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JEE • MATHEMATICS • LIMITS

Limits practice · Mixed difficulty

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541 Easy
541 of 930
Understanding +4 | -1
The value of lim⁡n→∞(11⋅3+13⋅5+⋯+\lim \limits_{{n \rightarrow \infty}}\left(\frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\cdots+\right. up to nn terms)) is
View answer and explanation
Correct option: B
1/21 / 2
Option B (1/21 / 2) is correct because the given series is a telescoping series, and its sum converges to 1/2 as n approaches infinity.
542 Easy
542 of 930
Analyzing +4 | -1
Let f(x)={tan⁡2[x]x2−[x]2for x>01∣x∣cot⁡∣x∣for x=0, where [x]for x<0f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{{{\tan }^2}\left[ x \right]}}{{{x^2} - {{\left[ x \right]}^2}}}}&{{\text{for}}\,x > 0} \\ {\frac{1}{{\sqrt {\left| x \right|\cot \left| x \right|} }}}&\begin{gathered} {\text{for}}\,x = 0,\, \quad \\ {\text{where }}\left[ x \right] \quad \\ {\text{for}}\,x < 0 \quad \\ \end{gathered} \end{array}} \right.
and {x} denote respectively the greatest integer less than equal to xx and the fractional part of xx, then
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because the left-hand limit, right-hand limit, and the cotangent inverse of the squared left-hand limit all satisfy the given conditions.
543 Easy
543 of 930
Application +4 | -1
The value of lim⁡x→ax−b−a−bx2−a2(a>b)\lim \limits_{{x \rightarrow a}} \frac{\sqrt{x-b}-\sqrt{a-b}}{x^2-a^2}(a>b) is
View answer and explanation
Correct option: D
14aa−b\frac{1}{4 a \sqrt{a-b}}
Option D (14aa−b\frac{1}{4 a \sqrt{a-b}}) is correct because after rationalizing and simplifying, the limit evaluates to 14aa−b\frac{1}{4a\sqrt{a-b}}.
544 Medium
544 of 930
Application +4 | -1
lim⁡x→03x−11+x−1\lim \limits_{{x \rightarrow 0}} \frac{3^x-1}{\sqrt{1+x-1}} is equal to-
View answer and explanation
Correct option: B
2ℓn32 \ell n 3
Option B (2 ln 3) is correct because the limit is evaluated by rationalizing the denominator and using the exponential limit.
545 Easy
545 of 930
Application +4 | -1
lim⁡x→0cos⁡2x−cos⁡xsin⁡2x\lim \limits_{{x \rightarrow 0}} \frac{\cos 2 x-\cos x}{\sin ^2 x} is
View answer and explanation
Correct option: D
−32-\frac{3}{2}
The correct answer is D (-32\frac{3}{2}) because applying L'Hôpital's rule twice to the limit yields the value -32\frac{3}{2}.
546 Easy
546 of 930
Application +4 | -1
lim⁡x→0(1−ax)1/x\lim\limits_{x \to 0} {\left( {1 - ax} \right)^{1/x}} is equal to
View answer and explanation
Correct option: A
e−a{e^{ - a}}
Option A (e−a{e^{-a}}) is correct because it is the direct application of the standard limit formula for exponential functions.
547 Easy
547 of 930
Analyzing +4 | -1
The value of lim⁡x→∞x3/2(x3+1−x3−1)\lim\limits_{x \to \infty } {x^{3/2}}\left( {\sqrt {{x^3} + 1} - \sqrt {{x^3} - 1} } \right), is
View answer and explanation
Correct option: A
1
Option A (1) is correct because after rationalizing and simplifying, the limit evaluates to 1.
548 Easy
548 of 930
Application +4 | -1
The value of cc for which the conclusion of Lagrange's theorem holds for the function f(x)=f(x)= a2−x2,a>1\sqrt{a^2-x^2}, a>1 on the interval [1,a][1, \mathrm{a}] is
View answer and explanation
Correct option: C
a(a+1)2\frac{\sqrt{a(a+1)}}{2}
Option C (a(a+1)2\frac{\sqrt{a(a+1)}}{2}) is correct because the value of c that satisfies Lagrange's theorem for the given function and interval is a(a+1)2\frac{\sqrt{a(a+1)}}{2}, found by equating the derivative at c to the average rate of change.
549 Easy
549 of 930
Understanding +4 | -1
If f(x)=∫1xsin⁡ttdtf(x)=\int\limits_{1}^x \frac{\sin t}{t} d t, then lim⁡x→∞f′(x)\lim \limits_{{x \rightarrow \infty}} f^{\prime}(x) is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because as x approaches infinity, the function f'(x) = sin⁡tt\frac{{\sin t}}{{t}} approaches 0, due to the denominator growing without bound while the numerator oscillates between -1 and 1.
550 Easy
550 of 930
Application +4 | -1
lim⁡x→01−cos⁡2xx=\lim \limits_{{x \rightarrow 0}} \frac{1-\cos 2 x}{x}=
View answer and explanation
Correct option: A
0
Option A (0) is correct because lim⁡x→01−cos⁡2xx=0\lim \limits_{{x \rightarrow 0}} \frac{1-\cos 2 x}{x} = 0.
551 Easy
551 of 930
Analyzing +4 | -1
If lim⁡x→0aex−bcos⁡x+ce−xxsin⁡x=2\lim \limits_{{x \rightarrow 0}} \frac{a e^x-b \cos x+c e^{-x}}{x \sin x}=2, then a+b+c=a+b+c=
View answer and explanation
Correct option: B
4
Option B (4) is correct because the limit exists only when certain conditions are met, which leads to a system of equations that can be solved for a+b+ca+b+c.
552 Easy
552 of 930
Application +4 | -1
lim⁡n→∞12+22+32+...+n2n3\lim\limits_{n \to \infty } \frac{{{1^2} + {2^2} + {3^2} + ... + {n^2}}}{{{n^3}}} is equal to
View answer and explanation
Correct option: C
13\frac{1}{3}
Option C (13\frac{1}{3}) is correct because the limit of the sum of squares divided by n3 is 1/3.
553 Medium
553 of 930
Analyzing +4 | -1
The value of lim⁡x→∞{log⁡(n−1)n.log⁡n(n+1).log⁡(n+1)(n+2)…{…log⁡(nk−1)(nk)}\lim\limits_{x \to \infty } \{ {\log _{\left( {n - 1} \right)}}n.{\log _n}\left( {n + 1} \right).{\log _{\left( {n + 1} \right)}}\left( {n + 2} \right) \ldots \left\{ { \ldots {{\log }_{\left( {{n^k} - 1} \right)}}\left( {{n^k}} \right)} \right\}, is
View answer and explanation
Correct option: C
kk
Option C (kk) is correct because after applying the change of base formula for logarithms and L'Hopital's rule, the limit evaluates to kk.
554 Easy
554 of 930
Analyzing +4 | -1
 Let f(θ)=1tan⁡2θ{(1+tan⁡θ)3+(2+tan⁡θ)3+….+(10+tan⁡θ)3}−10tan⁡θ\begin{aligned}\text { Let } f(\theta)=\frac{1}{\tan ^2 \theta}\left\{(1+\tan \theta)^3+(2+\tan \theta)^3+\ldots\right..\left.+(10+\tan \theta)^3\right\}-10 \tan \theta\end{aligned}

Then, lim⁡θ→π−2f(θ)\lim \limits_{\theta \rightarrow \frac{\pi^{-}}{2}} f(\theta) is equal to
View answer and explanation
Correct option: C
165
Option C (165) is correct because the limit is evaluated by simplifying the expression and applying limit properties as θ approaches π−2\frac{π^{-}}{2}.
555 Easy
555 of 930
Analyzing +4 | -1
lim⁡x→∞2+2x+sin⁡2x(2x+sin⁡2x)esin⁡x\lim \limits_{{x \rightarrow \infty}} \frac{2+2 x+\sin 2 x}{(2 x+\sin 2 x) e^{\sin x}} is equal to
View answer and explanation
Correct option: D
Non-existent
Option D (Non-existent) is correct because the limit oscillates between 1e\frac{1}{e} and e as x approaches infinity due to the esin⁡xe^{\sin x} term.

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556 Medium
556 of 930
Analyzing +4 | -1
lim⁡θ→04θ(tan⁡θ−2θtan⁡θ)(1−cos⁡2θ)\lim\limits_{\theta \to 0} \frac{{4\theta \left( {\tan \theta - 2\theta \tan \theta } \right)}}{{\left( {1 - \cos 2\theta } \right)}} is
View answer and explanation
Correct option: D
2
Option D (2) is correct because applying L'Hopital's rule twice to the limit results in a value of 2.
557 Easy
557 of 930
Analyzing +4 | -1
If [.] denotes the greatest integer function, then lim⁡x→0tan⁡([−2π2]x2)−x2tan⁡[−2π2]sin⁡2x\lim \limits_{{x \rightarrow 0}} \frac{\tan \left(\left[-2 \pi^2\right] x^2\right)-x^2 \tan \left[-2 \pi^2\right]}{\sin ^2 x} is equal to
View answer and explanation
Correct option: A
−20+tan⁡20-20+\tan 20
Option A (−20+tan⁡20-20+\tan 20) is correct because it is the result of applying limit rules and trigonometric identities to the given expression.
558 Medium
558 of 930
Application +4 | -1
If lim⁡x→5xk−5kx−5=500\lim \limits_{{x \rightarrow 5}} \frac{x^k-5^k}{x-5}=500, then kk is equal to:
View answer and explanation
Correct option: A
4
Option A (4) is correct because the value of k is 4, derived from the limit definition of the derivative.
559 Easy
559 of 930
Analyzing +4 | -1
The value of lim⁡x→∞1+x4−(1+x2)x2\lim\limits_{x \to \infty } \frac{{\sqrt {1 + {x^4}} - \left( {1 + {x^2}} \right)}}{{{x^2}}} is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because after dividing numerator and denominator by x2 and taking the limit, the expression evaluates to 0.
560 Medium
560 of 930
Analyzing +4 | -1
lim⁡x→∞∫0xtan⁡−1t dtx2+1\lim \limits_{{x \rightarrow \infty}} \frac{\int\limits_{0}^x \tan ^{-1} t \, d t}{\sqrt{x^2+1}} is equal to:
View answer and explanation
Correct option: A
π2\frac{\pi}{2}
Option A (π2\frac{\pi}{2}) is correct because after applying L'Hôpital's rule, the limit simplifies to π2\frac{\pi}{2}.
561 Hard
561 of 930
Analyzing +4 | -1
lim⁡x→π2acot⁡x−acos⁡xcot⁡x−cos⁡x,a>0\lim\limits_{x \to \frac{\pi }{2}} \frac{{{a^{\cot x}} - {a^{\cos x}}}}{{\cot x - \cos x}},a > 0 is equal to
View answer and explanation
Correct option: C
log⁡ea{\log _e}a
Option C (log⁡ea{\log _e}a) is correct because, by using L'Hopital's rule or series expansion, the limit evaluates to the natural logarithm of a.
562 Medium
562 of 930
Analyzing +4 | -1
The value of lim⁡x→2ax−2a+x−2ax2−4a2\lim \limits_{{x \rightarrow 2 a}} \frac{\sqrt{x-2 a}+\sqrt{x}-\sqrt{2 a}}{\sqrt{x^2-4 a^2}} is
View answer and explanation
Correct option: B
12a\frac{1}{2 \sqrt{a}}
Option B (12a\frac{1}{2 \sqrt{a}}) is correct because after simplifying and rationalizing, the limit evaluates to 12a\frac{1}{2 \sqrt{a}}.
563 Easy
563 of 930
Application +4 | -1
lim⁡x→+∞x−83−x+10x2=\lim _{x \rightarrow+\infty} \frac{x-8}{3-x+10 x^2}=
View answer and explanation
Correct option: A
0
Option A (0) is correct because after dividing by x2 and simplifying, the limit of the expression as x approaches infinity is 0.
564 Easy
564 of 930
Application +4 | -1
Value of lim⁡x→01+tan⁡x3−1−tan⁡x3x\lim \limits_{{x \rightarrow 0}} \frac{\sqrt[3]{1+\tan x}-\sqrt[3]{1-\tan x}}{x} is-
View answer and explanation
Correct option: D
23\frac{2}{3}
Option D (2/3) is correct because the limit is evaluated using L'Hopital's rule.
565 Easy
565 of 930
Analyzing +4 | -1
If l=lim⁡x→0tan⁡xn(tan⁡x)ml=\lim \limits_{{x \rightarrow 0}} \frac{\tan x^n}{(\tan x)^m}, where m,n∈Nm, n \in N, then
View answer and explanation
Correct option: C
l={1, if n=m0, if n>ml= \begin{cases}1, \text { if } n=m \\ 0, \text { if } n > m\end{cases}
The correct answer is l={1, if n=m0, if n>ml= \begin{cases}1, \text { if } n=m \\ 0, \text { if } n > m\end{cases} because the limit depends on the relationship between nn and mm. When n=mn=m, the limit is 1, and when n>mn>m, the limit is 0.
566 Easy
566 of 930
Analyzing +4 | -1
If f(x)={tan⁡−1(x+[x])[x]−2x,[x]≠00,[x]=0f(x)=\left\{\begin{array}{cl}\frac{\tan ^{-1}(x+[x])}{[x]-2 x}, {[x] \neq 0} \\ 0,[x]=0\end{array}\right. where [x][x] denotes the greatest integer less than or equal to xx, then lim⁡x→0f(x)\lim \limits_{{x \rightarrow 0}} f(x) is
View answer and explanation
Correct option: D
Non-existent
Option D (Non-existent) is correct because the left-hand limit and right-hand limit are not equal, hence the limit does not exist.
567 Easy
567 of 930
Application +4 | -1
Find lim⁡x→0(x+9)32+ex−28x\lim \limits_{{x \rightarrow 0}} \frac{(x+9)^{\frac{3}{2}}+e^x-28}{x}
View answer and explanation
Correct option: D
11/211 / 2
Option D (11/211 / 2) is correct because the limit can be evaluated by recognizing it as the derivative of the function at x=0.
568 Medium
568 of 930
Application +4 | -1
The value of Lim⁡x→0∫0xxet2dt1+x−ex\operatorname{Lim}_{x \rightarrow 0} \frac{\int_0^x x e^{t^2} d t}{1+x-e^x} is equal to
View answer and explanation
Correct option: D
-2
Option D (-2) is correct because applying L'Hôpital's Rule twice to the limit Lim⁡x→0∫0xxet2dt1+x−ex\operatorname{Lim}_{x \rightarrow 0} \frac{\int_0^x x e^{t^2} d t}{1+x-e^x} yields -2.
569 Medium
569 of 930
Application +4 | -1
If [.][.] denotes the greatest integer function, then lim⁡x→π/2[x−π2cos⁡x]\lim \limits_{{x \rightarrow \pi / 2}}\left[\frac{x-\frac{\pi}{2}}{\cos x}\right] is equal to
View answer and explanation
Correct option: D
-2
Option D (-2) is correct because by substituting x=π2+θx = \frac{\pi}{2} + \theta and using the limit lim⁡θ→0θsin⁡θ=1\lim_{\theta \to 0} \frac{\theta}{\sin \theta} = 1, the expression simplifies to -2.
570 Easy
570 of 930
Analyzing +4 | -1
The value of lim⁡n→∞{n2−n33+n}\lim \limits_{{n \rightarrow \infty}}\left\{\sqrt[3]{n^2-n^3}+n\right\}, is
View answer and explanation
Correct option: A
13\frac{1}{3}
Option A (1/3) is correct because, after rationalizing and simplifying, the limit evaluates to 1/3.
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