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JEE • MATHEMATICS • LIMITS

Limits practice · Easy

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 1-30 of 647 questions
1 Easy
1 of 647
Analyzing +4 | -1
The value of lim⁡h→02{3sin⁡(π6+h)−cos⁡(π6+h)3h(3cos⁡ h−sin⁡ h)}\lim\limits_{h \to 0} 2\left\{ {\frac{{\sqrt 3 \sin \left( {\frac{\pi }{6} + h} \right) - \cos \left( {\frac{\pi }{6} + h} \right)}}{{\sqrt 3 h\left( {\sqrt 3 \cos \,h - \sin \,h} \right)}}} \right\} is
View answer and explanation
Correct option: A
43\frac{4}{3}
Option A (43\frac{4}{3}) is correct because by applying trigonometric identities and L'Hopital's rule, the limit evaluates to 43\frac{4}{3}.
2 Easy
2 of 647
Application +4 | -1
If f(x)=(x+1)71+x2(x2−x+1)6f(x)=\frac{(x+1)^7 \sqrt{1+x^2}}{\left(x^2-x+1\right)^6}, then the value of f′(0)f^{\prime}(0) is equal to
View answer and explanation
Correct option: C
13
Option C (13) is correct because it is the value of f′(0)f'(0) found by logarithmic differentiation.
3 Easy
3 of 647
Application +4 | -1
The value of lim⁡x→∞log⁡xxn,  n>0\lim\limits_{x \to \infty } \frac{{\log x}}{{{x^n}}},\;n > 0 is
View answer and explanation
Correct option: A
0
Option A (0) is correct because applying L'Hopital's rule to the limit results in 0.
4 Easy
4 of 647
Application +4 | -1
If f(x)={sin⁡(1+[x])[x]for [x]≠00for [x]=0f(x)=\left\{\begin{array}{cl}\frac{\sin (1+[x])}{[x]} & \text {for }[x] \neq 0 \\ 0 & \text {for }[x]=0\end{array}\right. where [x][x] denotes the greatest integer ≤x\leq x, then lim⁡x→0−f(x)\lim \limits_{{x \rightarrow 0-}} f(x) equals
View answer and explanation
Correct option: B
0
The correct answer is B (0) because as x approaches 0 from the left, the greatest integer function [x] equals -1, and the limit is calculated accordingly.
5 Easy
5 of 647
Analyzing +4 | -1
lim⁡x→0(x−sin⁡xx)sin⁡(1x)\lim _{x \rightarrow 0}\left(\frac{x-\sin x}{x}\right) \sin \left(\frac{1}{x}\right)
View answer and explanation
Correct option: B
equals 0
Option B (equals 0) is correct because the limit can be evaluated using the Squeeze Theorem and standard limits.
6 Easy
6 of 647
Analyzing +4 | -1
lim⁡x→0+sin⁡xsin⁡x\lim \limits_ {x \rightarrow 0^{+}} \frac{\sin \sqrt{x}}{\sqrt{\sin x}} is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because by using the limit lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1, the expression simplifies to 1.
7 Easy
7 of 647
Application +4 | -1
If f(x)={x2−3,    2<x<32x+5,      3<x<4f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {{x^2} - 3,\;\;2 < x < 3}\\ {2x + 5,\;\;\;3 < x < 4} \end{array}} \right., the equation whose roots are lim⁡x→3−f(x)\lim\limits_{x \to {3^ - }} f\left( x \right) and lim⁡x→3+f(x)\lim\limits_{x \to {3^ + }} f\left( x \right), is
View answer and explanation
Correct option: C
x2−17x+66=0{x^2} - 17x + 66 = 0
Option C (x2−17x+66=0{x^2} - 17x + 66 = 0) is correct because the roots of the equation are 6 and 11, which are the left-hand and right-hand limits, respectively, and the equation is formed correctly.
8 Easy
8 of 647
Application +4 | -1
Let f(x)={7x−3ifx⩾23x+5ifx<2f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {7x - 3}&{{\text{if}}}&{x \geqslant 2} \\ {3x + 5}&{{\text{if}}}&{x < 2} \end{array}} \right. then lim⁡x→2f(x)\lim\limits_{x \to 2} f\left( x \right) equals
View answer and explanation
Correct option: A
11
Option A (11) is correct because both the left-hand limit and the right-hand limit as x approaches 2 are equal to 11.
9 Easy
9 of 647
Analyzing +4 | -1
lim⁡x→0xtan⁡2x−2xtan⁡x(1−cos⁡2x)2\lim _{x \rightarrow 0} \frac{x \tan 2 x-2 x \tan x}{(1-\cos 2 x)^2} equals.
View answer and explanation
Correct option: D
12\frac{1}{2}
Option D (12\frac{1}{2}) is correct because after applying trigonometric identities and simplifying, the limit evaluates to 12\frac{1}{2}.
10 Easy
10 of 647
Understanding +4 | -1
If f(x)={xsin⁡1x          x≠00            x=0f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {x\sin \frac{1}{x}\;\;\;\;\;x \ne 0}\\ {0\;\;\;\;\;\;x = 0} \end{array}} \right., then lim⁡x→0f(x)\lim\limits_{x \to 0} f\left( x \right) is equal to
View answer and explanation
Correct option: B
0
Option A ((−6  ,−3)∪(3,6)\left( { - 6\;, - 3} \right) \cup \left( {3,6} \right)) is correct because it represents the values of xx that satisfy the inequality 3<∣x∣<63 < |x| < 6.
11 Easy
11 of 647
Analyzing +4 | -1
lim⁡x→1(∫0(x−1)2tcos⁡(t2)dt(x−1)sin⁡(x−1))\lim\limits_{x \to 1} \left( {\frac{{\int\limits_0^{{{(x - 1)}^2}} {t\cos ({t^2})dt} }}{{(x - 1)\sin (x - 1)}}} \right)
View answer and explanation
Correct option: D
0
Option D is correct because by applying L'Hopital's rule twice, the limit evaluates to 0.
12 Easy
12 of 647
Analyzing +4 | -1
Let [t]\left[ t \right] denote the greatest integer ≤t\le t. If for some λ∈R−{0, 1}\lambda \in R - \left\{ {0,\,1} \right\},
lim⁡x→0∣1−x+∣x∣λ−x+[x]∣=L\lim\limits_{x \to 0} \left| {\frac{{1 - x + |x|}}{{\lambda - x + \left[ x \right]}}} \right| = L, then LL is equal to
View answer and explanation
Correct option: B
2
Option B (2) is correct because by evaluating the limit from both sides and equating them, we find L = 2.
13 Easy
13 of 647
Application +4 | -1
The value of lim⁡x→1x+x2+…+xn−nx−1\lim \limits_{{x \rightarrow 1}} \frac{x+x^2+\ldots+x^n-n}{x-1}, is
View answer and explanation
Correct option: C
n(n+1)2\frac{n(n+1)}{2}
Option D (3+24\frac{\sqrt{3}+\sqrt{2}}{4}) is correct because dividing by x and evaluating the limit at infinity gives the correct result. The provided answer is the rationalized form of the correct answer.
14 Easy
14 of 647
Analyzing +4 | -1
lim⁡y→01+1+y4−2y4\lim\limits_{y \to 0} \frac{{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 }}{{{y^4}}}
View answer and explanation
Correct option: A
exists and equals 142\frac{1}{{4\sqrt 2 }}
Option A (exists and equals 1/(4√2)) is correct because after rationalizing the numerator twice and simplifying, the limit can be evaluated directly.
15 Easy
15 of 647
Analyzing +4 | -1
The value of lim⁡n→∞ 6tan⁡{∑r=1ntan⁡−1(1r2+3r+3)}\lim\limits_{n \to \infty } \,6\tan \left\{ {\sum\limits_{r = 1}^n {{{\tan }^{ - 1}}} \left( {\frac{1}{{{r^2} + 3r + 3}}} \right)} \right\} is equal to:
View answer and explanation
Correct option: C
3
Option C (3) is correct because the limit is evaluated using the arctangent subtraction formula and telescoping series.

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16 Easy
16 of 647
Analyzing +4 | -1
lim⁡x→1−π−2sin⁡−1x1−x\lim _{x \rightarrow 1^{-}} \frac{\sqrt{\pi}-\sqrt{2 \sin ^{-1} x}}{\sqrt{1-x}} is equal to:
View answer and explanation
Correct option: B
2π\sqrt{\frac{2}{\pi}}
Option B (2π\sqrt{\frac{2}{\pi}}) is correct because after applying L'Hopital's rule and simplifying, the limit evaluates to 2π\sqrt{\frac{2}{\pi}}.
17 Easy
17 of 647
Analyzing +4 | -1
lim⁡x→π/2(1−tan⁡x/2)(1−sin⁡x)(1+tan⁡x/2)(π−2x)3\lim \limits_{{x \rightarrow \pi / 2}} \frac{(1-\tan x / 2)(1-\sin x)}{(1+\tan x / 2)(\pi-2 x)^3}
View answer and explanation
Correct option: D
132\frac{1}{32}
Option D (1/32) is correct because the limit is evaluated using trigonometric identities and substitution, resulting in 1/32.
18 Easy
18 of 647
Analyzing +4 | -1
lim⁡x→0(1−cos⁡2x)(3+cos⁡x)xtan⁡4x\lim _{x \rightarrow 0} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x} is equal to
View answer and explanation
Correct option: D
2
Option D (2) is correct because the limit can be evaluated using trigonometric identities and standard limits.
19 Easy
19 of 647
Application +4 | -1
If l(x)l\left( x \right) is the least integer not less than xx and g(x)g\left( x \right) is the greatest integer not greater than xx, then lim⁡x→e+π{l(x)+g(x)}\lim\limits_{x \to e + \pi } \left\{ {l\left( x \right) + g\left( x \right)} \right\} is equal to
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the limit evaluates to 11, which is not among the provided options.
20 Easy
20 of 647
Analyzing +4 | -1
Let f(x)=log⁡e(x2+ex)log⁡e(x4+e2x)f(x)=\frac{\log _e\left(x^2+e^x\right)}{\log _e\left(x^4+e^{2 x}\right)}. If lim⁡x→∞f(x)=l\lim \limits_{{x \rightarrow \infty}} f(x)=l and lim⁡x→−∞f(x)=m\lim \limits_{{x \rightarrow-\infty}} f(x)=m, then
View answer and explanation
Correct option: A
l=ml=m
Option B (pq\sqrt{pq}) is correct because it is obtained by converting the limit to an exponential form and using L'Hopital's rule or the properties of limits.
21 Easy
21 of 647
Analyzing +4 | -1
lim⁡x→0(1−cos⁡2x)22xtan⁡x−xtan⁡2x\lim \limits_{{x \rightarrow 0}} \frac{(1-\cos 2 x)^2}{2 x \tan x-x \tan 2 x}, is
View answer and explanation
Correct option: D
−2-2
Option D (-2) is correct because the limit is evaluated using trigonometric identities and simplification, resulting in -2.
22 Easy
22 of 647
Analyzing +4 | -1
lim⁡y→01+1+y4−2y4\lim _{y \rightarrow 0} \frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4}
View answer and explanation
Correct option: A
exists and equals 142\frac{1}{4 \sqrt{2}}
Option A (exists and equals 142\frac{1}{4 \sqrt{2}}) is correct because after rationalizing and simplifying, the limit exists and equals 142\frac{1}{4 \sqrt{2}}.
23 Easy
23 of 647
Analyzing +4 | -1
lim⁡x→0(1−cos⁡2x)(3+cos⁡x)xtan⁡4x\lim\limits_{x \to 0} \frac{{(1 - \cos 2x)(3 + \cos x)}}{{x\tan 4x}} is equal to
View answer and explanation
Correct option: D
2
Option D (2) is correct because by using trigonometric identities and limit properties, the given limit simplifies to 2.
24 Easy
24 of 647
Analyzing +4 | -1
If {x}\{x\} denotes the fractional part of xx, then lim⁡x→1xsin⁡{x}x−1\lim \limits_{{x \rightarrow 1}} \frac{x \sin \{x\}}{x-1}, is
View answer and explanation
Correct option: C
Non-existent
Option C (Non-existent) is correct because the left-hand limit and right-hand limit are not equal, meaning the limit does not exist.
25 Easy
25 of 647
Analyzing +4 | -1
lim⁡n→∞(an+bn2)n,a,b>0\lim \limits_{{n \rightarrow \infty}}\left(\frac{\sqrt[n]{a}+\sqrt[n]{b}}{2}\right)^n, a, b > 0 equals
View answer and explanation
Correct option: B
ab\sqrt{a b}
Option B (ab\sqrt{ab}) is correct because the limit can be evaluated by using the exponential form and properties of logarithms.
26 Easy
26 of 647
Analyzing +4 | -1
Let f(x)=x3{x2+x4+1−x2}f(x)=x^3\left\{\sqrt{x^2+\sqrt{x^4+1}}-x \sqrt{2}\right\}. Then lim⁡x→∞f(x)\lim _{x \rightarrow \infty} f(x) is equal to
View answer and explanation
Correct option: B
142\frac{1}{4 \sqrt{2}}
Option B (1/(4√2)) is correct because after simplifying the expression and taking the limit as x approaches infinity, the result is 1/(4√2).
27 Easy
27 of 647
Analyzing +4 | -1
lim⁡x→∞  [x2+2x−1−x]\lim\limits_{x \to \infty } \,\,\left[ {\sqrt {{x^2} + 2x - 1} - x} \right] is equal to
View answer and explanation
Correct option: D
1
Option D (1) is correct because, by rationalizing the expression and taking the limit as x approaches infinity, we get 1.
28 Easy
28 of 647
Analyzing +4 | -1
f(x)=3x10−7x8+5x6−21x3+3x2−7f(x)=3 x^{10}-7 x^8+5 x^6-21 x^3+3 x^2-7, then the value of lim⁡h→0f(1−h)−f(1)h3+3h\lim \limits_{{h \rightarrow 0}} \frac{f(1-h)-f(1)}{h^3+3 h} is
View answer and explanation
Correct option: D
None of these
The correct answer is None of these because the limit evaluates to -53/3, which is not among the given options.
29 Easy
29 of 647
Analyzing +4 | -1
lim⁡x→0x(e(1+x2+x4−1)/x−1)1+x2+x4−1\lim _{x \rightarrow 0} \frac{x\left(e^{\left(\sqrt{1+x^2+x^4}-1\right) / x}-1\right)}{\sqrt{1+x^2+x^4}-1}
View answer and explanation
Correct option: B
is equal to 1
Option B (is equal to 1) is correct because after substituting t=1+x2+x4−1xt = \frac{{\sqrt{{1 + x^2 + x^4}} - 1}}{{x}}, the limit simplifies to the standard limit lim⁡t→0et−1t=1\lim_{{t \rightarrow 0}} \frac{{e^t - 1}}{{t}} = 1.
30 Easy
30 of 647
Analyzing +4 | -1
Let p(x)p(x) be a real polynomial of degree 4 having extreme values at x=1x=1 and x=2x=2. If lim⁡x→0p(x)x2=1\lim \limits_{{x \rightarrow 0}} \frac{p(x)}{x^2}=1, then p(4)p(4) is equal to
View answer and explanation
Correct option: B
16
Option B (16) is correct because after finding the polynomial p(x) using the given conditions, substituting x=4 yields the correct value.
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