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JEE • MATHEMATICS • LIMITS

Limits practice · Mixed difficulty

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91 Medium
91 of 930
Analyzing +4 | -1
lim⁡x→0xtan⁡2x−2xtan⁡x(1−cos⁡2x)2\lim \limits_{{x \rightarrow 0}} \frac{x \tan 2 x-2 x \tan x}{(1-\cos 2 x)^2} is equal to
View answer and explanation
Correct option: C
12\frac{1}{2}
Option C (12\frac{1}{2}) is correct because after applying trigonometric identities and limit properties, the expression simplifies to 12\frac{1}{2}.
92 Easy
92 of 930
Application +4 | -1
The value of lim⁡x→1xn+xn−1+xn−2+⋯+x2+x−nx−1\lim \limits_{{x \rightarrow 1}} \frac{x^n+x^{n-1}+x^{n-2}+\cdots+x^2+x-n}{x-1} is
View answer and explanation
Correct option: A
n(n+1)2\frac{n(n+1)}{2}
Option A (n(n+1)2\frac{n(n+1)}{2}) is correct because the limit can be evaluated by rewriting the expression and using the limit definition of derivatives.
93 Easy
93 of 930
Analyzing +4 | -1
Let a be an integer such that lim⁡x→718−[1−x][x−3a]\lim\limits_{x \to 7} \frac{{18 - [1 - x]}}{{[x - 3a]}} exists, where [t][t] is greatest integer ⩽t.\leqslant t. Then a is equal to:
View answer and explanation
Correct option: A
-6
Option A (-6) is correct because for the limit to exist, the left-hand limit (LHL) must equal the right-hand limit (RHL), which leads to a = -6.
94 Easy
94 of 930
Application +4 | -1
lim⁡x→∞(x2+5x+3x2+x+2)x\lim\limits_{x \to \infty } {\left( {\frac{{{x^2} + 5x + 3}}{{{x^2} + x + 2}}} \right)^x} is equal to
View answer and explanation
Correct option: A
e4{e^4}
Option A (e4{e^4}) is correct because the limit can be evaluated by rewriting the expression and using the standard limit definition of ee.
95 Medium
95 of 930
Application +4 | -1
lim⁡x→π4sec⁡2x−2tan⁡x1+cos⁡4x\lim \limits_{x \rightarrow \frac{\pi}{4}} \frac{\sec ^2 x-2 \tan x}{1+\cos 4 x} is
View answer and explanation
Correct option: A
12\frac{1}{2}
Option A (12\frac{{1}}{{2}}) is correct because by applying L'Hôpital's rule twice, the limit simplifies to 12\frac{{1}}{{2}}.
96 Medium
96 of 930
Analyzing +4 | -1
lim⁡n→∞1+2−3+4+5−6+⋯(3n−2)+(3n−1)−3n2n4+3n+1−n4+n+3\lim \limits_{{n \rightarrow \infty}} \frac{1+2-3+4+5-6+\cdots(3 n-2)+(3 n-1)-3 n}{\sqrt{2 n^{4}+3 n+1}-\sqrt{n^{4}+n+3}}
View answer and explanation
Correct option: A
32(2+1)\frac{3}{2}(\sqrt{2}+1)
Option A (32\frac{{3}}{{2}}(2\sqrt{{2}}+1)) is correct because the limit can be evaluated by recognizing the pattern in the series and simplifying the expression.
97 Easy
97 of 930
Analyzing +4 | -1
The value of lim⁡x→0ex+log⁡(1+x)−(1−x)−2x2\lim \limits_{{x \rightarrow 0}} \frac{e^x+\log (1+x)-(1-x)^{-2}}{x^2} is equal to
View answer and explanation
Correct option: B
−3-3
Option B (-3) is correct because the limit is evaluated using the Taylor series expansions of exe^x, log⁡(1+x)\log(1+x), and (1−x)−2(1-x)^{-2}.
98 Easy
98 of 930
Application +4 | -1
lim⁡x→0ax−bxex−1\lim \limits_{{x \rightarrow 0}} \frac{a^x-b^x}{e^x-1} is equal to
View answer and explanation
Correct option: A
log⁡e(ab)\log _e\left(\frac{a}{b}\right)
Option A (log⁡e(ab)\log_e(\frac{a}{b})) is correct because the limit can be evaluated using the standard limit lim⁡x→0ax−1x=ln⁡a\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a.
99 Medium
99 of 930
Application +4 | -1
If lim⁡x→−nx7+n7x+n=7\lim \limits_{{x \rightarrow-n}} \frac{x^7+n^7}{x+n}=7, then the value of nn is-
View answer and explanation
Correct option: D
±1\pm 1
Option D (±\pm 1) is correct because the limit is evaluated using the formula for the limit of xn+anx+a\frac{{x^n + a^n}}{{x+a}} as x approaches -a.
100 Easy
100 of 930
Analyzing +4 | -1
The value of lim⁡x→0sin⁡xx2\lim\limits_{x \to 0} \frac{{\sin x}}{{\sqrt {{x^2}} }}, is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the limit does not exist as the left-hand limit is -1 and the right-hand limit is 1.
101 Easy
101 of 930
Analyzing +4 | -1
lim⁡x→0xcos⁡x+sin⁡xx2+tan⁡x\lim \limits_{{x \rightarrow 0}} \frac{x \cos x+\sin x}{x^2+\tan x} is equal to
View answer and explanation
Correct option: D
2
Option D (2) is correct because applying L'Hopital's rule yields a determinate form after the first application.
102 Easy
102 of 930
Application +4 | -1
The value of lim⁡n→∞13+23+…+n3n4\lim \limits_{{n \rightarrow \infty}} \frac{1^3+2^3+\ldots+n^3}{n^4}, is
View answer and explanation
Correct option: C
14\frac{1}{4}
Option C (14\frac{1}{4}) is correct because the limit of the given expression as n approaches infinity is equal to 14\frac{1}{4}.
103 Easy
103 of 930
Application +4 | -1
Value of lim⁡x→0sin⁡4xsin⁡2x\lim \limits_{{x \rightarrow 0}} \frac{\sin 4 x}{\sin 2 x} is
View answer and explanation
Correct option: A
2
Option A is correct because the limit evaluates to 2.
104 Medium
104 of 930
Analyzing +4 | -1
If f(x)=x−sin⁡xx+cos⁡2xf\left( x \right) = \sqrt {\frac{{x - \sin x}}{{x + {{\cos }^2}x}}}, then lim⁡x→∞f(x)\lim\limits_{x \to \infty } f\left( x \right) is
View answer and explanation
Correct option: C
1
Option C (1) is correct because as x approaches infinity, the terms sin⁡xx\frac{\sin x}{x} and cos⁡2xx\frac{\cos^2 x}{x} approach 0, leaving 11=1\sqrt{\frac{1}{1}} = 1.
105 Easy
105 of 930
Understanding +4 | -1
If f(x)={xsin⁡(1x),  x≠00,  x=0f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {x\sin \left( {\frac{1}{x}} \right),\;x \ne 0}\\ {0,\;x = 0} \end{array}} \right. Then, lim⁡x→0f(x)\lim\limits_{x \to 0} f\left( x \right)
View answer and explanation
Correct option: C
Is equal to 0
Option A (a2+b2+c2≥bc+ca+ab{a^2} + {b^2} + {c^2} \ge bc + ca + ab) is correct because it is a direct result of the AM-GM inequality or can be derived from the fact that the sum of squares is always non-negative.

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs
106 Medium
106 of 930
Application +4 | -1
The value of lim⁡x→0{tan⁡(π4+x)}1/x\lim \limits_{{x \rightarrow 0}}\left\{\tan \left(\frac{\pi}{4}+x\right)\right\}^{1 / x}, is
View answer and explanation
Correct option: B
e2e^2
Option B (e2e^2) is correct because the limit can be evaluated using the exponential form of limits and the tangent addition formula.
107 Medium
107 of 930
Analyzing +4 | -1
lim⁡x→α(tan⁡xcot⁡α)1x−α\lim \limits_{{x \rightarrow \alpha}}(\tan x \cot \alpha)^{\frac{1}{x-\alpha}} is equal to
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the correct limit evaluates to e2cosec⁡2αe^{2 \operatorname{cosec} 2 \alpha}, which is not among the given options.
108 Easy
108 of 930
Application +4 | -1
lim⁡x→0(1−cos⁡2x)sin⁡5xx2sin⁡3x\lim \limits_{{x \rightarrow 0}} \frac{(1-\cos 2 x) \sin 5 x}{x^2 \sin 3 x} equals
View answer and explanation
Correct option: A
103\frac{10}{3}
Option A (103\frac{10}{3}) is correct because the limit can be evaluated using the standard limit lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1 and the identity 1−cos⁡2x=2sin⁡2x1 - \cos 2x = 2 \sin^2 x.
109 Easy
109 of 930
Analyzing +4 | -1
The value of lim⁡x→∞x{1e−(xx+1)x}\lim \limits_{{x \rightarrow \infty}} x\left\{\frac{1}{e}-\left(\frac{x}{x+1}\right)^x\right\}, is
View answer and explanation
Correct option: C
−12e-\frac{1}{2 e}
Option C (-1/(2e)) is correct because it is obtained by using the Taylor series expansion of the logarithmic function and simplifying the limit.
110 Easy
110 of 930
Application +4 | -1
The value of lim⁡x→0x2+1−1x2+9−3\lim \limits_{{x \rightarrow 0}} \frac{\sqrt{x^2+1}-1}{\sqrt{x^2+9}-3}, is
View answer and explanation
Correct option: A
3
Option A (3) is correct because rationalizing both the numerator and denominator and simplifying leads to a limit of 3 as x approaches 0.
111 Easy
111 of 930
Analyzing +4 | -1
lim⁡n→∞[2log⁡(3n)−log⁡(n2+1)]\lim \limits_{{n \rightarrow\infty}}\left[2 \log (3 n)-\log \left(n^2+1\right)\right] is equal to
View answer and explanation
Correct option: C
2log⁡32 \log 3
Option C (2log⁡32 \log 3) is correct because after simplifying the limit using logarithm properties and evaluating as n approaches infinity, the result is 2log⁡32 \log 3.
112 Medium
112 of 930
Application +4 | -1
If [] denotes the greatest integer function, then lim⁡n→∞[x]+[2x]+…+[nx]n2\lim \limits_{{n \rightarrow \infty}} \frac{[x]+[2 x]+\ldots+[n x]}{n^2} is
View answer and explanation
Correct option: C
x2\frac{x}{2}
Option C (x2\frac{x}{2}) is correct because the limit can be evaluated using the properties of the greatest integer function and the sandwich theorem.
113 Medium
113 of 930
Analyzing +4 | -1
lim⁡x→0ax−1−xlog⁡ax2\lim \limits_{{x \rightarrow 0}} \frac{a^x-1-x \log a}{x^2} is
View answer and explanation
Correct option: B
12(log⁡a)2\frac{1}{2}(\log a)^2
Option B (12(log⁡a)2\frac{1}{2}(\log a)^2) is correct because after applying L'Hôpital's rule twice, the limit evaluates to 12(log⁡a)2\frac{1}{2}(\log a)^2.
114 Medium
114 of 930
Application +4 | -1
If f:R→Rf: R \rightarrow R is given by f(x)=x+1f(x)=x+1 then the value of n→∞lim⁡1n[f(0)+f(5n)+f(10n)+....+f(5(n−1)n)]_{n \to \infty }^{\lim }\frac{1}{n}\left[ {f\left( 0 \right) + f\left( {\frac{5}{n}} \right) + f\left( {\frac{{10}}{n}} \right) + .... + f\left( {\frac{{5\left( {n - 1} \right)}}{n}} \right)} \right] is
View answer and explanation
Correct option: D
72\frac{7}{2}
Option D (72\frac{7}{2}) is correct because it is the limit of the Riemann sum, which can be evaluated as a definite integral.
115 Easy
115 of 930
Understanding +4 | -1
lim⁡n→∞1n{1+e1/n+e2/n+e3/n+….+en−1n}\lim \limits_{{n \rightarrow \infty}} \frac{1}{n}\left\{1+e^{1 / n}+e^{2 / n}+e^{3 / n}+\ldots .+e^{\frac{n-1}{n}}\right\} is equal to
View answer and explanation
Correct option: C
e−1e-1
Option C (e−1e-1) is correct because the limit can be expressed as a Riemann sum, which evaluates to the integral of e^x from 0 to 1.
116 Easy
116 of 930
Analyzing +4 | -1
If I1=lim⁡x→∞(tan⁡−1πx−tan⁡−1x)cos⁡xI_1=\lim \limits_{{x \rightarrow \infty}}\left(\tan ^{-1} \pi x-\tan ^{-1} x\right) \cos x and I2=lim⁡x→0(tan⁡−1πx−tan⁡−1x)cos⁡xI_2=\lim \limits_{{x \rightarrow 0}}\left(\tan ^{-1} \pi x-\tan ^{-1} x\right) \cos x, then (I1,I2)\left(I_1, I_2\right) is
View answer and explanation
Correct option: A
(0,0)(0,0)
Option A ((0,0)) is correct because both limits I1I_1 and I2I_2 evaluate to 0.
117 Easy
117 of 930
Application +4 | -1
lim⁡x→0ex2−cos⁡xx2\lim \limits_{{x \rightarrow 0}} \frac{e^{x^2}-\cos x}{x^2} is equal to
View answer and explanation
Correct option: B
3/23 / 2
Option B (3/23 / 2) is correct because the limit is evaluated by using the series expansion of ex2e^{x^2} and cos⁡x\cos x or by applying L'Hopital's rule twice.
118 Medium
118 of 930
Application +4 | -1
lim⁡x→0  ∫0x2  tan⁡  t dtx3\lim\limits_{x \to 0} \,\,\frac{{\int \nolimits_0^{{x^2}} \,\,\tan \,\,\sqrt t \,dt}}{{{x^3}}} is equal to
View answer and explanation
Correct option: D
None of these
Option D is correct. The limit does not exist because the left-hand limit and right-hand limit are not equal.
119 Easy
119 of 930
Application +4 | -1
The value of lim⁡x→∞3x2−1+2x2−14x+3\lim \limits_{{x \rightarrow \infty}} \frac{\sqrt{3 x^2-1}+\sqrt{2 x^2-1}}{4 x+3}, is
View answer and explanation
Correct option: B
14(3−2)\frac{1}{4(\sqrt{3}-\sqrt{2})}
Option B (1/3) is correct because simplifying the expression using the sum of squares formula leads to a limit of 1/3 as n approaches infinity.
120 Medium
120 of 930
Application +4 | -1
The value of lim⁡h→0f(x+h)−f(x)h\lim \limits_{{h \rightarrow 0}} \frac{f(x+h)-f(x)}{h}, where f(x)=x+1x−1f(x)=\frac{x+1}{x-1}, is
View answer and explanation
Correct option: A
−2(x−1)2\frac{-2}{(x-1)^2}
Option A (−2(x−1)2\frac{-2}{(x-1)^2}) is correct because it is the result of differentiating f(x) using the quotient rule and then simplifying.
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