\lim }_{x\to 0}\left\{\frac{\sqrt{1+x}-\sqrt{1-x}}{x}\right\}$

$={\lim }_{x\to 0}\left\{\frac{\sqrt{1+x}-\sqrt{1-x}}{x}\cdot \frac{(\sqrt{1+x}+\sqrt{1-x})}{(\sqrt{1+x}+\sqrt{1-x})}\right\}$

$={\lim }_{x\to 0}\frac{\{(1+x)-(1-x)\}}{x(\sqrt{1+x}+\sqrt{1-x})}$

$={\lim }_{x\to 0}\frac{2x}{x(\sqrt{1+x}+\sqrt{1-x})}$

$={\lim }_{x\to 0}\frac{2}{(\sqrt{1+x}+\sqrt{1-x})}=1[puttingx=0]$

Example 8: Evaluate ${\lim }_{x\to 1}\frac{\left(x^2-4x+3\right)}{(x-1)}$

Solution: 

We have,

${\lim }_{x\to 1}\frac{\left(x^2-4x+3\right)}{(x-1)}$

$={\lim }_{x\to 1}\frac{(x-1)(x-3)}{(x-1)}$

$={\lim }_{x\to 1}(x-3)$

=-2

Example 9: Evaluate ${\lim }_{x\to 0}\left(\frac{e^{3x}-1}{x}\right)$

Solution: 

We have,

${\lim }_{x\to 0}\left(\frac{e^{3x}-1}{x}\right)$

$={\lim }_{3x\to 0}\left\{\left(\frac{e^{3x}-1}{3x}\right)\times 3\right\}$

$[\because (\mathrm{x}\to 0)\Rightarrow (3\mathrm{x}\to 0)]$

$=3\times {\lim }_{y\to 0}\left(\frac{e^y-1}{y}\right)$ , where y = 3x

= (3 × 1) = 3 $\left[⊠{\lim }_{\mathrm{y}\to 0}\left(\frac{{\mathrm{e}}^{\mathrm{y}}-1}{\mathrm{y}}\right)=1\right]$

Example 10: Evaluate ${\lim }_{x\to 0}\left(\frac{3^x-2^x}{x}\right)$

Solution: 

We have,

${\lim }_{x\to 0}\left(\frac{3^x-2^x}{x}\right)$

$={\lim }_{x\to 0}\left\{\frac{\left(3^x-1\right)-\left(2^x-1\right)}{x}\right\}$

${\lim }_{x\to 0}\left(\frac{3^x-1}{x}\right)-{\lim }_{x\to 0}\left(\frac{2^x-1}{x}\right)$

= (log 3 – log 2) 

$=\log \frac{3}{2}\left[{\lim }_{x\to 0}\left(\frac{a^x-1}{x}\right)=\log a\right].$

Example 11: Find the derivative of $f(x)=\sin (x)\cdot \cos (x).$

Solution:

This is a product of two functions, so use the product rule:

$f^{\prime }(x)=\frac{d}{dx}[\sin (x)\cdot \cos (x)]$

$f^{\prime }(x)=\sin (x)\cdot \frac{d}{dx}[\cos (x)]+\cos (x)\cdot \frac{d}{dx}[\sin (x)]$

$f^{\prime }(x)=\sin (x)\cdot (-\sin (x))+\cos (x)\cdot \cos (x)$

$f^{\prime }(x)=-{\sin }^2(x)+{\cos }^2(x)$

Example 12: Find the derivative of f(x) = e^3x.

Solution:

Given: f(x) = e^3x

Use the chain rule:

$f^{\prime }(x)=\frac{d}{dx}\left[e^{3x}\right]$

$f^{\prime }(x)=e^{3x}\cdot \frac{d}{dx}[3x]$

$f^{\prime }(x)=3e^{3x}$

Example 13: Find the derivative of $f(x)=\ln \left(2x^2+3x\right).$

Solution:

Given: $f(x)=\ln \left(2x^2+3x\right)$

Apply the chain rule:

$f^{\prime }(x)=\frac{1}{2x^2+3x}\cdot \frac{d}{dx}\left[2x^2+3x\right]$

$f^{\prime }(x)=\frac{1}{2x^2+3x}\cdot (4x+3)$

Simplify:

$f^{\prime }(x)=\frac{4x+3}{2x^2+3x}$

6.0Practice Questions of Limits and Derivatives

1. ${\lim }_{x\to 3}\left(\frac{x^2+9}{x+3}\right)$
2. ${\lim }_{x\to \frac{1}{2}}\left(\frac{4x^2-1}{2x-1}\right)$
3. ${\lim }_{x\to 0}\left(\frac{e^{4x}-1}{x}\right)$
4. Find the derivative of f(x) = 7x⁴ – 3x² + 5x – 9.
5. Find the derivative of $g(x)=x+\frac{1}{x}$ .
6. Find the derivative of f(x) = sin(3x).
7. Find the derivative of  g(x) = tan(x)·sec(x).
8. Find the derivative of $f(x)=e^{2x}\cdot \ln (x)$.
9. Find the derivative of $g(x)={\log }_{10}\left(x^2+1\right)$.

Answers:

1. 3
2. 2
3. 4
4. f’(x) = 28x³ − 6x + 5
5. ${\mathrm{g}}^{\prime }(\mathrm{x})=1-\frac{1}{{\mathrm{x}}^2}$
6. f’(x) = 3 cos (3x)
7. g’(x) = sec³ (x) + sec (x). tan² (x)
8. ${\mathrm{f}}^{\prime }(\mathrm{x})={\mathrm{e}}^{2\mathrm{x}}\left(2\ln (\mathrm{x})+\frac{1}{\mathrm{x}}\right)$
9. $g^{\prime }(x)=\frac{2x}{\left(x^2+1\right)\ln (10)}$

7.0Sample Questions on Limits and Derivatives

1. What are limits in mathematics?

Ans: A limit in mathematics describes the value that a function approaches as the input (or variable) approaches a certain point. For instance, if we say ${\lim }_{x\to a}f(x)=L$, it means that as x gets closer to a, the function f(x) gets closer to the value L.

2. How are limits and derivatives related?

Ans: The concept of a derivative is built on the idea of limits. The derivative of a function at a point is defined as the limit of the average rate of change of the function as the interval approaches zero. Mathematically, it's expressed as $f^{\prime }(x)={\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}$.

3. How do you find the derivative of a function using the limit definition?

Ans: To find the derivative of a function using the limit definition, you calculate:

$f^{\prime }(x)={\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}$

This formula calculates the slope of the secant line as the interval h approaches zero, effectively giving the slope of the tangent line at point x.