Limits for JEE: Standard Limits, Expansions and the 1^∞ Form

A limit answers one question: as $x$ gets closer and closer to a point, what value does $f(x)$ settle near? The value at the point itself does not matter, and often does not even exist. That is why limits can handle $\dfrac00$, $\dfrac{\infty}{\infty}$ and $1^\infty$, expressions that make no sense when you simply substitute.

This note builds the chapter in the order you use it in a paper: what a limit is and when it exists, the indeterminate forms, the standard limits, limits at infinity, the sandwich theorem, expansions, L'Hôpital's rule, the exponent forms $1^\infty$, $0^0$, $\infty^0$, the greatest integer and fractional part, and finally finding unknown constants from a given limit. Every worked example is a JEE Main 2025 or 2026 question unless it says otherwise, and every answer was re-derived from scratch.

What a limit means

We write

$$\lim_{x\to a}f(x)=L$$

when $f(x)$ can be made as close to $L$ as we like by taking $x$ close enough to $a$, with $x\ne a$.

Approaching from the left gives the left hand limit (LHL), from the right the right hand limit (RHL):

$$\text{LHL}=\lim_{x\to a^-}f(x),\qquad\text{RHL}=\lim_{x\to a^+}f(x)$$

The limit exists exactly when both exist, are finite, and are equal:

$$\lim_{x\to a}f(x)\text{ exists}\iff\text{LHL}=\text{RHL}$$ Two graphs side by side: on the left a curve that jumps at x equals a, with an open dot at the end of the left piece and a filled dot at the start of the right piece; on the right a smooth curve with a hollow dot at x equals a and a separate filled dot above it

The left picture has LHL $\ne$ RHL, so no limit. The right picture has a limit even though $f(a)$ is somewhere else: the limit only looks at nearby points.

Example 1 (IITianForum). Find, if it exists,

$$\lim_{x\to0}\frac{2^{1/x}-1}{2^{1/x}+1}$$

▸ SHOW SOLUTION

Step 1. Right side

$x\to0^+\Rightarrow\dfrac1x\to+\infty\Rightarrow2^{1/x}\to\infty$. Divide by $2^{1/x}$:

$$\text{RHL}=\lim_{x\to0^+}\frac{1-2^{-1/x}}{1+2^{-1/x}}=\frac{1-0}{1+0}=1$$

Step 2. Left side

$x\to0^-\Rightarrow\dfrac1x\to-\infty\Rightarrow2^{1/x}\to0$:

$$\text{LHL}=\frac{0-1}{0+1}=-1$$

Step 3. Compare

LHL $\ne$ RHL.

Answer: the limit does not exist.

Indeterminate forms, factorisation and rationalisation

If $\lim f=l$ and $\lim g=m$ are finite, limits obey ordinary algebra:

$$\lim(f\pm g)=l\pm m,\qquad\lim(fg)=lm$$ $$\lim\frac fg=\frac lm\quad(m\ne0)$$

The trouble starts when substitution gives one of the seven indeterminate forms:

$$\frac00,\quad\frac{\infty}{\infty},\quad0\cdot\infty,\quad\infty-\infty$$ $$1^\infty,\quad0^0,\quad\infty^0$$

Each one can end up as any number, so the expression has to be rewritten until substitution works. The three oldest rewrites are: cancel a common factor $(x-a)$, rationalise a root, and take out the highest power. The rest of this note supplies the others.

Example 2 (JEE Main 2025, 24 January Shift 1). Let $p=\sqrt{2\cos^2x+3\cos x}$ and $q=\sqrt{\cos^2x+\sin x+4}$. The value of

$$\lim_{x\to0}\operatorname{cosec}x\,(p-q)$$

is

▸ SHOW SOLUTION

Step 1. Rationalise

The difference of the two radicands is

$$\cos^2x+3\cos x-4-\sin x$$$$=(\cos x-1)(\cos x+4)-\sin x$$

Step 2. Divide by sin x

$$\frac{(\cos x-1)(\cos x+4)}{\sin x}-1$$

$\dfrac{\cos x-1}{\sin x}\to0$, so this tends to $-1$.

Step 3. The two roots

Each root tends to $\sqrt5$, so their sum tends to $2\sqrt5$.

$$\text{Limit}=\frac{-1}{2\sqrt5}$$

Answer: (4) $-\dfrac{1}{2\sqrt5}$

The standard limits

Almost every limit question reduces to one of these. Learn them with the shape of the argument, not just the answer.

In every row $x$ may be replaced by any expression that tends to $0$. For instance

$$\lim_{x\to0}\frac{\sin5x^2}{5x^2}=1$$

because $5x^2\to0$. This replacement is how most exam limits are cracked.

Why sin x over x tends to 1

For $0<x<\dfrac\pi2$, compare three areas in the unit circle: the triangle $OAP$, the sector $OAP$, and the triangle $OAT$ with $AT$ tangent.

Unit circle sector from O with radius OA along the x-axis, point P on the circle at angle x, the vertical tangent at A meeting OP extended at T, and the triangle OAP, the sector OAP and the triangle OAT outlined

$$\frac12\sin x<\frac12x<\frac12\tan x$$

Divide by $\dfrac12\sin x$ and take reciprocals:

$$\cos x<\frac{\sin x}{x}<1$$

As $x\to0$ both outer terms tend to $1$, so $\dfrac{\sin x}{x}\to1$. Both sides give the same value because $\dfrac{\sin x}{x}$ is even.

Graph of sin x over x for x between minus 3 pi and 3 pi, a central hump rising to 1 near the origin with a hollow dot at (0, 1) and smaller decaying waves on both sides

The other entries follow from this one and from $e$:

$$\frac{1-\cos x}{x^2}=\frac{2\sin^2(x/2)}{x^2}$$ $$=\frac12\left(\frac{\sin(x/2)}{x/2}\right)^2$$ $$\frac{\ln(1+x)}{x}=\ln(1+x)^{1/x}\to\ln e=1$$

Worked examples

Example 3 (JEE Main 2025, 7 April Shift 2). For $t>-1$, let $\alpha_t$, $\beta_t$ be the roots of

$$a_tx^2+b_tx+c_t=0$$

where $a_t=(t+2)^{1/7}-1$, $b_t=(t+2)^{1/6}-1$ and $c_t=(t+2)^{1/21}-1$.

If $\alpha_t\to a$ and $\beta_t\to b$ as $t\to-1^+$, then $72(a+b)^2$ is ______.

▸ SHOW SOLUTION

Step 1. Put s = t + 2

$s\to1$. Divide the whole equation by $s-1$ and use the first row of the table:

$$\frac{s^k-1}{s-1}\to k$$

with $k=\dfrac17,\ \dfrac16,\ \dfrac1{21}$.

Step 2. Limiting equation

$$\frac{x^2}{7}+\frac x6+\frac1{21}=0$$$$6x^2+7x+2=0$$

Step 3. Sum of roots

$$a+b=-\frac76$$$$72\cdot\frac{49}{36}=98$$

Answer: $98$

Example 4 (JEE Main 2026, 28 January Shift 1). The value of

$$\lim_{x\to0}\frac{\ln\left(\sec(ex)\sec(e^2x)\cdots\sec(e^{10}x)\right)}{e^2-e^{2\cos x}}$$

is

▸ SHOW SOLUTION

Step 1. Numerator

$\ln\sec\theta=-\ln\cos\theta$, and $\cos\theta=1-(1-\cos\theta)$, so

$$-\ln\cos\theta\approx1-\cos\theta\approx\frac{\theta^2}{2}$$$$\text{Numerator}\approx\frac{x^2}{2}\left(e^2+e^4+\cdots+e^{20}\right)$$

Step 2. Denominator

$$e^2-e^{2\cos x}=-e^2\left(e^{2\cos x-2}-1\right)$$

$2\cos x-2\approx-x^2\to0$, and $\dfrac{e^u-1}{u}\to1$, so

$$\text{Denominator}\approx e^2x^2$$

Step 3. Divide

$$\frac{e^2+e^4+\cdots+e^{20}}{2e^2}=\frac{1+e^2+\cdots+e^{18}}{2}$$$$=\frac{e^{20}-1}{2(e^2-1)}$$

Answer: (3)

Example 5 (JEE Main 2026, 5 April Shift 2). Let

$$f(x)=\lim_{y\to0}\frac{(1-\cos xy)\tan xy}{y^3}$$

The number of solutions of $f(x)=\sin x$, $x\in\mathbb R$, is

▸ SHOW SOLUTION

Step 1. Standard limits in y

$$1-\cos xy\approx\frac{x^2y^2}{2},\qquad\tan xy\approx xy$$$$f(x)=\frac{x^3}{2}$$

Step 2. Count roots of x³/2 = sin x

$g(x)=\dfrac{x^3}{2}-\sin x$ is odd and $g(0)=0$. For small $x>0$, $g<0$; for $x\ge2$, $\dfrac{x^3}{2}>1\ge\sin x$. Also $g''(x)=3x+\sin x>0$ for $x>0$, so $g$ is convex there and crosses zero only once after $0$. So one root in $(0,2)$, its mirror root, and $x=0$.

Answer: (3) three solutions

Example 6 (JEE Main 2026, 5 April Shift 1). Let $C(x)=\cos\alpha x\cos(\alpha+1)x\cos(\alpha+2)x$. The product of all values of $\alpha$ for which

$$\lim_{x\to0}\frac{1-C(x)}{\sin^2(\alpha+1)x}=2$$

is

▸ SHOW SOLUTION

Step 1. Each cosine to second order

$$\cos px\approx1-\frac{p^2x^2}{2}$$

so the product of the three cosines is

$$1-\frac{\alpha^2+(\alpha+1)^2+(\alpha+2)^2}{2}x^2$$

Step 2. The limit

$$\frac{\alpha^2+(\alpha+1)^2+(\alpha+2)^2}{2(\alpha+1)^2}=2$$

Step 3. Solve with u = α + 1

$$(u-1)^2+u^2+(u+1)^2=4u^2$$$$3u^2+2=4u^2\implies u^2=2$$

$\alpha=-1\pm\sqrt2$, product $1-2=-1$.

Answer: (3) $-1$

Example 7 (JEE Main 2025, 7 April Shift 1). The value of

$$\lim_{x\to0^+}\frac{\tan(5x^{1/3})\,\ln(1+3x^2)}{\left(\tan^{-1}3\sqrt x\right)^2\left(e^{5x^{4/3}}-1\right)}$$

is

▸ SHOW SOLUTION

Step 1. Replace each factor by its standard approximation

$$\tan5x^{1/3}\approx5x^{1/3},\qquad\ln(1+3x^2)\approx3x^2$$$$\left(\tan^{-1}3\sqrt x\right)^2\approx9x$$$$e^{5x^{4/3}}-1\approx5x^{4/3}$$

Step 2. Powers of x

Top: $15x^{7/3}$. Bottom: $45x^{7/3}$.

Answer: (3) $\dfrac13$

Example 8 (JEE Main 2025, 28 January Shift 2). Let

$$f(x)=\lim_{n\to\infty}\sum_{r=0}^{n}\frac{\tan\frac{x}{2^{r+1}}+\tan^3\frac{x}{2^{r+1}}}{1-\tan^2\frac{x}{2^{r+1}}}$$

Then the value of the following limit is ______.

$$\lim_{x\to0}\frac{e^x-e^{f(x)}}{x-f(x)}$$

▸ SHOW SOLUTION

Step 1. Each term is a difference

With $t=\tan\theta$,

$$\tan2\theta-\tan\theta=\frac{2t}{1-t^2}-t=\frac{t+t^3}{1-t^2}$$

so term $r$ is $\tan\dfrac{x}{2^r}-\tan\dfrac{x}{2^{r+1}}$.

Step 2. Telescope

$$\sum_{r=0}^{n}=\tan x-\tan\frac{x}{2^{n+1}}\to\tan x$$

Step 3. Standard exponential limit

$$\frac{e^x-e^{\tan x}}{x-\tan x}=e^{\tan x}\cdot\frac{e^{x-\tan x}-1}{x-\tan x}\to1\cdot1$$

Answer: $1$

Limits at infinity

As $x\to\infty$, divide numerator and denominator by the highest power of $x$ present. For a rational function this gives a rule by degree:

Graph of the rational function 2x squared plus 1 over x squared plus 3, starting low near the origin and rising towards the dashed horizontal line y equals 2 on both sides

Two more facts are used constantly. Every positive power of $x$ beats $\ln x$, and $e^x$ beats every power of $x$:

$$\frac{\ln x}{x^k}\to0,\qquad\frac{x^k}{e^x}\to0\qquad(k>0)$$ The curves y equals e to the x and y equals x cubed for x from 0 to 6, x cubed ahead for a while, then e to the x crossing it near x equals 4.5 and racing away

For a difference of roots, multiply by the conjugate. For cube roots use $a-b=\dfrac{a^3-b^3}{a^2+ab+b^2}$.

Example 9 (IITianForum). If

$$\lim_{x\to\infty}\left(\sqrt[3]{x^3+3x^2}-\sqrt[3]{x^3-2x^2}\right)=\frac pq$$

with $p,q$ coprime natural numbers, find $p+q$.

▸ SHOW SOLUTION

Step 1. Cube-root conjugate

Put $a=\sqrt[3]{x^3+3x^2}$, $b=\sqrt[3]{x^3-2x^2}$.

$$a^3-b^3=5x^2$$

Step 2. Size of the denominator

$a\approx x$, $b\approx x$, so

$$a^2+ab+b^2\approx3x^2$$

Step 3. Limit

$$a-b=\frac{5x^2}{a^2+ab+b^2}\to\frac53$$

Answer: $p+q=8$

Limits of the form xⁿ as n → ∞

As $n\to\infty$, $x^n\to0$ when $\lvert x\rvert<1$, $x^n\to\infty$ when $x>1$, and $x^{2n}\to\infty$ when $\lvert x\rvert>1$. So a limit in $n$ defines a function piece by piece, and the pieces meet at $\lvert x\rvert=1$.

Curves of x to the power 2n for n equal to 1, 2, 5 and 20 on the interval from minus 1.5 to 1.5, flattening to 0 inside minus 1 to 1 and shooting up outside, all passing through (1, 1) and (minus 1, 1)

Example 10 (IITianForum). Let

$$f(x)=\lim_{n\to\infty}\frac{x^{2n}-1}{x^{2n}+1}$$

Find $f(x)$, and decide whether the limit of $f(x)$ as $x\to1$ exists.

▸ SHOW SOLUTION

Step 1. Three cases

$\lvert x\rvert<1$: $x^{2n}\to0$, so $f(x)=-1$.

$\lvert x\rvert>1$: divide by $x^{2n}$, so $f(x)=1$.

$\lvert x\rvert=1$: $x^{2n}=1$, so $f(x)=0$.

Step 2. Near x = 1

LHL $=-1$ (inside), RHL $=1$ (outside).

Answer: $f$ is $-1$, $0$, $1$ as above, and the limit at $x=1$ does not exist.

Example 11 (JEE Main 2025, 4 April Shift 1). Let $f:\mathbb R\to\mathbb R$ be continuous with $f(0)=1$ and $f(2x)-f(x)=x$ for all $x$. If

$$G(x)=\lim_{n\to\infty}\left(f(x)-f\left(\frac{x}{2^n}\right)\right)$$

then $G(1)+G(4)+G(9)+\cdots+G(100)$ is

▸ SHOW SOLUTION

Step 1. Telescope with the given rule

Replacing $x$ by $\dfrac{x}{2^k}$ in $f(2x)-f(x)=x$:

$$f\left(\frac{x}{2^{k-1}}\right)-f\left(\frac{x}{2^k}\right)=\frac{x}{2^k}$$

Adding for $k=1$ to $n$:

$$f(x)-f\left(\frac{x}{2^n}\right)=x\left(1-\frac1{2^n}\right)$$

Step 2. Let n → ∞

$$G(x)=x$$

Step 3. Sum

$$1+4+9+\cdots+100=385$$

Answer: (2) $385$

Continuity of $f$ at $0$ is what lets $f\left(\frac{x}{2^n}\right)\to f(0)=1$, which also gives $f(x)=1+x$.

Sandwich theorem

If $g(x)\le f(x)\le h(x)$ near $a$, and $g$ and $h$ have the same limit $L$, then $f$ has limit $L$ too.

The curve x squared times sine of 1 over x oscillating faster and faster near the origin, trapped between the dashed parabolas y equals x squared and y equals minus x squared, all three meeting at the origin

The squeeze is the natural tool whenever a bounded quantity appears: $-1\le\sin\theta\le1$, or $t-1<[t]\le t$ for the greatest integer function.

Example 12 (IITianForum). For a real number $x$, find

$$\lim_{n\to\infty}\frac{[x]+[2x]+[3x]+\cdots+[nx]}{n^2}$$

▸ SHOW SOLUTION

Step 1. Bound each term

$$kx-1<[kx]\le kx$$

Step 2. Add for k = 1 to n

$$\frac{n(n+1)}{2}x-n<S\le\frac{n(n+1)}{2}x$$

Step 3. Divide by n² and squeeze

Both bounds tend to $\dfrac x2$.

Answer: $\dfrac x2$

Example 13 (JEE Main 2026, 21 January Shift 2). Let

$$f(x)=\lim_{n\to\infty}\frac1{n^3}\sum_{k=1}^{n}\left[\frac{k^2}{3^x}\right]$$

Then $12\left(f(1)+f(2)+f(3)+\cdots\right)$ is ______.

▸ SHOW SOLUTION

Step 1. Squeeze

$$\frac{k^2}{3^x}-1<\left[\frac{k^2}{3^x}\right]\le\frac{k^2}{3^x}$$

Sum, divide by $n^3$, and use $\displaystyle\sum k^2=\frac{n(n+1)(2n+1)}{6}$:

$$\frac{n(n+1)(2n+1)}{6n^3}\to\frac13,\qquad\frac{n}{n^3}\to0$$$$f(x)=\frac{1}{3^{x+1}}$$

Step 2. The series

$$\sum_{j=1}^{\infty}\frac{1}{3^{j+1}}=\frac{1/9}{1-1/3}=\frac16$$

Answer: $12\cdot\dfrac16=2$

Expansions

When a limit needs more than the first term, use these known expansions near $x=0$. Each line is exact up to the last term shown.

The rule: expand top and bottom to the first power that survives in the denominator, then compare.

Example 14 (JEE Main 2026, 6 April Shift 1). The value of

$$\lim_{x\to0}\frac{x^2\sin^2x}{x^2-\sin^2x}$$

is

▸ SHOW SOLUTION

Step 1. Factor the denominator

$$x^2-\sin^2x=(x-\sin x)(x+\sin x)$$

Step 2. Expand

$$x-\sin x\approx\frac{x^3}{6},\qquad x+\sin x\approx2x$$

Denominator $\approx\dfrac{x^4}{3}$, numerator $\approx x^4$.

Answer: (2) $3$

Example 15 (JEE Main 2025, 8 April Shift 2). Consider the two statements.

Statement I:

$$\lim_{x\to0}\frac{\tan^{-1}x+\ln\sqrt{\frac{1+x}{1-x}}-2x}{x^5}=\frac25$$

Statement II:

$$\lim_{x\to1}x^{\frac{2}{1-x}}=\frac1{e^2}$$

▸ SHOW SOLUTION

Step 1. Statement I

$$\ln\sqrt{\frac{1+x}{1-x}}=\frac12\left(\ln(1+x)-\ln(1-x)\right)$$$$=x+\frac{x^3}{3}+\frac{x^5}{5}$$

Add $\tan^{-1}x=x-\dfrac{x^3}{3}+\dfrac{x^5}{5}$ and subtract $2x$:

$$\frac{2x^5}{5}$$

True.

Step 2. Statement II

$1^\infty$ form (see the $1^\infty$ section):

$$e^{\lim\frac{2}{1-x}(x-1)}=e^{-2}$$

True.

Answer: (4) both true

Example 16 (IITianForum). Let $a>0$ and

$$L=\lim_{x\to0}\frac{x^4+2\left(a^2-\sqrt{a^4+x^4}\right)}{x^8}$$

If $L$ is finite, find $a$ and $L$.

▸ SHOW SOLUTION

Step 1. Expand the root

$$\sqrt{a^4+x^4}=a^2\left(1+\frac{x^4}{a^4}\right)^{1/2}$$$$=a^2+\frac{x^4}{2a^2}-\frac{x^8}{8a^6}$$

Step 2. Numerator

$$x^4\left(1-\frac1{a^2}\right)+\frac{x^8}{4a^6}$$

Step 3. Finite only if the x⁴ term vanishes

$a=1$, and then $L=\dfrac14$.

Answer: $a=1$, $L=\dfrac14$

Example 17 (IITianForum). Find

$$\lim_{n\to\infty}n^2\left(n^2\ln\left(1+\frac{2}{n^2}\right)-2\right)$$

▸ SHOW SOLUTION

Step 1. Put t = 1/n²

$t\to0$, and the expression is

$$\frac{1}{t}\left(\frac{\ln(1+2t)}{t}-2\right)=\frac{\ln(1+2t)-2t}{t^2}$$

Step 2. Expand

$$\ln(1+2t)=2t-2t^2+\cdots$$

Answer: $-2$

Example 18 (IITianForum). Find

$$\lim_{x\to0}\left(\frac{2}{\ln(1+2x)}-\frac{1}{\tan x}\right)$$

▸ SHOW SOLUTION

Step 1. First term

$$\ln(1+2x)=2x(1-x+\cdots)$$$$\frac{2}{\ln(1+2x)}=\frac1x(1+x+\cdots)$$$$=\frac1x+1+\cdots$$

Step 2. Second term

$$\frac1{\tan x}=\frac1{x\left(1+\frac{x^2}{3}\right)}=\frac1x-\frac x3+\cdots$$

Step 3. Subtract

The $\dfrac1x$ terms cancel.

Answer: $1$

An $\infty-\infty$ form is almost always settled this way: write each term as $\dfrac1x$ plus a constant plus smaller terms.

L'Hôpital's rule

If $\dfrac{f(x)}{g(x)}$ is of the form $\dfrac00$ or $\dfrac{\infty}{\infty}$ at $a$, and $f$, $g$ are differentiable near $a$ with $g'(x)\ne0$ there, then

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

whenever the right side exists. Differentiate the top and the bottom separately, never the fraction as a whole, and check the form again before every repeat.

Example 19 (IITianForum). Find

$$\lim_{x\to0}\frac{x-\sin x}{x^3}$$

▸ SHOW SOLUTION

Step 1. Three rounds, each 0/0

$$\frac{1-\cos x}{3x^2}\to\frac{\sin x}{6x}\to\frac{\cos x}{6}$$

Answer: $\dfrac16$

The expansion $x-\sin x=\dfrac{x^3}{6}-\cdots$ gives the same answer in one line. Use L'Hôpital when no expansion is at hand, typically when the function is only known through $f(a)$, $f'(a)$, $f''(a)$.

Example 20 (JEE Main 2026, 21 January Shift 1). Let $f:\mathbb R\to(0,\infty)$ be twice differentiable with $f(3)=18$, $f'(3)=0$, $f''(3)=4$. Then the following is ______.

$$\lim_{x\to1}\ln\left(\frac{f(2+x)}{f(3)}\right)^{\frac{18}{(x-1)^2}}$$

▸ SHOW SOLUTION

Step 1. Bring the power down

$$\frac{18}{(x-1)^2}\ln\frac{f(x+2)}{18}$$

This is $\dfrac00$.

Step 2. L'Hôpital once

$$\frac{18\cdot\frac{f'(x+2)}{f(x+2)}}{2(x-1)}$$

Still $\dfrac00$ because $f'(3)=0$.

Step 3. L'Hôpital again, at x = 1

$$\frac{18}{2}\cdot\frac{f''(3)f(3)-f'(3)^2}{f(3)^2}=9\cdot\frac{4\cdot18}{18^2}=2$$

Answer: $2$

The 1^∞ form

If $f(x)\to1$ and $g(x)\to\infty$, then

$$\lim f(x)^{g(x)}=e^{\lim g(x)\left(f(x)-1\right)}$$

Why. Write $f=1+h$ with $h\to0$. Then $f^g=\left((1+h)^{1/h}\right)^{hg}$, the inner part tends to $e$, and $hg=g(f-1)$.

Dots showing (1 + 1/n) to the power n for n from 1 to 40, starting at 2 and rising quickly then slowly towards a dashed horizontal line at e, about 2.718, never crossing it

Example 21 (JEE Main 2025, 3 April Shift 2). If

$$\lim_{x\to0}\left(\frac{\tan x}{x}\right)^{1/x^2}=p$$

then $96\ln p$ is ______.

▸ SHOW SOLUTION

Step 1. Formula

$$\ln p=\lim\frac1{x^2}\left(\frac{\tan x}{x}-1\right)$$$$=\lim\frac{\tan x-x}{x^3}$$

Step 2. Expansion

$$\tan x-x=\frac{x^3}{3}+\cdots\implies\ln p=\frac13$$

Answer: $96\cdot\dfrac13=32$

Example 22 (JEE Main 2025, 4 April Shift 2). Let $f$ be differentiable on $\mathbb R$ with $f(2)=1$, $f'(2)=4$, and

$$\lim_{x\to0}\left(f(2+x)\right)^{3/x}=e^\alpha$$

Then the number of times the curve $y=4x^3-4x^2-4(\alpha-7)x-\alpha$ meets the $x$-axis is

▸ SHOW SOLUTION

Step 1. 1^∞ form

$$\alpha=\lim_{x\to0}\frac{3\left(f(2+x)-1\right)}{x}=3f'(2)=12$$

Step 2. The cubic

$$4x^3-4x^2-20x-12$$$$=4(x+1)^2(x-3)$$

A double root at $-1$ (touching) and a root at $3$.

Answer: (1) $2$ points

Example 23 (JEE Main 2025, 22 January Shift 2). If

$$\lim_{x\to\infty}\left(\frac{e}{1-e}\left(\frac1e-\frac{x}{1+x}\right)\right)^x=\alpha$$

then $\dfrac{\ln\alpha}{1+\ln\alpha}$ is

▸ SHOW SOLUTION

Step 1. Simplify the base

$$\frac{e}{1-e}\cdot\frac{1+x-ex}{e(1+x)}=\frac{x+\frac1{1-e}}{x+1}\to1$$

Step 2. Formula

$$\ln\alpha=\lim x\cdot\frac{\frac1{1-e}-1}{x+1}=\frac{e}{1-e}$$

Step 3. Ratio

$$\frac{\frac{e}{1-e}}{1+\frac{e}{1-e}}=\frac{\frac{e}{1-e}}{\frac{1}{1-e}}=e$$

Answer: (1) $e$

Example 24 (JEE Main 2025, 23 January Shift 2). The value of

$$\lim_{x\to\infty}\frac{(2x^2-3x+5)(3x-1)^{x/2}}{(3x^2+5x+4)\sqrt{(3x+2)^x}}$$

is

▸ SHOW SOLUTION

Step 1. Split

$$\frac{2x^2-3x+5}{3x^2+5x+4}\to\frac23$$$$\left(\frac{3x-1}{3x+2}\right)^{x/2}\quad(1^\infty)$$

Step 2. The power part

$$\lim\frac x2\left(\frac{3x-1}{3x+2}-1\right)$$$$=\lim\frac x2\cdot\frac{-3}{3x+2}=-\frac12$$

Answer: (4) $\dfrac23e^{-1/2}=\dfrac{2}{3\sqrt e}$

Example 25 (JEE Main 2025, 29 January Shift 2). If

$$\lim_{t\to0}\left(\int_0^1(3x+5)^t\,dx\right)^{1/t}=\frac{\alpha}{5e}\left(\frac85\right)^{2/3}$$

then $\alpha$ is ______.

▸ SHOW SOLUTION

Step 1. The integral

$$I(t)=\frac{8^{t+1}-5^{t+1}}{3(t+1)},\qquad I(0)=1$$

Step 2. 1^∞, then the exponent as a derivative

$$\lim_{t\to0}\frac{I(t)-1}{t}=I'(0)=\frac{8\ln8-5\ln5}{3}-1$$

Step 3. Exponentiate

$$e^{I'(0)}=\frac{8^{8/3}}{5^{5/3}\,e}=\frac{256}{5^{5/3}\,e}$$

Step 4. Compare

$$\frac{\alpha}{5e}\cdot\frac{4}{5^{2/3}}=\frac{4\alpha}{5^{5/3}\,e}$$

Answer: $4\alpha=256$, so $\alpha=64$

Example 26 (IITianForum). Let $f(x)=Ax^4+Bx^3+Cx^2+Dx+E$ with $A\ne0$. If

$$\lim_{x\to0}\left(\frac{f(x)}{3x^3}\right)^{1/x}=e^2$$

find $f(1)+f'(1)$.

▸ SHOW SOLUTION

Step 1. The base must tend to 1

$\dfrac{f(x)}{3x^3}\to1$ forces $C=D=E=0$ and $B=3$.

$$\frac{f(x)}{3x^3}=1+\frac{A}{3}x$$

Step 2. 1^∞

$$\lim\frac1x\cdot\frac{A}{3}x=\frac A3=2\implies A=6$$

Step 3. Evaluate

$f(x)=6x^4+3x^3$, $f(1)=9$, $f'(1)=24+9=33$.

Answer: $42$

The 0^0 and ∞^0 forms

Take logarithms. If $y=f(x)^{g(x)}$, then

$$\ln y=g(x)\ln f(x)$$

which is a $0\cdot\infty$ form, turned into $\dfrac00$ or $\dfrac{\infty}{\infty}$ and finished by the earlier methods. Two limits do most of the work:

$$\lim_{t\to0^+}t\ln t=0,\qquad\lim_{x\to\infty}\frac{\ln x}{x}=0$$

The first gives $x^x\to1$ as $x\to0^+$; the second gives $x^{1/x}\to1$ as $x\to\infty$.

Example 27 (IITianForum). Find

$$\lim_{x\to0^+}(\sin x)^{\tan x}$$

▸ SHOW SOLUTION

Step 1. Logarithm

$$\ln y=\tan x\,\ln\sin x$$$$=\frac{\tan x}{\sin x}\cdot\sin x\ln\sin x$$

Step 2. Each factor

$\dfrac{\tan x}{\sin x}=\dfrac{1}{\cos x}\to1$, and with $t=\sin x\to0^+$, $t\ln t\to0$.

$$\ln y\to0$$

Answer: $1$

Limits with [x] and {x}

$[x]$ is constant between integers and jumps by $1$ at each integer; $\{x\}=x-[x]$ rises from $0$ towards $1$ and drops back. So at an integer $n$:

$$\lim_{x\to n^-}[x]=n-1,\qquad\lim_{x\to n^+}[x]=n$$ $$\lim_{x\to n^-}\{x\}=1,\qquad\lim_{x\to n^+}\{x\}=0$$ Two graphs side by side: the step graph of the greatest integer of x from minus 2 to 3 with a filled dot at the left end and a hollow dot at the right end of each step, and the sawtooth graph of the fractional part of x rising from 0 to 1 on each unit interval with filled dots at the bottom and hollow dots at the top

Two facts save time. For non-integer $x$, $[x]+[-x]=-1$. And near $0^+$, $x\left[\dfrac kx\right]\to k$, because $\dfrac kx-1<\left[\dfrac kx\right]\le\dfrac kx$.

Example 28 (IITianForum). For an integer $n$, find

$$\lim_{x\to n}\left((-1)^{[x]}-(-1)^{[-x]}\right)$$

▸ SHOW SOLUTION

Step 1. Left of n

$[x]=n-1$, and $-x$ is just above $-n$, so $[-x]=-n$.

$$(-1)^{n-1}-(-1)^{n}=-2(-1)^n$$

Step 2. Right of n

$[x]=n$, and $-x$ is just below $-n$, so $[-x]=-n-1$.

$$(-1)^{n}-(-1)^{n+1}=2(-1)^n$$

Step 3. Compare

The two sides differ in sign.

Answer: the limit does not exist.

With a plus sign instead of a minus, both sides would give $0$ and the limit would exist. The sign of the combination decides everything here.

Example 29 (JEE Main 2025, 29 January Shift 1). The least $p\in\mathbb N$ for which

$$\lim_{x\to0^+}\left(x\sum_{k=1}^{p}\left[\frac kx\right]-x^2\sum_{k=1}^{9}\left[\frac{k^2}{x^2}\right]\right)\ge1$$

is ______.

▸ SHOW SOLUTION

Step 1. Each bracket

$$x\left[\frac kx\right]\to k,\qquad x^2\left[\frac{k^2}{x^2}\right]\to k^2$$

Step 2. The limit

$$\frac{p(p+1)}{2}-(1^2+2^2+\cdots+9^2)$$$$=\frac{p(p+1)}{2}-285$$

Step 3. Inequality

$$p(p+1)\ge572$$

$23\cdot24=552$ fails, $24\cdot25=600$ works.

Answer: $24$

Finding constants from a given limit

This is the single most common JEE pattern. The logic is always the same.

1. If the denominator tends to $0$ and the limit is finite, the numerator must also tend to $0$. That gives the first equation.
2. Expand the numerator in powers of $(x-a)$. Every power lower than the one in the denominator must have zero coefficient.
3. The first surviving coefficient, divided by the denominator's, equals the given limit.

Example 30 (JEE Main 2026, 2 April Shift 1). If

$$\lim_{x\to2}\frac{\sin\left(x^3-5x^2+ax+b\right)}{\left(\sqrt{x-1}-1\right)\ln(x-1)}=m$$

then $a+b+m$ is

▸ SHOW SOLUTION

Step 1. Denominator near x = 2

$$\sqrt{x-1}-1\approx\frac{x-2}{2}$$$$\ln(x-1)\approx x-2$$

So the denominator $\approx\dfrac{(x-2)^2}{2}$.

Step 2. The cubic must have a double root at 2

Write $p(x)=x^3-5x^2+ax+b$. Then $p(2)=0$ and $p'(2)=0$:

$$2a+b=12,\qquad a=8$$

So $b=-4$ and $p(x)=(x-2)^2(x-1)$.

Step 3. m

$$m=\lim\frac{(x-2)^2(x-1)}{(x-2)^2/2}=2$$

Answer: (2) $8-4+2=6$

Example 31 (JEE Main 2026, 22 January Shift 2). If

$$\lim_{x\to0}\frac{e^{(a-1)x}+2\cos bx+(c-2)e^{-x}}{x\cos x-\ln(1+x)}=2$$

then $a^2+b^2+c^2$ is

▸ SHOW SOLUTION

Step 1. Denominator

$$x\cos x-\ln(1+x)=\frac{x^2}{2}+\cdots$$

Step 2. Numerator coefficients

Constant: $1+2+(c-2)=c+1=0$, so $c=-1$.

$x$: $(a-1)-(c-2)=a+2=0$, so $a=-2$.

$x^2$: $\dfrac{(a-1)^2}{2}-b^2+\dfrac{c-2}{2}=3-b^2$.

Step 3. Ratio

$$\frac{3-b^2}{1/2}=2\implies b^2=2$$

Answer: (3) $4+2+1=7$

Example 32 (JEE Main 2025, 2 April Shift 2). If

$$\lim_{x\to0}\frac{\cos2x+a\cos4x-b}{x^4}$$

is finite, then $a+b$ is

▸ SHOW SOLUTION

Step 1. Expand

$$\cos2x=1-2x^2+\cdots$$$$\cos4x=1-8x^2+\cdots$$

Step 2. Kill the constant and the x² term

$$1+a-b=0,\qquad-2-8a=0$$

$a=-\dfrac14$, $b=\dfrac34$.

Answer: (1) $\dfrac12$

Example 33 (JEE Main 2025, 2 April Shift 1). If

$$\lim_{x\to0}\frac{x^2\sin\alpha x+(\gamma-1)e^{x^2}}{\sin2x-\beta x}=3$$

then $\beta+\gamma-\alpha$ is

▸ SHOW SOLUTION

Step 1. Numerator must vanish at 0

$\gamma-1=0$, so $\gamma=1$, and the numerator is $x^2\sin\alpha x\approx\alpha x^3$.

Step 2. Denominator must be of order x³

$$\sin2x-\beta x=(2-\beta)x-\frac{4}{3}x^3+\cdots$$

If $\beta\ne2$ the limit is $0$, so $\beta=2$.

Step 3. Ratio

$$\frac{\alpha}{-4/3}=3\implies\alpha=-4$$

Answer: (1) $2+1+4=7$

Example 34 (JEE Main 2025, 4 April Shift 1). Let $N(x)=(x-1)\left(6+\lambda\cos(x-1)\right)+\mu\sin(1-x)$. If

$$\lim_{x\to1^+}\frac{N(x)}{(x-1)^3}=-1$$

then $\lambda+\mu$ is

▸ SHOW SOLUTION

Step 1. Put x = 1 + t

$$t(6+\lambda\cos t)-\mu\sin t$$$$=(6+\lambda-\mu)t+\left(\frac{\mu}{6}-\frac{\lambda}{2}\right)t^3+\cdots$$

Step 2. Coefficients

$$6+\lambda-\mu=0,\qquad\frac{\mu}{6}-\frac{\lambda}{2}=-1$$

$\lambda=6$, $\mu=12$.

Answer: (1) $18$

Example 35 (JEE Main 2026, 6 April Shift 2). Let $g(x)=rx^2+(p-2)x-2p$ and

$$\lim_{x\to2}\frac{g(x)\tan(x-2)}{(x-2)^2}=5$$

If the set of all $q$ for which both roots of $rx^2-px+q=0$ lie in $(0,2)$ is $(\alpha,\beta]$, then $4(\alpha+\beta)$ is

▸ SHOW SOLUTION

Step 1. The quadratic must vanish at 2

$$4r+2(p-2)-2p=0\implies r=1$$

Then $x^2+(p-2)x-2p=(x-2)(x+p)$.

Step 2. The limit

$$\frac{\tan(x-2)}{x-2}\cdot(x+p)\to2+p$$

$2+p=5$, so $p=3$.

Step 3. Both roots of the quadratic in (0, 2)

The quadratic is $x^2-3x+q$. Real roots: $9-4q\ge0$. Value at $0$: $q>0$. Value at $2$: $q-2>0$. Vertex $\dfrac32$ lies inside.

$$2<q\le\frac94$$

Answer: (3) $4\left(2+\dfrac94\right)=17$

Example 36 (JEE Main 2026, 2 April Shift 2). Let $f(x)$ be a polynomial of degree $5$ with extrema at $x=1$ and $x=-1$. If

$$\lim_{x\to0}\frac{f(x)}{x^3}=-5$$

then $f(2)-f(-2)$ is

▸ SHOW SOLUTION

Step 1. Shape from the limit

The limit is finite and non-zero, so $f$ has no terms below $x^3$ and the $x^3$ coefficient is $-5$:

$$f(x)=ax^5+bx^4-5x^3$$

Step 2. Extrema at ±1

$$f'(x)=5ax^4+4bx^3-15x^2$$$$5a+4b=15,\qquad5a-4b=15$$

$b=0$, $a=3$.

Step 3. Evaluate

$f(x)=3x^5-5x^3$ is odd, so $f(2)-f(-2)=2f(2)=2(96-40)$.

Answer: (4) $112$

Example 37 (JEE Main 2025, 24 January Shift 1). Let $f:\mathbb R-\{0\}\to\mathbb R$ satisfy

$$f(x)-6f\left(\frac1x\right)=\frac{35}{3x}-\frac52$$

If

$$\lim_{x\to0}\left(\frac{1}{\alpha x}+f(x)\right)=\beta$$

then $\alpha+2\beta$ is

▸ SHOW SOLUTION

Step 1. Replace x by 1/x and solve

$$f\left(\frac1x\right)-6f(x)=\frac{35x}{3}-\frac52$$

Solving the two equations:

$$f(x)=-2x+\frac12-\frac1{3x}$$

Step 2. Finite limit

$$\frac1{\alpha x}-\frac1{3x}+\frac12-2x$$

The $\dfrac1x$ terms must cancel: $\alpha=3$, and then $\beta=\dfrac12$.

Answer: (3) $3+1=4$

Which method to use

A decision chart: from what substitution gives, arrows lead to factor or rationalise, standard limits, divide by the highest power, expansions, the 1 to the infinity formula, logarithms, and checking left and right limits

Quick recall

Practise these methods on real exam questions in the doMath Limits chapter, and then move on to Continuity and Differentiability, which is built on the LHL and RHL ideas of the first section.