Limits Questions for JEE Main 2027 - PYQs, Practice Questions & Answers

Limits: JEE Last 5 Years' Most Asked Topics

Below is the table which indicates the topics which have repeated the most times in the past five years in the JEE Main for the subjects of Limits and Calculus. This table would help the students in knowing what the important topics are.

Also Check: JEE Main Maths: 20 Most Repeated Topic-Based Questions 

Limits JEE Main Questions 

Solving questions from JEE Main Limits assists students in learning the real application of theories in exam problems. Some of the JEE Main exam questions along with answers are provided below for your revision purposes.

Question 1: The value of x0x22xx2-2x

(1) 2

(2) 3

(3) 4

(4) 6

Correct Answer: Option (2)

Solution: 

x0x22xx2-2x

Applying expansion

$\begin{aligned}
 & \lim _{x \rightarrow 0} \frac{x^2 \sin ^2 x}{x^2-\left(x-\frac{x^3}{3!}+\ldots\right)^2} \\
 & \lim _{x \rightarrow 0} \frac{x^2 \sin ^2 x}{x^2-x^2+\frac{2 x^4}{3!} \ldots}=\frac{3!}{2}=3
 \end{aligned}$

Question 2: For the function f(x)=e|x|-|x|,xR, consider the following statements:

Statement I : f is differentiable for all xR.

Statement II : f is increasing in -,-2.

In the light of the above statements, choose the correct answer from the options given below :

(1) Both Statement I and Statement II are true

(2) Both Statement I and Statement II are false

(3) Statement I is true but Statement II is false

(4) Statement I is false but Statement II is true

Correct Answer: Option (4)

Solution: 

$\mathrm{f}(\mathrm{x})=\mathrm{e}^{|\sin x|}-|\mathrm{x}| \text { is }$

non-differentiable at x=
 and for x-,-2

$\begin{aligned}
 & f(x)=e^{-\operatorname{sinx}}+x \\
 & f^{\prime}(x)=-e^{-\operatorname{sinx}} \cos x+1 \\
 & \text { for } x \in\left(-\pi, \frac{-\pi}{2}\right) \quad \cos x<0 \\
 & \therefore f^{\prime}(x)>0 \\
 & \therefore f(x) \text { is increasing }
 \end{aligned}$

Question 3: Let f(x) and g(x) be twice differentiable functions satisfying $\mathrm{f}^{\prime \prime}(\mathrm{x})=\mathrm{g}^{\prime \prime}(\mathrm{x})$ for all xR, $f^{\prime}(1)=2 g^{\prime}(1)=4$ and g(2)=3f(2)=9. Then f(25)-g(25) is equal to :

(1) 20

(2) 40

(3) -20

(4) -40

Correct Answer: Option (2)

Solution: 

$\mathrm{f}^{\prime \prime}(\mathrm{x})=\mathrm{g}^{\prime \prime}(\mathrm{x})$

Integrate

$\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{g}^{\prime}(\mathrm{x})+\mathrm{C}_1$

Put x=1

$\begin{aligned}
 & f^{\prime}(1)=g^{\prime}(1)+C_1 \\
 & 4=2+C_1 \\
 & C_1=2 \\
 & \Rightarrow f^{\prime}(x)=g^{\prime}(x)+2 \\
 & f^{\prime}(x)=g(x)+2 x+C_2
 \end{aligned}$

Put x=2

$\begin{aligned}
 & f(2)=g(2)+4+C_2 \\
 & 3=9+4+C_2 \\
 & C_2=-10 \\
 & f(x)=g(x)+2 x-10
 \end{aligned}$

Put x=25

f(x)-g(25)=40

Question 4: Let f:RR be a differentiable function such that fx+y3=f(x)+f(y)3 for all x,yR and $f^{\prime}(0)=3$. Then the minimum value of the function g(x)=3+exf(x), is

(1) 3e+1e

(2) 3e-1e

(3) 3-ee

(4) 3 e

Correct Answer: Option (2)

Solution: 

$\begin{aligned} & f\left(\frac{x+y}{3}\right)=\frac{f(x)+f(y)}{3}, \text { put } x=y=0 \\ & f(0)=\frac{2 f(0)}{3} \\ & \Rightarrow f(0)=0 \ldots \ldots(1) \\ & f^{\prime}\left(\frac{x+y}{3}\right) \cdot \frac{1}{3}=\frac{1}{3} f^{\prime}(x) \\ & \text { put } x=0 \\ & f^{\prime}\left(\frac{y}{3}\right) \frac{1}{3}=\frac{1}{3} \times 3 \\ & f^{\prime}\left(\frac{y}{3}\right)=3 \\ & \text { put } y=3 x \\ & f^{\prime}(x)=3 \\ & \text { integrate both sides } \\ & f(x)=3 x+c \\ & \text { from eq. (1) } \\ & f(0)=0 \\ & \Rightarrow f(x)=3 x\end{aligned}$

Now

$\begin{aligned}
 & \mathrm{g}(\mathrm{x})=3+\mathrm{e}^{\mathrm{x}} \cdot 3 \mathrm{x} \\
 & \mathrm{~g}^{\prime}(\mathrm{x})=3\left[\mathrm{e}^{\mathrm{x}}+\mathrm{x} \cdot \mathrm{e}^{\mathrm{x}}\right] \\
 & =3 \mathrm{e}^{\mathrm{x}}(\mathrm{x}+1) \\
 & \mathrm{g}^{\prime}(\mathrm{x})=0, \text { at } \mathrm{x}=-1 \\
 & (\mathrm{~g}(\mathrm{x}))_{\min }=3+\mathrm{e}^{-1}(-3) \\
 & =3\left[1-\frac{1}{\mathrm{e}}\right]=\frac{3(\mathrm{e}-1)}{\mathrm{e}}
 \end{aligned}$

Question 5: The number of points, at which the fucntion

$\begin{aligned}
 & f(x)=\max \left\{6 x, 2+3 x^2\right\}+|x-1| \cos \\
 & \left|x^2-\frac{1}{4}\right|, x \in(-\pi, \pi), \text { is not differentiable, is______ }
 \end{aligned}$

(1) 3

(2) 4

(3) 8

(4) 9

Correct Answer: Option (1)

Solution: 

2+3x2=6x3x2-6x+2=0

x=6126=323

|x-1|x2-14 is differentiable at x=1
 Total number of points where f(x) is nonDifferentiable is 3

Question 6: The number of points in the interval [ 2,4 ], at which the function f(x)=x2-x-12, where [] denotes the greatest integer function, is discontinuous, is ____ .

(1) 10

(2) 12

(3) 24

(4) 25

Correct Answer: Option (1)

Solution: In x[2,4], range of x2-x-12 is [1.5,11.5]f(x) is discontinuous at points where x2-x-12 becomes integers.

∴ Integral values of x2-x-12 are 2,3,4,5,6,7, 8,910,11

∴ Number of points where is discontinuous is 10

Question 7: If x2x3-5x2+ax+b(x-1-1)e(x-1)=m, then a+b+m is equal to:

(1) 5

(2) 6

(3) 8

(4) 10

Correct Answer: Option (2)

Solution: $\begin{aligned}
 & \because \text { Denominator }=0 \text { at } \mathrm{x}=2 \\
 & \Rightarrow \text { Numerator }=0 \text { at } \mathrm{x}=2 \\
 & \Rightarrow 2^3-5(2)^2+\mathrm{a}(2)+\mathrm{b}=0 \\
 & \Rightarrow 2 \mathrm{a}+\mathrm{b}=12 \ldots(1) \\
 & \therefore \mathrm{m}=\operatorname{Lim}_{\mathrm{x} \rightarrow 2} \frac{\sin \left(\mathrm{x}^3-5 \mathrm{x}^2+\mathrm{ax}+\mathrm{b}\right)}{\left(\mathrm{x}^3-5 \mathrm{x}^2+\mathrm{ax}+\mathrm{b}\right)} \cdot\left(\mathrm{x}^3-5 \mathrm{x}^2+\mathrm{ax}+\mathrm{b}\right) \\
 & \Rightarrow \mathrm{m}=\operatorname{Lim}_{\mathrm{x} \rightarrow 2} 2\left(\frac{\mathrm{x}^3-1-1}{\sqrt{\mathrm{x}-1}+1}\right) \cdot \frac{\log _{\mathrm{e}}(1+(\mathrm{x}-2))}{(\mathrm{x}-2)} \cdot(\mathrm{x}-2) \\
 & \Rightarrow \mathrm{m}=\operatorname{Lim}_{\mathrm{x} \rightarrow 2} 2\left(\frac{3 \mathrm{x}^2-10 \mathrm{x}+\mathrm{a}}{2(\mathrm{x}-2)}\right) \\
 & \because \text { Denominator }=0 \text { at } \mathrm{x}=2 \\
 & \Rightarrow \text { Numerator }=0 \text { at } \mathrm{x}=2 \\
 & \Rightarrow 3(2)^2-10(2)+\mathrm{a}=0 \Rightarrow \mathrm{a}=8 \\
 & \text { put a }=8 \text { in }(1), \text { we get } \mathrm{b}=-4 \\
 & \therefore \mathrm{~m}=\operatorname{Lim}_{\mathrm{x} \rightarrow 2} 2\left(\frac{6 \mathrm{x}-10}{2}\right) \Rightarrow \mathrm{m}=2 \\
 & \therefore \mathrm{a}+\mathrm{b}+\mathrm{m}=6
 \end{aligned}$

Question 8: Consider the following three statements for the function f:(0,)R defined by f(x)=ex-|x-1| :

(I) f is differentiable at all x>0
 (II) f is increasing in (0,1).
 (III) f is decreasing in (1,).

Then,

(1) Only (I) and (III) are TRUE.

(2) Only (I) is TRUE.

(3) Only (II) and (III) are TRUE.

(4) All (I), (II) and (III) are TRUE.

Correct Answer: Option (1)

Solution: 

f(x)=ex-|x-1|+5

$\begin{aligned} & f(x)= \begin{cases}\log _e x-(x-1)+5, & x \geq 1 \\ -\log _e x+(x-1)+5, & 0<x \leq 1\end{cases} \\ & f^{\prime}(x)= \begin{cases}\frac{1}{x}-1, & x \geq 1 \\ -\frac{1}{x}+1, & 0<x \leq 1\end{cases} \\ & f^{\prime}(x)<0 \text { for } \in(1, \infty) \\ & f^{\prime}(x)<0 \text { for } \in(0,1) \\ & \therefore f(x) \text { is decreasing in }(1, \infty) \text { and also in }(0,1) \\ & \therefore f(x) \text { is differentiable in }(0, \infty)\end{aligned}$

Hence, the correct answer is option (1).

Question 10: Let $f(x)=x^3+x^2 f^{\prime}(1)+2 x f^{\prime \prime}(2)+f^{\prime \prime \prime}(3), x \in R$. Then the value of $f^{\prime}(5)$ is :

(1) 25

(2) 1175

(3) 6575

(4) 625

Correct Answer: Option (2)

Solution: 

Let $f^{\prime}(1)=a$

$\begin{aligned}
 & f^{\prime}(2)=b \\
 & f^{\prime}(3)=c \\
 & f(x)=x^3+a x^2+b x+c \\
 & f^{\prime}(x)=3 x^2+2 a x+b \\
 & f^{\prime}(x)=a=3+2 a+b \Rightarrow a+b=3 \ldots . . (1)\\
 & f^{\prime}(x)=6 x+2 a \\
 & f^{\prime \prime}(2)=12+2 a=\frac{b}{2} \Rightarrow 4 a-b=-24 \\
 & f^{\prime \prime}(x)=6 \\
 & f^{\prime \prime}(3)=c=6 \\
 & \Rightarrow a=\frac{-27}{5}, b=\frac{12}{5} \\
 & f^{\prime}(5)=75+10 a+b \\
 & =75-54+\frac{12}{5}=21+\frac{12}{5}=\frac{117}{5}
 \end{aligned}$

Hence, the correct answer is option (2).

Question 11: Let f:R(0,) be a twice differentiable function such that f(3)=18,fl(3)=0 and f(3)=4, then $\operatorname{Lim}\left[\log _e\left(\frac{f(2+x)}{f(3)}\right)^{\frac{18}{(x-1)^2}}\right]$ is equal to:

(1) 1

(2) 12

(3) 2

(4) 18

Correct Answer: Option (3)

Solution: 

Given:
 $
 \lim _{x \rightarrow 1} \log \left(\frac{f(2+x)}{f(3)}\right)^{\frac{18}{(x-1)^2}}
 $
 Since f(3)=18,
 $
 =\lim _{x \rightarrow 1} \frac{18}{(x-1)^2} \ln \left(\frac{f(2+x)}{18}\right)
 $
 As x1,

$
 \ln \left(\frac{f(2+x)}{18}\right) \rightarrow 0 \text { and }(x-1)^2 \rightarrow 0
 $
 So form is 00.
 Applying the L'Hospital rule, we get,

$
 \begin{aligned}
 & =\lim _{x \rightarrow 1} \frac{18 \cdot \frac{f^{\prime}(2+x)}{f(2+x)}}{2(x-1)} \\
 = & \lim _{x \rightarrow 1} \frac{9 f^{\prime}(2+x)}{(x-1) f(2+x)}
 \end{aligned}
 $
 Since $f^{\prime}(3)=0$, still 00.
 Applying the L'Hospital rule again, we get,
 Numerator derivative:
 $9 f^{\prime \prime}(2+x)$
 Denominator derivative:
 $f(2+x)+(x-1) f^{\prime}(2+x)$
 So,
 $=\lim _{x \rightarrow 1} \frac{9 f^{\prime \prime}(2+x)}{f(2+x)+(x-1) f^{\prime}(2+x)}$
 Substitute x=1 :

$
 =\frac{9 f^{\prime \prime}(3)}{f(3)}=\frac{9 \cdot 4}{18}=\frac{36}{18}=2
 $

Hence, the correct answer is option (3).

Question 12: Let f:(0,)(0,),f(x)=xe1/x. Then which of the following option is INCORRECT?

(1) f(x) is not monotonic

(2) f(x) is not bounded

(3) There exist atleast one c(1,) for which $f^{\prime}(c)>e$

(4) f(x)=1e has no solution in x.

Correct Answer: Option (3)

Solution: 

$\begin{aligned}
 f(x) & =x \cdot e^{1 / x} \\
 f^{\prime}(x) & =e^{\frac{1}{x}}\left(1-\frac{1}{x}\right)
 \end{aligned}$

and $f^{\prime \prime}(x)=\frac{e^{\frac{1}{x}}}{x^3}$
 So, f(x) is decreasing in (0,1) and increasing is (1,)

Range of f(x) is [e,).
 $\begin{aligned} f^{\prime}(x) \text { is also increasing in }(1, \infty) & \text { and } f^{\prime}(x)<\lim _{x \rightarrow \infty} f^{\prime}(x) \\ & \Rightarrow f^{\prime}(x)<1\end{aligned}$

Question 13: Absolute Maximum Value of f(x)=1|x-4|+1+1|x+8|+1 is :

(1) 1413

(2) 1513

(3) 85

(4) 1314

Correct Answer: Option (1)

Solution: 

$\text { Say } g(x)=\frac{1}{|x-4|+1} \Rightarrow \frac{1}{|x+8|+1}=g(x+12)$

Now for x(-,-8) both g(x) and g(x+12) are increasing hence maximum value can't occur in this interval.
 Similarly for x(4,) both g(x) and g(x+12) are decreasing hence maximum value can't occur in this interval.
 So, now for all values of x(-8,4)

f(x)=1x+9+15-x=14(5-x)(x+9)

which will have maximum either at x=4 or x=-8 or at minima of (5-x)(x+9). Also minima of (5-x)(x+9) does not exist.

f(4)=1+113=1413=f(-8).

So, the absolute maximum is 1413.

Question 14: For which value of ' a ' the function a+41-a-1x5-3x+5 doesnt decreases for all xR ?

(1) -3

(2) -1

(3) 0

(4) 2

Correct Answer: Option (3)

Solution: 

We have f(x)=a+41-a-1x5-3x+5 so that $f^{\prime}(x)=5\left[\frac{\sqrt{a+4}}{1-a}-1\right] x^4-3<0$ for all xR. This holds for all a>1.

Also a+40a-4.

For a<1,a+41-a-10

a+41-a

or a2-3a-30a3+212 or a3-212.

Hence the given function is a decreasing function for all xR, when a-4,3-212(1,).

Hence, the correct answer is option (3).

Question 15: Evaluate: x0x-xx2

(1) 1

(2) 12

(3) 0

(4) -12

Correct Answer: Option (3)

Solution: 

$\begin{aligned}
 & \text { Let } A=\lim _{x \rightarrow 0} \frac{\tan x-x}{x^2}=\lim _{2 x \rightarrow 0} \frac{\tan 2 x-2 x}{4 x^2} \\
 & \Rightarrow A=\lim _{x \rightarrow 0} \frac{\tan x-x\left(1-\tan ^2 x\right)}{2 x^2}=\lim _{x \rightarrow 0} \frac{\tan x-x}{2 x^2}+\lim _{x \rightarrow 0} \frac{x \tan ^2 x}{2 x^2} \\
 & \Rightarrow A=\frac{A}{2}+0 \Rightarrow A=0
 \end{aligned}$

Question 16: If x0(2x)+a(4x)-bx4 is finite, then (a+b) is equal to :

(1) 12

(2) 0

(3) 34

(4) -1

Correct Answer: Option (1)

Solution: 

The Maclaurin series expansion of a function f(x) about x=0 is given by:

f(x)=f(0)+f'(0)x1!+f''(0)x22!+f(3)(0)x33!+

For f(x)=(ax):

•  
  f(0)=(0)=1
   

•  
  $f'(x) = -a \sin(ax) \implies f'(0) = 0$
   

•  
  $f''(x) = -a^2 \cos(ax) \implies f''(0) = -a^2$
   

•  
  $f^{(3)}(x) = a^3 \sin(ax) \implies f^{(3)}(0) = 0$
   

•  
  $f^{(4)}(x) = a^4 \cos(ax) \implies f^{(4)}(0) = a^4$
   

Hence, the series is:

(ax)=1-a2x22!+a4x44!-a6x66!+

Question 17: The number of points of discontinuity of the function f(x)=x22-[x],x[0,4], where [] denotes the greatest integer function is 

(1) 8

(2) 9

(3) 10

(4) 12

Correct Answer: Option (1)

Solution: 

Given: 
 $f(x) = \left\lfloor \frac{x^2}{2} \right\rfloor - \left\lfloor \sqrt{x} \right\rfloor,\quad x \in [0, 4]$

Discontinuity in greatest integer function occurs at points where the inner function becomes an integer.

So, we consider:
 x22=nx=2n
 x=mx=m2

Within x[0,4], check relevant n and m values:

From x22 being integer:
 $x = \sqrt{0}, \sqrt{2}, \sqrt{4}, \sqrt{6}, \sqrt{8}, \sqrt{10}, \sqrt{12}, \sqrt{14}, \sqrt{16}
 \Rightarrow x = 0, \sqrt{2}, 2, \sqrt{6}, \sqrt{8}, \sqrt{10}, \sqrt{12}, \sqrt{14}, 4$

From x being integer:
 x=02,12,22x=0,1,4

Combine all points:
 {0,1,2,2,6,8,10,12,14,4}

Total candidates for discontinuity: 10 points

Check endpoints.

At x=0 and x=4, check continuity:

- At x=0: right-hand limit equals f(0) → continuous  
 - At x=4: left-hand limit equals f(4) → continuous

So, remove 2 points (0 and 4).

Final count of discontinuities:
 10-2=8

$\therefore$ The function is discontinuous at 8 points.

Hence, the answer is 8 points.

Question 18: Let a>0. If the function f(x)=6x3-45ax2+108a2x+1 attains its local maximum and minimum values at the points x1 and x2 respectively such that x1x2= 54 , then a+x1+x2 is equal to :-

(1) 15

(2) 18

(3) 24

(4) 13

Correct Answer: Option (2)

Solution: 

$\begin{aligned}
 & f^{\prime}(\mathrm{x})=18 \mathrm{x}^2-90 \mathrm{ax}+108 \mathrm{a}^2=0 \\
 & \mathrm{x}=2 \mathrm{a} \space \& \space \mathrm{x}=3 \mathrm{a} \\
 & \mathrm{x}_1=2 \mathrm{a} \quad \mathrm{x}_2=3 \mathrm{a} \\
 & \mathrm{x}_1 \mathrm{x}_2=54 \\
 & 6 \mathrm{a}^2=54 \\
 & \mathrm{a}=3 \\
 & \mathrm{a}+\mathrm{x}_1+\mathrm{x}_2 \\
 & 3+2 \times 3+3 \times 3=18
 \end{aligned}$

Hence, the answer is option (2).

Question 19: Let m and n be the number of points at which the function f(x)=x,x3,x5,,x21,xR, is not differentiable and not continuous, respectively. Then m+n is equal to ___________ .

(1) 3

(2) 4

(3) 2

(4) 1

Correct Answer: Option (1)

Solution: 

$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}\mathrm{x}, & \mathrm{x}<-1 \\ \mathrm{x}^{21}, & -1 \leq \mathrm{x}<0 \\ \mathrm{x}, & 0 \leq \mathrm{x}<1 \\ \mathrm{x}^{21}, & \mathrm{x} \geq 1\end{array}\right.$

f(x) is continuous everywhere.

$\therefore \mathrm{n}=0$

$\mathrm{f}^{\prime}(\mathrm{x})=\left\{\begin{array}{cc}1, & \mathrm{x}<-1 \\ 21 \mathrm{x}^{20}, & -1 \leq \mathrm{x}<0 \\ 1, & 0<\mathrm{x}<1 \\ 21 \mathrm{x}^{20}, & \mathrm{x} \geq 1\end{array}\right.$

$\therefore \mathrm{f}(\mathrm{x})$ is non-differentiable at x=-1,0,1

$\begin{aligned}
 & \therefore \mathrm{m}=3 \\
 & \mathrm{~m}+\mathrm{n}=3
 \end{aligned}$

Question 20: Let f:RR be a function defined by $f(x)=\|x+2|-2| x\|$. If m is the number of points of local minima and n is the number of points of local maxima of f, then m+n is:

(1) 5

(2) 3

(3) 2

(4) 4

Correct Answer: Option (2)

Solution: 

For the given piecewise function,

$f(x)=\left\{\begin{array}{cc}|-x-2+2 x| & x \leq-2 \\ |x+2+2 x| & -2 \leq x \leq 0 \\ |x+2-2 x| & x \geq 0\end{array}\right.$

Similplify each interval,

$f(x)=\left\{\begin{array}{lc}|-x-2+2 x| & x \leq-2 \\ |x+2+2 x| & -2 \leq x \leq 0 \\ |x+2-2 x| & x \geq 0\end{array}\right.$ $f(x)=\left\{\begin{array}{cc}|x-2| & x \leq-2 \\ |3 x+2| & -2<x \leq 0 \\ |2-x| & x>0\end{array}\right.$
 Now, simplify the modulus,

$f(x)=\left\{\begin{array}{cc}2-x & x \leq-2 \\ -3 x-2 & -2<x \leq-\frac{2}{3} \\ 3 x+2 & -\frac{2}{3}<x \leq 0 \\ 2-x & 0<x<2 \\ x-2 & x \geq 2\end{array}\right.$

No. of maxima$ =  1$  (Since there is one peak at top)

No. of minima =2 (Since there are two crests which are minimum and are at the same level)

$\begin{aligned}
 & m=2 \\
 & n=1 \\
 & m+n=3
 \end{aligned}$

Question 21: If t001(3x+5)tdx1t=5e8523, then  is equal to______.

(1) 64

(2) 25

(3) 78

(4) 89

Correct Answer: Option (1)

Solution: 

Let us denote the limit as:
 $
 L = \lim_{t \to 0} \left( \int_0^1 (3x + 5)^t \, dx \right)^{1/t}.
 $

This is an indeterminate form of type 1. To resolve it, take the natural logarithm:
 $
 \ln L = \lim_{t \to 0} \frac{1}{t} \ln \left( \int_0^1 (3x + 5)^t \, dx \right).
 $

Let:
 $
 I(t) = \int_0^1 (3x + 5)^t \, dx.
 $

Now compute this integral:
 $
 I(t) = \int_0^1 (3x + 5)^t \, dx.
 $

Substitute u=3x+5x=u-53,dx=du3. When x=0, u=5; when x=1, u=8.

$
 I(t) = \int_5^8 u^t \cdot \frac{1}{3} \, du = \frac{1}{3} \int_5^8 u^t \, du.
 $

Now evaluate the integral:
 $
 I(t) = \frac{1}{3} \cdot \left[ \frac{u^{t+1}}{t+1} \right]_5^8 = \frac{1}{3(t+1)} (8^{t+1} - 5^{t+1}).
 $

So:
 $
 \ln L = \lim_{t \to 0} \frac{1}{t} \ln \left( \frac{1}{3(t+1)} (8^{t+1} - 5^{t+1}) \right).
 $

Split the logarithm:
 $
 \ln L = \lim_{t \to 0} \frac{1}{t} \left[ \ln(8^{t+1} - 5^{t+1}) - \ln(3(t+1)) \right].
 $

Using approximation:
 $
 a^t \approx 1 + t \ln a \quad \text{as } t \to 0.
 $

So:
 $
 8^{t+1} = 8 \cdot 8^t \approx 8(1 + t \ln 8), \quad 5^{t+1} = 5 \cdot 5^t \approx 5(1 + t \ln 5),
 $
 $
 \Rightarrow 8^{t+1} - 5^{t+1} \approx 8 + 8t \ln 8 - 5 - 5t \ln 5 = 3 + t(8 \ln 8 - 5 \ln 5).
 $

Thus:
 $
 \ln(8^{t+1} - 5^{t+1}) \approx \ln \left( 3 + t(8 \ln 8 - 5 \ln 5) \right) \approx \ln 3 + \frac{t(8 \ln 8 - 5 \ln 5)}{3}.
 $

Also, (3(t+1))=3+(1+t)3+t.

Putting these into L:
 $
 \ln L = \lim_{t \to 0} \frac{1}{t} \left[ \ln 3 + \frac{t(8 \ln 8 - 5 \ln 5)}{3} - (\ln 3 + t) \right] = \frac{8 \ln 8 - 5 \ln 5 - 3}{3}.
 $

Therefore:
 $
 L = e^{\frac{8 \ln 8 - 5 \ln 5 - 3}{3}}.
 $

Now write:
 $
 L = \frac{\alpha}{5e} \left( \frac{8}{5} \right)^{2/3}.
 $

Try simplifying the RHS:
 $
 \frac{1}{5e} \cdot \alpha \cdot \left( \frac{8}{5} \right)^{2/3} = \alpha \cdot \frac{1}{5e} \cdot \left( \frac{64}{25} \right)^{1/3}.
 $

So:
 $
 L = \frac{\alpha}{5e} \cdot \frac{64^{1/3}}{25^{1/3}} = \frac{\alpha}{5e} \cdot \frac{4}{5^{2/3}} = \frac{4\alpha}{5^{5/3} e}.
 $

Now match it with:
 $
 L = e^{\frac{8 \ln 8 - 5 \ln 5 - 3}{3}} = \frac{4 \cdot 16}{5^{5/3} e} = \frac{64}{5^{5/3} e}.
 $

So,
 $
 \frac{4\alpha}{5^{5/3} e} = \frac{64}{5^{5/3} e} \Rightarrow \alpha = 64.
 $

Hence, the answer is 64.

Question 22: Consider the region R=(x,y):xy9-113x2,x0. The area of the largest rectangle of sides parallel to the coordinate axes and inscribed in R is :

(1) 625111

(2) 730119

(3) 567121

(4) 821123

Correct Answer: Option (3)

Solution: 

Use, 
 $
 A = t \left( 9 - \frac{11t^2}{3} - t \right) = 9t - t^2 - \frac{11}{3}t^3
 $

Now differentiate:
 $
 \frac{dA}{dt} = 9 - 2t - 11t^2
 $

Set derivative to zero to find critical points:
 $
 11t^2 + 2t - 9 = 0
 $

Solve the quadratic:
 $
 11t^2 + 11t - 9t - 9 = 0
 \Rightarrow (11t + 9)(t - 1) = 0
 \Rightarrow t = -1 \quad \text{or} \quad t = \frac{9}{11}
 $

We consider the positive root for maximum area:
 $
 t = \frac{9}{11}
 $

Now compute the area:
 $
 A = \frac{9}{11} \left( 9 - \frac{11}{3} \cdot \left( \frac{81}{121} \right) - \frac{9}{11} \right)
 $

First compute each part:
 $
 \frac{11}{3} \cdot \frac{81}{121} = \frac{891}{363} = \frac{297}{121}
 \quad \text{and} \quad \frac{9}{11} = \frac{99}{121}
 $

So,
 $
 A = \frac{9}{11} \left( 9 - \frac{297}{121} - \frac{99}{121} \right)
 = \frac{9}{11} \left( \frac{1089 - 396}{121} \right)
 = \frac{9}{11} \cdot \frac{693}{121}
 = \frac{6237}{1331}
 = \frac{567}{121}
 $

Question 23: If $\displaystyle\lim _{x \rightarrow \infty}\left(\left(\frac{e}{1-e}\right)\left(\frac{1}{e}-\frac{x}{1+x}\right)\right)^x=\alpha$, then the value of e1+e equals:

(1) e

(2) e-2

(3) e2

(4) e-1

Correct Answer: Option (1)

Solution: 

Consider the expression inside the limit:

$
 A_x = \left(\frac{e}{1 - e}\right) \left( \frac{1}{e} - \frac{x}{1+x} \right)
 $

First simplify the term inside the parentheses:

$
 \frac{1}{e} - \frac{x}{1+x} = \frac{1}{e} - \left(1 - \frac{1}{1+x}\right) = \frac{1}{e} - 1 + \frac{1}{1+x} = \left(\frac{1}{e} - 1\right) + \frac{1}{1+x}
 $

Since

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

The entire term is:

$
 A_x = \frac{e}{1 - e} \left( \frac{1}{e} - 1 + \frac{1}{1+x} \right) = \frac{e}{1 - e} \left( \frac{1}{e} - 1 \right) + \frac{e}{1 - e} \cdot \frac{1}{1+x}
 $

Calculate the constant part:

$
 \frac{e}{1 - e} \left( \frac{1}{e} - 1 \right) = \frac{e}{1 - e} \cdot \frac{1 - e}{e} = 1
 $

So,

$
 A_x = 1 + \frac{e}{(1 - e)(1 + x)}
 $

As $ x \to \infty $,

$
 A_x \to 1
 $

Now write the limit as:

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

Rewrite the term inside the power:

$
 1 + \frac{e}{(1 - e)(1 + x)} = 1 + \frac{c}{x + 1} \quad \text{where} \quad c = \frac{e}{1 - e}
 $

For large $ x $,

$
 \left( 1 + \frac{c}{x+1} \right)^x \approx \left( 1 + \frac{c}{x} \right)^x \to e^c
 $

Therefore,

$
 \alpha = e^{\frac{e}{1 - e}}
 $

Next, compute

$
 \log_e \alpha = \frac{e}{1 - e}
 $

Then,

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

Question 24: Let a>0 be a root of the equation 2x2+x-2=0.
 If x1a161-2+x-2x21-ax2=+17, where ,Z then + is equal to

(1) 170

(2) 180

(3) 210

(4) 150

Correct Answer: Option (1)

Solution: 

$\begin{aligned} & \lim _{x \rightarrow \frac{1}{a}} 16 \cdot \frac{\left(1-\cos 2\left(x-\frac{1}{a}\right)\left(x-\frac{1}{b}\right)\right)}{4\left(x-\frac{1}{b}\right)^2} \times \frac{4\left(x-\frac{1}{b}\right)^2}{a^2\left(x-\frac{1}{a}\right)^2} \\ & =16 \times \frac{2}{a^2}\left(\frac{1}{a}-\frac{1}{b}\right)^2 \\ & =\frac{32}{a^2}\left(\frac{17}{4}\right)=\frac{17.8}{a^2}=\frac{17 \times 8 \times 16}{(-1+\sqrt{117})^2} \\ & =\frac{136.16}{18.2 \sqrt{7}} \times \frac{18+2 \sqrt{7}}{18+2 \sqrt{7}} \\ & =\frac{136}{256}(18+2 \sqrt{7}) \cdot 16 \\ & =153+17 \sqrt{17}=\alpha+\beta \sqrt{17} \\ & \alpha+\beta=153+17=170\end{aligned}$

Question 25: $\lim _{n \rightarrow \infty} \frac{\left(1^2-1\right)(n-1)+\left(2^2-2\right)(n-2)+\ldots .+\left((n-1)^2-(n-1)\right) \cdot 1}{\left(1^3+2^3+\ldots . .+n^3\right)-\left(1^2+2^2+\ldots . .+n^2\right)}
 $ is equal to:

(1) 23

(2) 13

(3) 34

(4) 12

Correct Answer: Option (2)

Solution: 

$\begin{aligned} & \lim _{n \rightarrow \infty} \frac{\sum_{r=1}^{n-1}\left(r^2-r\right)(n-r)}{\sum_{r=1}^n r^3-\sum_{r=1}^n r^2} \\\\ & \lim _{n \rightarrow \infty} \frac{\sum_{r=1}^{n-1}\left(-r^3+r^2(n+1)-n r\right)}{\left(\frac{n(n+1)}{2}\right)^2-\frac{n(n+1)(2 n+1)}{6}}\end{aligned}$

 $\begin{aligned} & \lim _{n \rightarrow \infty} \frac{\left(\frac{((n-1) n)}{2}\right)^2+\frac{(n+1)(n-1) n(2 n-1)}{6}-\frac{n^2(n-1)}{2}}{\frac{n(n+1)}{2}\left(\frac{n(n+1)}{2}-\frac{2 n+1}{3}\right)} \\\\ & \lim _{n \rightarrow \infty} \frac{\frac{n(n-1)}{2}\left(\frac{-n(n-1)}{2}+\frac{(n+1)(2 n-1)}{3}-n\right)}{\frac{n(n+1)}{2} \frac{3 n^2+3 n-4 n-2}{6}}\end{aligned}$

 $\begin{aligned} & \lim _{n \rightarrow \infty} \frac{(n-1)\left(-3 n^2+3 n+2\left(2 n^2+n-1\right)-6\right)}{(n+1)\left(3 n^2-n-2\right)} \\\\ & \lim _{n \rightarrow \infty} \frac{(n-1)\left(n^2+5 n-8\right)}{(n+1)\left(3 n^2-n-2\right)}=\frac{1}{3}\end{aligned}$

JEE Main 2026 Limits Most Asked Concepts

The table that is presented below highlights the concepts that have been asked in JEE Main 2026. This table will assist students in understanding which concepts have been asked more times.

Important Limit Formulas for JEE Mains

Knowing the formulas for Limits will certainly prove beneficial when attempting problems. Students must make it a point to revise these formulas from time to time and apply them to JEE Main problems.

1. Standard trigonometric limit: x0xx=1

2. Another standard limit: x0xx=1

3. Cosine limit: x01-xx2=12

4. Exponential limit: x0ex-1x=1

5. Logarithmic limit: x0(1+x)x=1

6. Important power limit: x0ax-1x=a

7. Basic form of L'Hospital's rule: For indeterminate forms such as 0/0 or /, differentiate the numerator and denominator and then evaluate the limit, when the rule's conditions are satisfied.

8. Maclaurin expansion of exponential function: ex=1+x+x22!+x33!+

9. Maclaurin expansion of sine function: x=x-x33!+x55!-

10. Maclaurin expansion of cosine function: x=1-x22!+x44!-

11. Maclaurin expansion of the logarithm: (1+x)=x-x22+x33-, for the appropriate range of x.

12. Important exponential form: x1+1xx=e

Best Books for JEE Limits Questions Preparation

While preparing limits questions for JEE, it is very important to follow the best books because they help properly cover all the concepts. Refer to the Best Books for JEE Main 2027 given below for solving limits questions for JEE PDF. 

Also Check: 

• JEE Main Physics: 20 Most Repeated Topic-Based Questions

• JEE Main Chemistry: 20 Most Repeated Topic-Based Questions