Questions and Answers Section

Part 1: Evaluation of Limits

Example 1: Factorization Method

Question: Evaluate $\lim_{x \to 2} \frac{x^3-8}{x-2}$.

Solution:

Step 1: Identify the Indeterminate Form

Direct substitution gives $\frac{2^3-8}{2-2} = \frac{0}{0}$.

Step 2: Factor the Expression

Using the difference of cubes formula, $a^3 - b^3 = (a-b)(a^2+ab+b^2)$, we get:

$x^3 - 8 = (x-2)(x^2 + 2x + 4)$.

Step 3: Simplify and Evaluate

$\lim_{x \to 2} \frac{(x-2)(x^2+2x+4)}{x-2} = \lim_{x \to 2} (x^2+2x+4) = 2^2 + 2(2) + 4 = 12$.

Answer: 12.

Example 2: Rationalization Method

Question: Evaluate $\lim_{x \to 0} \frac{\sqrt{1+x}-1}{x}$.

Solution:

This is a $\frac{0}{0}$ form. Multiply the numerator and denominator by the conjugate, $\sqrt{1+x}+1$.

$\lim_{x \to 0} \frac{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}{x(\sqrt{1+x}+1)} = \lim_{x \to 0} \frac{(1+x)-1}{x(\sqrt{1+x}+1)} = \lim_{x \to 0} \frac{x}{x(\sqrt{1+x}+1)}$.

Cancel $x$ and substitute: $\lim_{x \to 0} \frac{1}{\sqrt{1+x}+1} = \frac{1}{\sqrt{1}+1} = \frac{1}{2}$.

Answer: 1/2.

Example 3: L'Hôpital's Rule

Question: Evaluate $\lim_{x \to 0} \frac{e^{2x}-1}{x}$.

Solution:

This is a $\frac{0}{0}$ form. Applying L'Hôpital's Rule, we differentiate the numerator and denominator:

$\lim_{x \to 0} \frac{\frac{d}{dx}(e^{2x}-1)}{\frac{d}{dx}(x)} = \lim_{x \to 0} \frac{2e^{2x}}{1} = 2e^0 = 2$.

Answer: 2.

Example 4: Repeated L'Hôpital's Rule

Question: Evaluate $\lim_{x \to 0} \frac{e^x - x - 1}{x^2}$.

Solution:

This is a $\frac{0}{0}$ form. Applying the rule once gives $\lim_{x \to 0} \frac{e^x - 1}{2x}$. This is still $\frac{0}{0}$. Applying the rule again gives $\lim_{x \to 0} \frac{e^x}{2} = \frac{1}{2}$.

Answer: 1/2.

Example 5: Algebraic Manipulation

Question: Evaluate the limit: $\lim_{x \to 0} \frac{e^x - e^{\sin x}}{x - \sin x}$

Solution:

Factor out $e^{\sin x}$ from the numerator: $L = \lim_{x \to 0} \frac{e^{\sin x}(e^{x - \sin x} - 1)}{x - \sin x}$.

This splits into two limits: $L = ( \lim_{x \to 0} e^{\sin x} ) \cdot ( \lim_{x \to 0} \frac{e^{x - \sin x} - 1}{x - \sin x} )$.

Let $u = x - \sin x$. As $x \to 0$, $u \to 0$. The second limit becomes $\lim_{u \to 0} \frac{e^u-1}{u} = 1$.

The first limit is $e^{\sin 0} = 1$. So, $L = 1 \cdot 1 = 1$.

Answer: 1.

Example 6: The $1^{\infty}$ Form

Question: Evaluate $\lim_{x \to 0} (1+2x)^{1/x}$.

Solution:

This is a $1^{\infty}$ form. The limit can be found using the formula $e^{\lim_{x \to c} g(x)[f(x)-1]}$.

The limit is $e^{\lim_{x \to 0} \frac{1}{x}[(1+2x)-1]} = e^{\lim_{x \to 0} \frac{2x}{x}} = e^2$.

Answer: $e^2$.

Example 7: Series Expansions

Question: Evaluate $\lim_{x \to 0} \frac{x - \sin x}{x^3}$.

Solution:

We replace $\sin x$ with its series expansion $x - \frac{x^3}{3!} + O(x^5)$.

The limit becomes $\lim_{x \to 0} \frac{x - (x - x^3/6 + ...)}{x^3} = \lim_{x \to 0} \frac{x^3/6}{x^3} = \frac{1}{6}$.

Answer: 1/6.

Example 8: Squeeze Theorem

Question: Evaluate $\lim_{x \to 0} x^2 \cos(1/x)$.

Solution:

We know that $-1 \le \cos(1/x) \le 1$. Multiplying by $x^2$ gives $-x^2 \le x^2\cos(1/x) \le x^2$.

Since $\lim_{x \to 0} -x^2 = 0$ and $\lim_{x \to 0} x^2 = 0$, by the Squeeze Theorem, the limit is 0.

Answer: 0.

Example 9: Limit at Infinity

Question: Evaluate $\lim_{x \to \infty} \frac{3x^2-x-2}{5x^2+4x+1}$.

Solution:

Divide the numerator and denominator by the highest power of x, which is $x^2$.

$\lim_{x \to \infty} \frac{3-1/x-2/x^2}{5+4/x+1/x^2} = \frac{3-0-0}{5+0-0} = \frac{3}{5}$.

Answer: 3/5.

Example 10: Using Leibniz Rule and L'Hôpital's Rule

Question: Evaluate: $\lim_{x \to 0} \frac{1}{x} [\int_{0}^{x} (1+\sin(2t))^{1/t} dt ]$

Solution:

This is a $\frac{0}{0}$ form. We use L'Hôpital's Rule and Leibniz rule. The limit becomes $\lim_{x \to 0} \frac{\frac{d}{dx}\int_{0}^{x} (1+\sin(2t))^{1/t} dt}{1}$.

This simplifies to $\lim_{x \to 0} (1+\sin(2x))^{1/x}$, which is a $1^\infty$ form.

$L = e^{\lim_{x \to 0} \frac{1}{x}(\sin(2x))} = e^{\lim_{x \to 0} 2 \frac{\sin(2x)}{2x}} = e^2$.

Answer: $e^2$.

Example 11: Logarithms and L'Hôpital's Rule

Question: Evaluate: $\lim_{x \to 0} (\frac{a^x+b^x+c^x}{3})^{1/x}$ where $a, b, c > 0$.

Solution:

This is a $1^\infty$ form. Let $L$ be the limit and take the natural log: $\ln L = \lim_{x \to 0} \frac{1}{x} \ln(\frac{a^x+b^x+c^x}{3})$.

This is a $0/0$ form. Apply L'Hôpital's Rule:

$\ln L = \lim_{x \to 0} \frac{d/dx(\ln(a^x+b^x+c^x) - \ln 3)}{d/dx(x)} = \lim_{x \to 0} \frac{a^x\ln a + b^x\ln b + c^x\ln c}{a^x+b^x+c^x}$.

Substitute $x=0$: $\ln L = \frac{\ln a + \ln b + \ln c}{3} = \frac{\ln(abc)}{3} = \ln((abc)^{1/3})$.

Thus, $L = (abc)^{1/3}$.

Answer: $(abc)^{1/3}$.

Part 2: Continuity and Discontinuity

Example 12: Finding Parameters for Continuity

Question: Find 'k' for which $f(x) = \begin{cases} kx+1, & x \le 5 \\ 3x-5, & x > 5 \end{cases}$ is continuous at $x=5$.

Solution:

For continuity, LHL = RHL at $x=5$.

LHL = $\lim_{x \to 5^-} (kx+1) = 5k+1$.

RHL = $\lim_{x \to 5^+} (3x-5) = 10$.

Equating them: $5k+1 = 10 \implies 5k=9 \implies k=9/5$.

Answer: k = 9/5.

Example 13: Finding Parameters for Continuity (Piecewise)

Question: Find the values of $a$ and $b$ such that the function given by $f(x) = \begin{cases} 5, & \text{if } x \leq 2 \\ ax+b, & \text{if } 2 < x < 10 \\ 21, & \text{if } x \geq 10 \end{cases}$ is a continuous function.

Solution:

For continuity, limits must match at $x=2$ and $x=10$.

At $x=2$: $\lim_{x \to 2^-} 5 = \lim_{x \to 2^+} (ax+b) \implies 5 = 2a+b$. (1)

At $x=10$: $\lim_{x \to 10^-} (ax+b) = \lim_{x \to 10^+} 21 \implies 10a+b=21$. (2)

Subtracting (1) from (2): $8a = 16 \implies a=2$.

Substituting into (1): $2(2)+b=5 \implies b=1$.

Answer: a=2 and b=1.

Example 14: Removable Discontinuity

Question: Discuss the continuity of $f(x)=\frac{x^2-4}{x-2}$ at $x=2$.

Solution:

The function is not defined at x=2. However, $\lim_{x \to 2} \frac{(x-2)(x+2)}{x-2} = 4$. Since the limit exists but the function is not defined at the point, this is a removable discontinuity. It can be made continuous by defining $f(2)=4$.

Answer: Removable discontinuity at x=2.

Example 15: Jump Discontinuity

Question: Find the points of discontinuity for the function $f(x) = [x]$ (Greatest Integer Function).

Solution:

At any integer 'n', LHL = $\lim_{x \to n^-} [x] = n-1$ and RHL = $\lim_{x \to n^+} [x] = n$. Since LHL $\n e$ RHL, the function has a jump discontinuity at every integer.

Answer: Jump discontinuity at every integer.

Example 16: Sum of Continuous Functions

Question: Discuss the continuity of $f(x)=|x|+|x-1|$ at x=0 and x=1.

Solution:

The function $|x|$ is continuous everywhere. The function $|x-1|$ is also continuous everywhere. The sum of two continuous functions is always continuous. Therefore, $f(x)$ is continuous everywhere, including at x=0 and x=1.

Answer: Continuous everywhere.

Part 3: Differentiability

Example 17: Checking Differentiability of Modulus Function

Question: Check the differentiability of the function $f(x) = |x-2|$ at $x=2$.

Solution:

LHD = $\lim_{h \to 0^-} \frac{|(2+h)-2|-0}{h} = \lim_{h \to 0^-} \frac{|h|}{h} = -1$.

RHD = $\lim_{h \to 0^+} \frac{|h|}{h} = 1$.

Since LHD $\n e$ RHD, the function is not differentiable at x=2.

Answer: Not differentiable at x=2.

Example 18: Finding Parameters for Differentiability

Question: If $f(x) = \begin{cases} x^2+1, & x \le 1 \\ ax+b, & x > 1 \end{cases}$ is differentiable at $x=1$, find a and b.

Solution:

1. Continuity: $1^2+1=a(1)+b \implies a+b=2$.

2. Differentiability: LHD = RHD. The derivative of $x^2+1$ is $2x$. The derivative of $ax+b$ is $a$. At x=1, LHD=2(1)=2 and RHD=a. So, $a=2$.

From $a+b=2$, we get $2+b=2 \implies b=0$.

Answer: a=2, b=0.

Example 19: Differentiability of Piecewise Modulus Function

Question: Check the differentiability of the function $f(x) = |x| + |x-1|$ at $x=0$ and $x=1$.

Solution:

Writing $f(x)$ as a piecewise function: $f(x) = \begin{cases} -2x+1, & x < 0 \\ 1, & 0 \leq x < 1 \\ 2x-1, & x \geq 1 \end{cases}$.

At $x=0$: LHD = $\frac{d}{dx}(-2x+1) = -2$. RHD = $\frac{d}{dx}(1) = 0$. Not differentiable.

At $x=1$: LHD = $\frac{d}{dx}(1) = 0$. RHD = $\frac{d}{dx}(2x-1) = 2$. Not differentiable.

Answer: Not differentiable at x=0 and x=1.

Example 20: Differentiability of a Special Function

Question: Show that $f(x) = \begin{cases} x^2 \sin(1/x), & x \n eq 0 \\ 0, & x=0 \end{cases}$ is differentiable at $x=0$, but its derivative is not continuous at $x=0$.

Solution:

Differentiability: $f'(0) = \lim_{h \to 0} \frac{h^2\sin(1/h)-0}{h} = \lim_{h \to 0} h\sin(1/h)$. By Squeeze Theorem, this limit is 0. So, $f'(0)=0$.

Continuity of derivative: For $x \n eq 0$, $f'(x) = 2x\sin(1/x) - \cos(1/x)$.

As $x \to 0$, the term $2x\sin(1/x) \to 0$, but $\lim_{x \to 0} \cos(1/x)$ does not exist. Thus $\lim_{x \to 0} f'(x)$ does not exist and is not equal to $f'(0)$.

Answer: Proven.

Part 4: Techniques of Differentiation

Example 21: Derivative from First Principle (Polynomial)

Question: Find the derivative of $f(x) = x^3$ using the first principle.

Solution:

$f'(x) = \lim_{h \to 0} \frac{(x+h)^3 - x^3}{h} = \lim_{h \to 0} \frac{x^3+3x^2h+3xh^2+h^3 - x^3}{h} = \lim_{h \to 0} (3x^2+3xh+h^2) = 3x^2$.

Answer: $3x^2$.

Example 22: Derivative from First Principle (Trigonometric)

Question: Find the derivative of $f(x) = \tan x$ from the first principle.

Solution:

$f'(x) = \lim_{h \to 0} \frac{1}{h} [ \frac{\sin(x+h)}{\cos(x+h)} - \frac{\sin x}{\cos x} ] = \lim_{h \to 0} \frac{1}{h} [ \frac{\sin(x+h-x)}{\cos(x+h)\cos x} ]$.

The numerator is $\sin(h)$. So, $f'(x) = \lim_{h \to 0} \frac{\sin(h)}{h} \cdot \frac{1}{\cos(x+h)\cos x} = 1 \cdot \frac{1}{\cos^2 x} = \sec^2 x$.

Answer: $\sec^2 x$.

Example 23: Product and Quotient Rules

Question (Product Rule): Find the derivative of $f(x) = e^x \tan x$.

Solution: $f'(x) = (e^x)'(\tan x) + (e^x)(\tan x)' = e^x\tan x + e^x\sec^2x$.

Question (Quotient Rule): Find the derivative of $f(x) = \frac{\sin x}{x^2}$.

Solution: $f'(x) = \frac{(\cos x)(x^2) - (\sin x)(2x)}{(x^2)^2} = \frac{x\cos x - 2\sin x}{x^3}$.

Example 24: Chain Rule

Question: Find the derivative of $y = \log(\sin(x^2))$.

Solution:

This requires applying the chain rule multiple times.

$\frac{dy}{dx} = \frac{1}{\sin(x^2)} \cdot \frac{d}{dx}(\sin(x^2)) = \frac{1}{\sin(x^2)} \cdot \cos(x^2) \cdot \frac{d}{dx}(x^2) = \cot(x^2) \cdot 2x = 2x\cot(x^2)$.

Answer: $2x\cot(x^2)$.

Example 25: Logarithmic Differentiation

Question: Find the derivative of $y=x^{\sin x}$.

Solution:

Take logs: $\ln y = \sin x \ln x$. Differentiate implicitly: $\frac{1}{y}\frac{dy}{dx} = (\cos x)(\ln x) + (\sin x)(\frac{1}{x})$.

So, $\frac{dy}{dx} = y[\cos x \ln x + \frac{\sin x}{x}] = x^{\sin x}[\cos x \ln x + \frac{\sin x}{x}]$.

Answer: $x^{\sin x}[\cos x \ln x + \frac{\sin x}{x}]$.

Example 26: Implicit Differentiation

Question: If $x^3+y^3=3axy$, find $\frac{dy}{dx}$.

Solution:

Differentiate each term with respect to x: $3x^2 + 3y^2\frac{dy}{dx} = 3a(y + x\frac{dy}{dx})$.

Rearranging the terms: $(3y^2 - 3ax)\frac{dy}{dx} = 3ay - 3x^2$.

Solving for $\frac{dy}{dx}$ gives $\frac{dy}{dx} = \frac{3ay-3x^2}{3y^2-3ax} = \frac{ay-x^2}{y^2-ax}$.

Answer: $\frac{ay-x^2}{y^2-ax}$.

Example 27: Parametric Differentiation

Question: If $x = a(t-\sin t)$ and $y = a(1-\cos t)$, find $\frac{dy}{dx}$.

Solution:

$\frac{dx}{dt} = a(1-\cos t)$ and $\frac{dy}{dt} = a(\sin t)$.

Then $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a\sin t}{a(1-\cos t)} = \frac{2\sin(t/2)\cos(t/2)}{2\sin^2(t/2)} = \cot(t/2)$.

Answer: $\cot(t/2)$.

Example 28: Derivative of Inverse Trigonometric Functions

Question: Find the derivative of $y = \tan^{-1}(\frac{2x}{1-x^2})$.

Solution:

Use the substitution $x=\tan\theta$. The expression becomes $y = \tan^{-1}(\frac{2\tan\theta}{1-\tan^2\theta}) = \tan^{-1}(\tan(2\theta)) = 2\theta = 2\tan^{-1}x$. Now, $\frac{dy}{dx} = \frac{2}{1+x^2}$.

Answer: $\frac{2}{1+x^2}$.

Example 29: Derivative as a function of another derivative

Question: Find the derivative of $\sin^{-1}(2x\sqrt{1-x^2})$ with respect to $\cos^{-1}(2x^2-1)$.

Solution:

Let $u = \sin^{-1}(2x\sqrt{1-x^2})$ and $v = \cos^{-1}(2x^2-1)$. We need to find $\frac{du}{dv}$.

Let $x=\sin\theta$. Then $u = \sin^{-1}(\sin 2\theta) = 2\theta = 2\sin^{-1}x$.

Let $x=\cos\phi$. Then $v = \cos^{-1}(\cos 2\phi) = 2\phi = 2\cos^{-1}x$.

We need $\frac{d(2\sin^{-1}x)}{d(2\cos^{-1}x)} = \frac{d(\sin^{-1}x)}{d(\cos^{-1}x)} = \frac{1/\sqrt{1-x^2}}{-1/\sqrt{1-x^2}} = -1$.

Answer: -1.

Example 30: Implicit differentiation and substitution

Question: If $x\sqrt{1+y} + y\sqrt{1+x} = 0$ for $x \n eq y$, find $\frac{dy}{dx}$.

Solution:

Rearrange and square: $x^2(1+y) = y^2(1+x)$.

$x^2-y^2 = y^2x - x^2y \implies (x-y)(x+y) = -xy(x-y)$.

Since $x \n eq y$, we have $x+y = -xy \implies x+y+xy=0$.

Differentiating implicitly: $1 + y' + y + xy' = 0 \implies y'(1+x) = -(1+y)$.

This gives $y' = -\frac{1+y}{1+x}$. From $y(1+x)=-x$, we get $y = -x/(1+x)$. Substituting this in gives $y' = -\frac{1}{(1+x)^2}$.

Answer: $-\frac{1}{(1+x)^2}$.

Part 5: Applications of Derivatives & Mean Value Theorems

Example 31: Second Derivative

Question: If $y=e^x \sin x$, find $\frac{d^2y}{dx^2}$.

Solution:

First derivative: $\frac{dy}{dx} = e^x\sin x + e^x\cos x = e^x(\sin x + \cos x)$.

Second derivative: $\frac{d^2y}{dx^2} = e^x(\sin x + \cos x) + e^x(\cos x - \sin x) = 2e^x\cos x$.

Answer: $2e^x\cos x$.

Example 32: Rolle's Theorem

Question: Verify Rolle's Theorem for the function $f(x)=x^2-4x+3$ on the interval [1,3].

Solution:

1. $f(x)$ is a polynomial, so it's continuous on [1,3] and differentiable on (1,3).

2. $f(1) = 0$ and $f(3)=0$. So, $f(1)=f(3)$.

All conditions are met. We find c such that $f'(c)=0$. $f'(x)=2x-4$. Setting $f'(c)=0$ gives $2c-4=0 \implies c=2$. Since $c=2$ is in (1,3), Rolle's Theorem is verified.

Answer: Verified, c=2.

Example 33: Lagrange's Mean Value Theorem

Question: Verify LMVT for $f(x)=x^2$ in the interval [2,4].

Solution:

$f(x)$ is continuous and differentiable. We need c in (2,4) such that $f'(c) = \frac{f(4)-f(2)}{4-2}$.

$f'(x)=2x$, so $f'(c)=2c$. The slope of the secant is $\frac{16-4}{2}=6$. So, $2c=6 \implies c=3$. Since $c=3$ is in (2,4), LMVT is verified.

Answer: Verified, c=3.

Example 34: Second Parametric Derivative

Question: If $x=a(\theta - \sin\theta)$ and $y=a(1-\cos\theta)$, find $\frac{d^2y}{dx^2}$ at $\theta=\pi/2$.

Solution:

First, $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a\sin\theta}{a(1-\cos\theta)} = \cot(\theta/2)$.

Now, $\frac{d^2y}{dx^2} = \frac{d}{d\theta}(\cot(\theta/2)) \cdot \frac{d\theta}{dx} = [-\csc^2(\theta/2) \cdot \frac{1}{2}] \cdot [\frac{1}{a(1-\cos\theta)}]$.

At $\theta=\pi/2$: $\frac{d^2y}{dx^2} = [-\csc^2(\pi/4) \cdot \frac{1}{2}] \cdot [\frac{1}{a(1-0)}] = [-2 \cdot \frac{1}{2}] \cdot \frac{1}{a} = -\frac{1}{a}$.

Answer: -1/a.

Example 35: Functional Equations

Question: If $f(x+y)=f(x)f(y)$, and $f(x)$ is differentiable with $f(0)=1, f'(0)=2$, find $f'(x)$.

Solution:

From first principles, $f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h \to 0} \frac{f(x)f(h)-f(x)}{h}$.

$f'(x) = f(x) \lim_{h \to 0} \frac{f(h)-1}{h} = f(x) \lim_{h \to 0} \frac{f(h)-f(0)}{h-0}$.

The limit is the definition of $f'(0)$, which is given as 2.

Therefore, $f'(x) = f(x) \cdot 2 = 2f(x)$.

Answer: $2f(x)$.