How to find limits — direct substitution, factoring, L'Hôpital, squeeze theorem

Question

Given a limit problem, how do we decide which method to use — direct substitution, factoring, rationalisation, L’Hopital’s rule, or the squeeze theorem? Walk through the decision process with examples.

(CBSE 11/12 + JEE Main — appears in almost every paper)

Solution — Step by Step

Plug in the value of $x$ . If you get a finite number (not $\frac{0}{0}$ or $\frac{\infty}{\infty}$ ), that IS the limit.

Example: $\displaystyle\lim_{x \to 2} (3x + 1) = 3(2) + 1 = \mathbf{7}$

No tricks needed. Most students overcomplicate limits that are just plug-and-play.

Factor the numerator and denominator, then cancel the common factor.

Example: $\displaystyle\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = \lim_{x \to 3} \frac{(x-3)(x+3)}{x-3} = \lim_{x \to 3} (x+3) = \mathbf{6}$

For expressions with square roots, use rationalisation — multiply by the conjugate.

When you still get $\frac{0}{0}$ or $\frac{\infty}{\infty}$ after simplification, differentiate numerator and denominator separately:

\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}

Example: $\displaystyle\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\cos x}{1} = \mathbf{1}$

If $g(x) \leq f(x) \leq h(x)$ near $x = a$ , and $\lim g(x) = \lim h(x) = L$ , then $\lim f(x) = L$ .

Example: $\displaystyle\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = \mathbf{0}$

Since $-x^2 \leq x^2 \sin(1/x) \leq x^2$ and both bounds go to 0.

flowchart TD
    A["Given: Find limit as x → a"] --> B["Substitute x = a"]
    B --> C{"Result?"}
    C -- "Finite number" --> D["That is the answer"]
    C -- "0/0 form" --> E{"Can you factor or rationalise?"}
    E -- Yes --> F["Cancel common factor, substitute again"]
    F --> D
    E -- No --> G["Apply L'Hopital: differentiate top and bottom"]
    G --> H["Substitute again"]
    H --> D
    C -- "∞/∞ form" --> G
    C -- "Oscillating / bounded × vanishing" --> I["Try Squeeze Theorem"]
    I --> D
    C -- "∞ - ∞ form" --> J["Rearrange to 0/0 or ∞/∞"]
    J --> G

Why This Works

Every limit method is really about resolving an indeterminate form. Direct substitution works when there is no indeterminacy. Factoring removes the factor causing $0/0$ . L’Hopital replaces the original ratio with the ratio of rates-of-change, which often resolves the indeterminacy. The squeeze theorem handles cases where the function oscillates but is pinched to a single value.

The key insight: always identify the indeterminate form FIRST ( $0/0$ , $\infty/\infty$ , $0 \cdot \infty$ , $\infty - \infty$ , $1^\infty$ , $0^0$ , $\infty^0$ ). The form tells you which tool to reach for.

Alternative Method

For standard limits, memorise these results directly:

\lim_{x \to 0} \frac{\sin x}{x} = 1, \quad \lim_{x \to 0} \frac{\tan x}{x} = 1, \quad \lim_{x \to 0} \frac{e^x - 1}{x} = 1

\lim_{x \to 0} \frac{\ln(1+x)}{x} = 1, \quad \lim_{x \to 0} \frac{a^x - 1}{x} = \ln a

\lim_{x \to 0} (1 + x)^{1/x} = e, \quad \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e

In JEE Main, about 70% of limit questions can be solved using standard limits combined with algebraic manipulation. L’Hopital is the backup plan, not the first choice — it is slower and more error-prone under time pressure.

Common Mistake

Students apply L’Hopital’s rule when the form is NOT indeterminate. If substitution gives $\frac{5}{0}$ , that is not $\frac{0}{0}$ — it means the limit is $\pm\infty$ or does not exist. L’Hopital only applies for $\frac{0}{0}$ or $\frac{\infty}{\infty}$ . Misapplying it to a non-indeterminate form gives a completely wrong answer.