JEE Corner — Limits and Derivatives for JEE Main and Advanced

Limits at Infinity and Rationalization

The rationalised syllabus stops at finite points and rational functions; JEE adds four standard extensions.

1. Limits at infinity of rational functions. Divide top and bottom by the highest power of $x$ present. With numerator degree $p$ and denominator degree $q$:

Key Point (Degree comparison): $p < q$ gives limit 0; $p = q$ gives the ratio of leading coefficients; $p > q$ gives no (finite) limit — the expression grows without bound.

So $\lim_{x \to \infty} \frac{3x^2 + 2x + 1}{5x^2 - 4} = \frac{3}{5}$, while $\frac{x^3 + 1}{x^2 + 1}$ blows up.

2. Rationalization for $\frac{0}{0}$ with surds. Multiply by the conjugate to release the hidden factor:

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

The same trick handles $\infty - \infty$ forms: $\lim_{x \to \infty}(\sqrt{x^2 + x} - x) = \lim \frac{x}{\sqrt{x^2+x} + x} = \frac{1}{2}$.

3. Sandwich applications. Anything of the form (bounded) × (vanishing) dies: $\lim_{x \to 0} x\sin\frac{1}{x} = 0$ because $-|x| \le x\sin\frac{1}{x} \le |x|$ — even though $\sin\frac{1}{x}$ itself oscillates forever and has no limit.

[JEE Tip] At infinity, resist substituting $\infty$: divide by the top power first and let each $\frac{1}{x^k}$ term visibly go to 0. The one-line "ratio of leading coefficients" quote is safe only after that habit is solid.

Exponential-Log Limits, Differentiability, and Tangent Applications

4. The exponential-logarithmic standard limits. Four limits power a huge share of JEE Main limit questions:

Key Point: $\lim_{x \to 0}\frac{e^x - 1}{x} = 1$; $\lim_{x \to 0}\frac{a^x - 1}{x} = \ln a$; $\lim_{x \to 0}\frac{\log(1+x)}{x} = 1$; $\lim_{x \to 0}(1 + x)^{1/x} = e = \lim_{x \to \infty}\left(1 + \frac{1}{x}\right)^x$.

Scaling works exactly as for $\frac{\sin x}{x}$: $\frac{e^{3x} - 1}{x} = 3 \cdot \frac{e^{3x} - 1}{3x} \to 3$, and $\frac{a^x - b^x}{x} = \frac{(a^x - 1) - (b^x - 1)}{x} \to \ln a - \ln b = \ln\frac{a}{b}$.

5. Differentiability is stronger than continuity. $f(x) = |x|$ is continuous at 0, but the one-sided derivative limits disagree: from the right $\frac{|h|}{h} = 1$, from the left $-1$. So $|x|$ is not differentiable at 0 — a corner. Differentiable $\Rightarrow$ continuous, never the converse.

6. The derivative in action. Tangent-slope questions: the slope of $y = f(x)$ at $x = a$ is $f'(a)$, so "where is the tangent parallel to a given line?" means solving $f'(x) = m$. Rate questions: if $A = \pi r^2$ then $\frac{dA}{dr} = 2\pi r$ — the derivative is the instantaneous rate of one quantity with respect to another, with the falling body's $v = \frac{ds}{dt}$ as the prototype.

[JEE Tip] "$\lim$ exists", "continuous", "differentiable" is a strict ladder — each rung implies the ones below it and not above. Exam statements love testing the gaps: $|x|$ sits exactly one rung short of the top at 0.

JEE-Style Solved Examples

Example 1: Degrees equal

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

Solution:

Step 1 — Divide by the top power $x^2$. $\frac{3 + \frac{2}{x} + \frac{1}{x^2}}{5 - \frac{4}{x^2}}$.

Step 2 — Let each reciprocal die. $\frac{1}{x}, \frac{1}{x^2} \to 0$.

Step 3 — Read off. $\frac{3}{5}$ — the ratio of leading coefficients, as the equal degrees promise.

Takeaway: Divide-by-top-power makes the "leading coefficients" shortcut VISIBLE — do it until the shortcut is reflex, then quote it.

Example 2: An $\infty - \infty$ rationalization

Evaluate $\lim_{x \to \infty} \left(\sqrt{x^2 + x} - x\right)$.

Solution:

Step 1 — Multiply by the conjugate. $\frac{(x^2 + x) - x^2}{\sqrt{x^2+x} + x} = \frac{x}{\sqrt{x^2+x} + x}$.

Step 2 — Divide by $x$. $\frac{1}{\sqrt{1 + \frac{1}{x}} + 1}$.

Step 3 — Limit. $\frac{1}{1 + 1} = \frac{1}{2}$.

Takeaway: Never subtract infinities — rationalise so the difference becomes a ratio, then divide by the top power.

Example 3: A double-surd rationalization

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

Solution:

Step 1 — Conjugate. $\frac{(1+x) - (1-x)}{x(\sqrt{1+x} + \sqrt{1-x})} = \frac{2x}{x(\sqrt{1+x} + \sqrt{1-x})}$.

Step 2 — Cancel $x$. $\frac{2}{\sqrt{1+x} + \sqrt{1-x}}$.

Step 3 — Substitute. $\frac{2}{1 + 1} = 1$.

Takeaway: The conjugate releases the hidden factor of $x$ — a $\frac{0}{0}$ with surds is a factoring problem wearing a disguise.

Example 4: Exponential scaling

Evaluate $\lim_{x \to 0} \frac{e^{3x} - 1}{x}$.

Solution:

Step 1 — Match the argument. $\frac{e^{3x} - 1}{x} = \frac{e^{3x} - 1}{3x} \cdot 3$.

Step 2 — Standard limit. $\frac{e^{3x} - 1}{3x} \to 1$.

Step 3 — Conclude. $1 \times 3 = 3$.

Takeaway: $\frac{e^{kx} - 1}{x} \to k$ — the exponential scales exactly like $\frac{\sin kx}{x}$.

Example 5: Two exponentials

Evaluate $\lim_{x \to 0} \frac{a^x - b^x}{x}$ ($a, b > 0$).

Solution:

Step 1 — Add and subtract 1. $\frac{(a^x - 1) - (b^x - 1)}{x}$.

Step 2 — Split. $\frac{a^x - 1}{x} - \frac{b^x - 1}{x} \to \ln a - \ln b$.

Step 3 — Combine. $\ln\frac{a}{b}$.

Takeaway: "Insert $\pm 1$" converts any $a^x - b^x$ into two standard limits — a two-second manoeuvre worth automating.

Example 6: A log limit

Evaluate $\lim_{x \to 0} \frac{\log(1 + 4x)}{x}$.

Solution:

Step 1 — Match the argument. $\frac{\log(1+4x)}{4x} \cdot 4$.

Step 2 — Standard limit. $\frac{\log(1+4x)}{4x} \to 1$.

Step 3 — Conclude. $4$.

Takeaway: $\frac{\log(1 + kx)}{x} \to k$ — same scaling law, third family.

Example 7: The $1^\infty$ form

Evaluate $\lim_{x \to \infty} \left(1 + \frac{2}{x}\right)^x$.

Solution:

Step 1 — Force the $e$-shape. $\left(1 + \frac{2}{x}\right)^x = \left[\left(1 + \frac{2}{x}\right)^{x/2}\right]^2$.

Step 2 — Inner limit. With $t = \frac{x}{2} \to \infty$: $\left(1 + \frac{1}{t}\right)^t \to e$.

Step 3 — Conclude. $e^2$.

Takeaway: $\left(1 + \frac{k}{x}\right)^x \to e^k$ — the exponent collects whatever the fraction carries.

Example 8: Bounded times vanishing

Show that $\lim_{x \to 0} x\sin\frac{1}{x} = 0$, although $\lim_{x \to 0} \sin\frac{1}{x}$ does not exist.

Solution:

Step 1 — Bound. $\left|\sin\frac{1}{x}\right| \le 1$ gives $-|x| \le x\sin\frac{1}{x} \le |x|$.

Step 2 — Squeeze. Both bounds $\to 0$: the product $\to 0$. ∎

Step 3 — Contrast. Alone, $\sin\frac{1}{x}$ sweeps $[-1, 1]$ infinitely often near 0 — no limit; the vanishing factor $x$ is what tames it.

Takeaway: Bounded × vanishing → 0, by sandwich — and neither factor needs a limit of its own for the product to have one.

Example 9: A corner point

Show that $f(x) = |x|$ is continuous but not differentiable at 0.

Solution:

Step 1 — Continuity. LHL $= \lim_{x \to 0^-}(-x) = 0$; RHL $= \lim_{x \to 0^+} x = 0$; $f(0) = 0$ — all agree ✓.

Step 2 — The derivative quotient. $\frac{f(0+h) - f(0)}{h} = \frac{|h|}{h}$: from the right $+1$, from the left $-1$.

Step 3 — Conclude. The one-sided limits disagree: $f'(0)$ does not exist — a corner. ∎

Takeaway: Differentiable $\Rightarrow$ continuous, never conversely — $|x|$ is the one-line counterexample every JEE statement question expects you to know.

Example 10: Where is the tangent parallel to a line?

Find the point on $y = x^2 - 3x + 2$ where the tangent has slope 1.

Solution:

Step 1 — Differentiate. $y' = 2x - 3$.

Step 2 — Set the slope. $2x - 3 = 1$: $x = 2$.

Step 3 — Find the point. $y(2) = 4 - 6 + 2 = 0$: the point is $(2, 0)$.

Takeaway: A "tangent parallel to a given line" question always means: differentiate, equate slopes, solve, then substitute BACK for the point.

Example 11: A rate of change

The area of a circle is $A = \pi r^2$. Find the rate of change of $A$ with respect to $r$ at $r = 3$.

Solution:

Step 1 — Differentiate with respect to $r$. $\frac{dA}{dr} = 2\pi r$.

Step 2 — Evaluate. At $r = 3$: $6\pi$.

Step 3 — Interpret. $2\pi r$ is the circumference — growing a disc adds a rim, so area's rate IS the boundary length.

Takeaway: Derivatives against any variable work the same way — and good rate answers admit a geometric reading.

Example 12: Mixing families

Evaluate $\lim_{x \to 0} \frac{3^x - 1}{\sqrt{1 + x} - 1}$.

Solution:

Step 1 — Divide top and bottom by $x$. $\frac{(3^x - 1)/x}{(\sqrt{1+x} - 1)/x}$.

Step 2 — Limit each family. Top $\to \ln 3$ (exponential standard); bottom $\to \frac{1}{2}$ (workhorse with $n = \frac{1}{2}$).

Step 3 — Divide. $\frac{\ln 3}{1/2} = 2\ln 3$.

Takeaway: Divide-by-$x$ is the universal splitter — each family contributes its standard constant, and the answer is their ratio.