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JEE • MATHEMATICS • LIMITS

Limits practice · Easy

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391 Easy
391 of 647
Application +4 | -1
If lim⁡x→1x4−1x−1=lim⁡x→kx3−k3x2−k2\lim _{x \rightarrow 1} \frac{x^4-1}{x-1}=\lim _{x \rightarrow k} \frac{x^3-k^3}{x^2-k^2}, then kk is:
View answer and explanation
Correct option: A
83\frac{8}{3}
Option A (83\frac{8}{3}) is correct because it is the value of k that satisfies the given limit equation after applying L'Hopital's Rule.
392 Easy
392 of 647
Application +4 | -1
lim⁡x→3x2−3x2+33x−12\lim \limits_{x \rightarrow \sqrt{3}} \frac{x^2-3}{x^2+3 \sqrt{3} x-12} is equal to
View answer and explanation
Correct option: B
25\frac{2}{5}
Option B (25\frac{2}{5}) is correct because it is the result of factoring and simplifying the limit expression as x approaches 3\sqrt{3}.
393 Easy
393 of 647
Analyzing +4 | -1
lim⁡x→−1(x4+x2+x+1x2−x+1)1−cos⁡(x+1)(x+1)2\lim \limits_{{x \rightarrow-1}}\left(\frac{x^4+x^2+x+1}{x^2-x+1}\right)^{\frac{1-\cos (x+1)}{(x+1)^2}} is equal to
View answer and explanation
Correct option: B
2/3\sqrt{2 / 3}
Option B (2/3\sqrt{2 / 3}) is correct because by substituting x=−1x = -1 into the base and using the limit lim⁡x→01−cos⁡xx2=12\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} for the exponent, the limit evaluates to 2/3\sqrt{2/3}.
394 Easy
394 of 647
Analyzing +4 | -1
lim⁡n→∞3.2n+1−4.5n+15.2n+7.5n\lim\limits_{n \to \infty } \frac{{{{3.2}^{n + 1}} - {{4.5}^{n + 1}}}}{{{{5.2}^n} + {{7.5}^n}}} is equal to
View answer and explanation
Correct option: C
−207- \frac{{20}}{7}
The correct answer is C (−207\frac{{ - 20}}{7}) because the limit is evaluated by dividing by the dominant term and simplifying the expression.
395 Easy
395 of 647
Analyzing +4 | -1
lim⁡x→0−[x]+[x2]+[x3]+….+[x2n+1]+n+11+[x2]+∣x∣+2x,n∈N\lim \limits_ {x \rightarrow 0^{-}} \frac{[x]+\left[x^2\right]+\left[x^3\right]+\ldots .+\left[x^{2 n+1}\right]+n+1}{1+\left[x^2\right]+|x|+2 x}, n \in N is equal to
View answer and explanation
Correct option: D
0
Option D (0) is correct because as x approaches 0 from the left, the numerator becomes 0 and the denominator approaches 1, resulting in a limit of 0.
396 Easy
396 of 647
Analyzing +4 | -1
lim⁡x→0sin⁡−1x−xx3cos⁡x\lim\limits_{x \to 0} \frac{{{{\sin }^{ - 1}}x - x}}{{{x^3}\cos x}} is equal to
View answer and explanation
Correct option: C
16\frac{1}{6}
Option D (None of these) is correct because the limit does not exist since the function oscillates between 0 and 1 as x approaches 0.
397 Easy
397 of 647
Analyzing +4 | -1
lim⁡t→01−(1+t)tln⁡(1+t)−t\lim \limits_{{t \rightarrow 0}} \frac{1-(1+t)^t}{\ln (1+t)-t} is equal to
View answer and explanation
Correct option: C
2
Option C (2) is correct because after applying L'Hopital's rule or using series expansion, the limit evaluates to 2.
398 Easy
398 of 647
Analyzing +4 | -1
lim⁡x→1[tan⁡(π4+log⁡x)]1log⁡x\lim \limits_{{x \rightarrow 1}}\left[\tan \left(\frac{\pi}{4}+\log x\right)\right]^{\frac{1}{\log x}} is
View answer and explanation
Correct option: D
e2\mathrm{e}^2
Option D (e2e^2) is correct because by applying L'Hôpital's Rule after transforming the limit into an exponential form, the limit evaluates to e2e^2.
399 Easy
399 of 647
Understanding +4 | -1
If f(x)={2x+3,x≤03(x+1),x>0f(x)=\left\{\begin{array}{cc}2 x+3, & x \leq 0 \\ 3(x+1), & x>0\end{array}\right. then the value of lim⁡x→0f(x)\lim \limits_{{x \rightarrow 0}} f(x) is
View answer and explanation
Correct option: A
3
Option A (3) is correct because both the left-hand limit and the right-hand limit as x approaches 0 are equal to 3.
400 Easy
400 of 647
Analyzing +4 | -1
lim⁡x→0atan⁡x−asin⁡xtan⁡x−sin⁡x\lim _{x \rightarrow 0} \frac{a^{\tan x}-a^{\sin x}}{\tan x-\sin x} is equal to (a>0)(a>0)
View answer and explanation
Correct option: A
log⁡ea\log _e a
Option A (log⁡ea\log _e a) is correct because the limit can be evaluated by recognizing it as the derivative of axa^x and applying L'Hopital's rule or a substitution.
401 Easy
401 of 647
Analyzing +4 | -1
If lim⁡x→1ax2+bx+c(x−1)2=2\lim \limits_{{x \rightarrow 1}} \frac{a x^2+b x+c}{(x-1)^2}=2, then lim⁡x→1(x−a)(x−b)(x−c)x+1\lim \limits_{{x \rightarrow 1}} \frac{(x-a)(x-b)(x-c)}{x+1}, is
View answer and explanation
Correct option: B
52\frac{5}{2}
Option B (52\frac{5}{2}) is correct because by using L'Hopital's rule and the given limit, we can find the values of a, b, and c and then evaluate the required limit.
402 Easy
402 of 647
Analyzing +4 | -1
If f(x)=xm,mf(x)=x^m, m being a non-negative integer, then the value of mm for which f′(α+β)=f′(α)+f′(β)f^{\prime}(\alpha+\beta)=f^{\prime}(\alpha)+f^{\prime}(\beta), for all α,β>0\alpha, \beta > 0, is
View answer and explanation
Correct option: B
2
Option B (2) is correct. By substituting the given condition into the derivative of f(x) and comparing terms, we find that m = 2 satisfies the equation.
403 Easy
403 of 647
Application +4 | -1
lim⁡x→π/63sin⁡x−3cos⁡x6x−π\lim \limits_{{x \rightarrow \pi / 6}} \frac{3 \sin x-\sqrt{3} \cos x}{6 x-\pi} equals
View answer and explanation
Correct option: B
13\frac{1}{\sqrt{3}}
Option B (13\frac{1}{\sqrt{3}}) is correct because the limit can be evaluated using L'Hopital's rule or by rewriting the expression using trigonometric identities.
404 Easy
404 of 647
Analyzing +4 | -1
If α\alpha is a repeated root of ax2+bx+c=0a x^2+b x+c=0, then lim⁡x→αsin⁡(ax2+bx+c)(x−α)2\lim \limits_{{x \rightarrow \alpha}} \frac{\sin \left(a x^2+b x+c\right)}{(x-\alpha)^2} is equal to
View answer and explanation
Correct option: B
aa
Option A (aa) is correct because by factoring the quadratic and using the limit lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1, the expression simplifies to aa.
405 Easy
405 of 647
Application +4 | -1
Let [ ] represents the greatest integer function less than or equal to xx. The value of lim⁡x→0([nsin⁡xx]+[ntan⁡xx])\lim \limits_{{x \rightarrow 0}}\left(\left[\frac{n \sin x}{x}\right]+\left[\frac{n \tan x}{x}\right]\right) is
View answer and explanation
Correct option: D
2n−12 n-1
Option D (2n−12 n-1) is correct because as x approaches 0, nsin⁡xx\frac{n \sin x}{x} approaches n from below and ntan⁡xx\frac{n \tan x}{x} approaches n from above, leading to the greatest integer function values of n-1 and n respectively.

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406 Easy
406 of 647
Application +4 | -1
If a1=1{a_1} = 1 and an+1=4+3an3+2an,  n≥1{a_{n + 1}} = \frac{{4 + 3{a_n}}}{{3 + 2{a_n}}},\;n \ge 1 and if lim⁡n→∞an=a\lim\limits_{n \to \infty } {a_n} = a, then the value of aa is
View answer and explanation
Correct option: A
2\sqrt 2
Option A (2\sqrt 2) is correct because the limit of the sequence is found by setting an+1=an=aa_{n+1} = a_n = a and solving the resulting quadratic equation, discarding the negative root since all terms are positive.
407 Easy
407 of 647
Application +4 | -1
lim⁡x→π6(2sin⁡2x+sin⁡x−12sin⁡2x−3sin⁡x+1)=\lim _{x \rightarrow \frac{\pi}{6}}\left(\frac{2 \sin ^2 x+\sin x-1}{2 \sin ^2 x-3 \sin x+1}\right)=
View answer and explanation
Correct option: C
-3
Option C (-3) is correct because by factoring the numerator and denominator and simplifying, the limit evaluates to -3.
408 Easy
408 of 647
Application +4 | -1
If the function
f(x)={2x for ∣x∣≤1x2+ax+b for ∣x∣>1f(x)= \begin{cases}2 x & \text { for }|x| \leq 1 \\ x^2+a x+b & \text { for }|x| > 1\end{cases}
is continuous, then
View answer and explanation
Correct option: C
a=2,b=−1a=2, b=-1
Option C (a=2,b=−1a=2, b=-1) is correct because for the function to be continuous, the left-hand limit and right-hand limit must be equal at the points where the function definition changes, which leads to these values.
409 Easy
409 of 647
Analyzing +4 | -1
lim⁡x→0−sin⁡xx\lim\limits_{x \to {0^ - }} \frac{{\sin x}}{{\sqrt x }} is equal to
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the limit does not exist in the real number system due to the square root of a negative number.
410 Easy
410 of 647
Analyzing +4 | -1
lim⁡x→02∣x∣e∣x∣−∣x∣−∣x∣log⁡e2−1xtan⁡x\lim \limits_{{x \rightarrow 0}} \frac{2^{|x|} e^{|x|}-|x|-|x| \log _e 2-1}{x \tan x} is equal to
View answer and explanation
Correct option: D
12(ln⁡2)2+(ln⁡2)+12\frac{1}{2}(\ln 2)^2+(\ln 2)+\frac{1}{2}
Option D (12(ln⁡2)2+(ln⁡2)+12\frac{1}{2}(\ln 2)^2+(\ln 2)+\frac{1}{2}) is correct because it is obtained by using the Taylor series expansion of a∣x∣a^{|x|} and simplifying the limit.
411 Easy
411 of 647
Analyzing +4 | -1
If lt⁡x→∞(xnex+(cos⁡1x)x2)\operatorname{lt}_{x \rightarrow \infty}\left(\frac{x^n}{e^x}+\left(\cos \frac{1}{x}\right)^{x^2}\right) is finite and is equal to k\mathrm{k}, then which of the following is false ?
View answer and explanation
Correct option: C
k=12k=\frac{1}{2}
Option C (k=12k=\frac{1}{2}) is false because the limit evaluates to e−12e^{-\frac{1}{2}}, thus k=e−12k = e^{-\frac{1}{2}}.
412 Easy
412 of 647
Application +4 | -1
lim⁡x→9−[x2+1]+[x2−1]+{x}\lim \limits_ {x \rightarrow 9^{-}}\left[x^2+1\right]+\left[x^2-1\right]+\{x\} [Here [. ] is GIF and {\{.}\} is Fractional part function]
View answer and explanation
Correct option: C
161
Option C (161) is correct because the limit of the expression as x approaches 9 from the left is 2[81] + 9 - 8 = 160 + 1 = 161.
413 Easy
413 of 647
Analyzing +4 | -1
If lim⁡x→0cos⁡4x+acos⁡2x+bx4\lim \limits_{{x \rightarrow 0}} \frac{\cos 4 x+a \cos 2 x+b}{x^4} is finite, then (a,b)=(a, b)=
View answer and explanation
Correct option: C
(−4,3)(-4,3)
Option C ((-4,3)) is correct because the limit is finite only when a=−4a = -4 and b=3b = 3, which are found by applying L'Hospital's rule and setting up equations.
414 Easy
414 of 647
Analyzing +4 | -1
lim⁡n→∞(121−n3+221−n3+...+n21−n3)\lim\limits_{n \to \infty } \left( {\frac{{{1^2}}}{{1 - {n^3}}} + \frac{{{2^2}}}{{1 - {n^3}}} + ... + \frac{{{n^2}}}{{1 - {n^3}}}} \right) is equal to
View answer and explanation
Correct option: D
−16- \frac{1}{6}
Option D (-1/6) is correct because after summing the series and taking the limit, the expression evaluates to -1/6.
415 Easy
415 of 647
Analyzing +4 | -1
If Sn=(1−tan⁡4π23)(1−tan⁡4π24)…(1−tan⁡4π2n).S_n=\left(1-\tan ^4 \frac{\pi}{2^3}\right)\left(1-\tan ^4 \frac{\pi}{2^4}\right) \ldots\left(1-\tan ^4 \frac{\pi}{2^n}\right) .

The value of lim⁡n→∞Sn\lim \limits_{{n \rightarrow \infty}} S_n, is
View answer and explanation
Correct option: C
π332\frac{\pi^3}{32}
Option C (π332\frac{\pi^3}{32}) is correct because the limit of the given product is evaluated using trigonometric identities and limit properties.
416 Easy
416 of 647
Application +4 | -1
If S1=Σn,S2=Σn2,S3=Σn3S_1=\Sigma n, S_2=\Sigma n^2, S_3=\Sigma n^3, then the value of lim⁡n→∞S1(1+S38)S22\lim \limits_{{n \rightarrow \infty}} \frac{S_1\left(1+\frac{S_3}{8}\right)}{{S_2}^2} is equal to
View answer and explanation
Correct option: D
964\frac{9}{64}
Option D (964\frac{9}{64}) is correct because after substituting the summation formulas and simplifying, the limit evaluates to 964\frac{9}{64}.
417 Easy
417 of 647
Application +4 | -1
lim⁡x→πsin⁡3xsin⁡2x\lim \limits_{{x \rightarrow \pi}} \frac{\sin 3 x}{\sin 2 x} is equal to
View answer and explanation
Correct option: D
−32-\frac{3}{2}
The correct answer is D (−32-\frac{3}{2}) because it is the limit calculated using L'Hopital's rule or trigonometric identities.
418 Easy
418 of 647
Analyzing +4 | -1
lim⁡x→0(1−2x−5x+10xxsin⁡x)=\lim _{x \rightarrow 0}\left(\frac{1-2^x-5^x+10^x}{x \sin x}\right)=
View answer and explanation
Correct option: C
log⁡e5⋅log⁡e2\log _e 5 \cdot \log _e 2
Option C (log⁡e5⋅log⁡e2\log _e 5 \cdot \log _e 2) is correct because the limit can be simplified by factoring and using standard limit formulas.
419 Easy
419 of 647
Analyzing +4 | -1
If lim⁡n→∞(n2−n−1+nα+β)=0\lim _{n \rightarrow \infty}\left(\sqrt{n^2-n-1}+n \alpha+\beta\right)=0, then 8(α+β)8(\alpha+\beta) is equal to
View answer and explanation
Correct option: C
−4-4
Option C (-4) is correct because, after rationalizing and comparing coefficients, 8(α+β)8(\alpha + \beta) equals -4.
420 Easy
420 of 647
Analyzing +4 | -1
Let f(x)=[x]+[−x]f(x)=[x]+[-x], where [x][x] denotes the greatest integer less than or equal to xx. Then, for any integer mm
View answer and explanation
Correct option: B
lim⁡x→mf(x)≠f(m)\lim \limits_{{x \rightarrow m}} f(x) \neq f(m)
Option B (lim⁡x→mf(x)≠f(m)\lim \limits_{{x \rightarrow m}} f(x) \neq f(m)) is correct because the limit as x approaches m is -1, while f(m) = 0.
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