Techniques for Evaluating Limits

Hey JEE aspirants! Limits form the foundation of calculus, and mastering them is crucial for acing your JEE Main exam. You'll encounter limits directly in questions and indirectly within continuity, differentiability, and integration. Let's dive into the techniques you need to solve limit problems efficiently.

1. Direct Substitution Method

The simplest approach! If the function $f(x)$ is well-defined at $x = a$ , and doesn't result in an indeterminate form (like $0/0$ , $\infty/\infty$ ), then:

$\lim_{x \rightarrow a} f(x) = f(a)$

Example: Evaluate $\lim_{x \rightarrow 2} (x^2 + 3x - 1)$ . Since polynomials are continuous everywhere, we can directly substitute:

$\lim_{x \rightarrow 2} (x^2 + 3x - 1) = (2)^2 + 3(2) - 1 = 4 + 6 - 1 = 9$

2. Factoring Technique for 0/0 Forms

Encountering $0/0$ ? This means both the numerator and denominator have a common factor that's causing the expression to approach zero. Factoring helps reveal and cancel this factor.

Example: Find $\lim_{x \rightarrow 3} \frac{x^2 - 9}{x - 3}$ . Direct substitution gives $0/0$ . Factor the numerator:

$\lim_{x \rightarrow 3} \frac{x^2 - 9}{x - 3} = \lim_{x \rightarrow 3} \frac{(x - 3)(x + 3)}{x - 3} = \lim_{x \rightarrow 3} (x + 3) = 3 + 3 = 6$

3. Rationalizing Technique for Indeterminate Forms

When dealing with square roots (or other radicals) that lead to indeterminate forms, rationalizing the numerator or denominator can simplify the expression. Multiply by the conjugate!

Example: Evaluate $\lim_{x \rightarrow 0} \frac{\sqrt{1 + x} - 1}{x}$ . Direct substitution results in $0/0$ . Rationalize the numerator:

$\lim_{x \rightarrow 0} \frac{\sqrt{1 + x} - 1}{x} = \lim_{x \rightarrow 0} \frac{(\sqrt{1 + x} - 1)(\sqrt{1 + x} + 1)}{x(\sqrt{1 + x} + 1)} = \lim_{x \rightarrow 0} \frac{(1 + x) - 1}{x(\sqrt{1 + x} + 1)} = \lim_{x \rightarrow 0} \frac{x}{x(\sqrt{1 + x} + 1)} = \lim_{x \rightarrow 0} \frac{1}{\sqrt{1 + x} + 1} = \frac{1}{\sqrt{1 + 0} + 1} = \frac{1}{2}$

4. Limits of Piecewise Functions

For piecewise functions, you need to evaluate the left-hand limit (LHL) and the right-hand limit (RHL) separately. The limit exists only if LHL = RHL.

Example: Consider the function:

$f(x) = \begin{cases} x + 1, & x < 2 \\ 3x - 3, & x \ge 2 \end{cases}$

Find $\lim_{x \rightarrow 2} f(x)$ .

LHL: $\lim_{x \rightarrow 2^-} f(x) = \lim_{x \rightarrow 2^-} (x + 1) = 2 + 1 = 3$

RHL: $\lim_{x \rightarrow 2^+} f(x) = \lim_{x \rightarrow 2^+} (3x - 3) = 3(2) - 3 = 3$

Since LHL = RHL = 3, $\lim_{x \rightarrow 2} f(x) = 3$ .

5. Standard Limits Formulas

These are your bread and butter! Memorize them and recognize situations where you can apply or manipulate an expression to fit these forms.

Formula 1: $\lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1$

Explanation: Geometrically, as $x$ approaches 0, the arc length of a circle with radius 1 is approximately equal to the sine of the angle. This is a cornerstone of trigonometric limits. You can prove it using the Squeeze Theorem.

Formula 2: $\lim_{x \rightarrow 0} \frac{1 - \cos(x)}{x} = 0$

Explanation: Multiply by the conjugate: $\lim_{x \rightarrow 0} \frac{1 - \cos(x)}{x} \cdot \frac{1 + \cos(x)}{1 + \cos(x)} = \lim_{x \rightarrow 0} \frac{1 - \cos^2(x)}{x(1 + \cos(x))} = \lim_{x \rightarrow 0} \frac{\sin^2(x)}{x(1 + \cos(x))} = \lim_{x \rightarrow 0} \frac{\sin(x)}{x} \cdot \lim_{x \rightarrow 0} \frac{\sin(x)}{1 + \cos(x)} = 1 \cdot \frac{0}{2} = 0$

Formula 3: $\lim_{x \rightarrow 0} \frac{e^x - 1}{x} = 1$

Explanation: This is the derivative of $e^x$ evaluated at $x = 0$ . Alternatively, let $y = e^x - 1$ , so $x = \ln(1 + y)$ . As $x \rightarrow 0$ , $y \rightarrow 0$ . Thus, $\lim_{x \rightarrow 0} \frac{e^x - 1}{x} = \lim_{y \rightarrow 0} \frac{y}{\ln(1 + y)} = \lim_{y \rightarrow 0} \frac{1}{\frac{\ln(1 + y)}{y}} = \frac{1}{1} = 1$ (using Formula 4, see below, in reverse).

Formula 4: $\lim_{x \rightarrow 0} \frac{\ln(1 + x)}{x} = 1$

Explanation: Similar to Formula 3, this is related to the derivative of $\ln(x)$ at $x = 1$ . Let $y = \ln(1 + x)$ , then $e^y = 1 + x$ , so $x = e^y - 1$ . As $x \rightarrow 0$ , $y \rightarrow 0$ . Thus, $\lim_{x \rightarrow 0} \frac{\ln(1 + x)}{x} = \lim_{y \rightarrow 0} \frac{y}{e^y - 1} = \lim_{y \rightarrow 0} \frac{1}{\frac{e^y - 1}{y}} = \frac{1}{1} = 1$ (using Formula 3).

Formula 5: $\lim_{x \rightarrow \infty} \left(1 + \frac{1}{x}\right)^x = e$

Explanation: This defines the number $e$ . Let $y = 1/x$ , then as $x \rightarrow \infty$ , $y \rightarrow 0$ . So, $\lim_{x \rightarrow \infty} \left(1 + \frac{1}{x}\right)^x = \lim_{y \rightarrow 0} (1 + y)^{1/y} = e$ . A related form is $\lim_{x \rightarrow 0} (1 + x)^{1/x} = e$ .

Formula 6: $\lim_{x \rightarrow a} \frac{x^n - a^n}{x - a} = n \cdot a^{n-1}$

Explanation: This is essentially the definition of the derivative of $x^n$ at $x = a$ . You can prove it by factoring $x^n - a^n = (x - a)(x^{n-1} + x^{n-2}a + ... + a^{n-1})$ . Then, $\lim_{x \rightarrow a} \frac{x^n - a^n}{x - a} = \lim_{x \rightarrow a} (x^{n-1} + x^{n-2}a + ... + a^{n-1}) = a^{n-1} + a^{n-1} + ... + a^{n-1}$ ( $n$ terms) $= n a^{n-1}$ .

Tip: Always try to manipulate the given expression into a form that resembles one of the standard limits. Use algebraic identities and trigonometric transformations to simplify the expression.

Common Mistake: Forgetting to check LHL and RHL for piecewise functions. A limit exists ONLY if they are equal.

JEE Trick: When dealing with limits at infinity involving rational functions (polynomials divided by polynomials), divide both the numerator and denominator by the highest power of $x$ present in the denominator. This helps simplify the expression and evaluate the limit easily.

Example (JEE Style): Evaluate $\lim_{x \rightarrow \infty} \frac{3x^2 + 5x - 2}{2x^2 - x + 7}$ .

Divide both numerator and denominator by $x^2$ :

$\lim_{x \rightarrow \infty} \frac{3 + \frac{5}{x} - \frac{2}{x^2}}{2 - \frac{1}{x} + \frac{7}{x^2}} = \frac{3 + 0 - 0}{2 - 0 + 0} = \frac{3}{2}$

Keep practicing, and you'll become a limits ninja! Good luck with your JEE prep!