Module 4: Limits and Derivatives

Dear Class 12 Student! Limits and Derivatives is the gatekeeper of Class 12. The entire first half of your Class 12 syllabus (Continuity, Differentiability, Application of Derivatives, and Integrals) is built on these two concepts. If you cannot solve a limit or differentiate a basic function using the Product Rule, Class 12 will feel like a nightmare. Let's fix that now!

1. Concept of a Limit

A limit describes how a function behaves as its input approaches a specific value, even if the function is not defined at that exact value. We write this as $\lim_{x \to a} f(x) = L$.

Condition for Existence A limit exists at $x = a$ if and only if the Left Hand Limit (LHL) and Right Hand Limit (RHL) are equal and finite: $$\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L$$

LHL: Approaching $a$ from values slightly less than $a$. ($x = a - h$, where $h \to 0$)
RHL: Approaching $a$ from values slightly more than $a$. ($x = a + h$, where $h \to 0$)

Topic 1 Check Question: Given $f(x) = \begin{cases} 2x+1, & x < 3 \\ 10, & x=3 \\ x^2 - 2, & x > 3 \end{cases}$. Does $\lim_{x \to 3} f(x)$ exist?

Solution: Check LHL and RHL.
LHL: $\lim_{x \to 3^-} (2x+1) = 2(3)+1 = 7$.
RHL: $\lim_{x \to 3^+} (x^2-2) = (3^2)-2 = 7$.
Since LHL = RHL = 7, the limit exists and equals 7 (even though $f(3)=10$).

2. Algebra of Limits

Assuming $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$ exist:

Sum/Difference: $\lim [f(x) \pm g(x)] = L \pm M$
Product: $\lim [f(x) \cdot g(x)] = L \cdot M$
Quotient: $\lim \frac{f(x)}{g(x)} = \frac{L}{M}$, provided $M \neq 0$.

Topic 2 Check Question: Evaluate $\lim_{x \to 2} \frac{x^2 - 4}{x+2}$.

Solution: Since the denominator does not equal zero at $x=2$, we can apply the quotient theorem directly via direct substitution.
$= \frac{2^2 - 4}{2+2} = \frac{0}{4} = \mathbf{0}$.

3. Standard Limits to Memorize

The Golden Formulas

Polynomial Type: $$\lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1}$$
Trigonometric Type: $$\lim_{x \to 0} \frac{\sin x}{x} = 1 \quad \text{and} \quad \lim_{x \to 0} \frac{\tan x}{x} = 1$$
Exponential Type: $$\lim_{x \to 0} \frac{e^x - 1}{x} = 1$$
Logarithmic Type: $$\lim_{x \to 0} \frac{\ln(1+x)}{x} = 1$$

Practice Problem 1 Question: Evaluate $\lim_{x \to 0} \frac{\sin 5x}{3x}$.

Solution:
$\frac{1}{3} \lim_{x \to 0} \frac{\sin 5x}{x} = \frac{5}{3} \lim_{x \to 0} \frac{\sin 5x}{5x} = \frac{5}{3} \times 1 = \mathbf{\frac{5}{3}}$.

4. Derivatives: The Rate of Change

A derivative represents the instantaneous rate of change of a function. Geometrically, it is the slope of the tangent to the curve at a given point.

The First Principle of Differentiation

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

Topic 4 Check Question: True or False: Geometrically, the derivative is the slope of the secant line connecting two points.

Solution: False. The secant line connects two distinct points. The derivative is the slope of the tangent line at a single exact point, which occurs when the two points converge ($h \to 0$).

5. Rules of Differentiation

Memorize These Rules

Power Rule: $\frac{d}{dx}(x^n) = nx^{n-1}$
Product Rule: $\frac{d}{dx}(uv) = uv' + vu'$
Quotient Rule: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{vu' - uv'}{v^2}$

Topic 5 Check Question: Identify the mathematical error in this differentiation block: $\frac{d}{dx}(x^3 \sin x) = (3x^2)(\cos x)$.

Solution: The error is assuming the derivative of a product is the product of derivatives. You cannot just multiply the individual derivatives.
You must use the Product Rule: $uv' + vu'$.
Correct derivative is: $\mathbf{(x^3)(\cos x) + (\sin x)(3x^2)}$.

6. Derivatives of Standard Functions

Function $f(x)$	Derivative $f'(x)$
$x^n$	$nx^{n-1}$
$\sin x$	$\cos x$
$\cos x$	$-\sin x$
$\tan x$	$\sec^2 x$
$\sec x$	$\sec x \tan x$
$e^x$	$e^x$
$\ln x$	$1/x$

Topic 6 Check Question: What is $\frac{d}{dx}(x^2 - 3\tan x + 5e^x)$?

Solution: Apply the sum/difference rule and the standard formulas.
$\frac{d}{dx}(x^2) = 2x$
$\frac{d}{dx}(-3\tan x) = -3\sec^2 x$
$\frac{d}{dx}(5e^x) = 5e^x$
Answer: $\mathbf{2x - 3\sec^2 x + 5e^x}$.

MEGA PRACTICE BANK

Problem 2: Limit Factorization (0/0 Form) Question: Evaluate $\lim_{x \to 3} \frac{x^2 - 9}{x^2 - 5x + 6}$.

Solution: Direct substitution yields $\frac{0}{0}$. We must factorize.
Numerator: $x^2 - 9 = (x - 3)(x + 3)$
Denominator: $x^2 - 5x + 6 = (x - 3)(x - 2)$
Cancel the $(x - 3)$ term:
$\lim_{x \to 3} \frac{x + 3}{x - 2} = \frac{3 + 3}{3 - 2} = \frac{6}{1} = \mathbf{6}$.

Problem 3: Product Rule Mechanics Question: Differentiate $y = x^2 \sin x$.

Solution: Let $u = x^2$ and $v = \sin x$.
Apply Product Rule: $y' = u v' + v u'$
$y' = (x^2)(\cos x) + (\sin x)(2x) = \mathbf{x^2 \cos x + 2x \sin x}$.

Problem 4: Quotient Rule Mechanics Question: Differentiate $y = \frac{x+1}{x-1}$.

Solution: Let $u = x+1, u' = 1$ and $v = x-1, v' = 1$.
Apply Quotient Rule: $y' = \frac{v u' - u v'}{v^2}$
$y' = \frac{(x-1)(1) - (x+1)(1)}{(x-1)^2}$
$y' = \frac{x - 1 - x - 1}{(x-1)^2} = \mathbf{\frac{-2}{(x-1)^2}}$.

Problem 5: First Principle Derivation Question: Use the First Principle of differentiation to find the derivative of $f(x) = x^2$.

Solution: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
$f'(x) = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h}$
Expand the square: $f'(x) = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h}$
$f'(x) = \lim_{h \to 0} \frac{2xh + h^2}{h} = \lim_{h \to 0} (2x + h)$
As $h \to 0, f'(x) = \mathbf{2x}$. This mathematically proves the Power Rule!

Problem 6: Mixed Derivative Question: Differentiate $y = e^x \tan x$.

Solution: Apply Product Rule ($uv' + vu'$).
$u = e^x \implies u' = e^x$
$v = \tan x \implies v' = \sec^2 x$
$y' = e^x(\sec^2 x) + \tan x(e^x)$
Factoring out $e^x$: $y' = \mathbf{e^x (\sec^2 x + \tan x)}$.