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JEE • MATHEMATICS • LIMITS

Limits practice · Mixed difficulty

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331 Easy
331 of 930
Application +4 | -1
lim⁡x→0(1+x)12−1(1+x)13−1\lim\limits_{x \to 0} \frac{{{{\left( {1 + x} \right)}^{\frac{1}{2}}} - 1}}{{{{\left( {1 + x} \right)}^{\frac{1}{3}}} - 1}} is equal to
View answer and explanation
Correct option: B
32\frac{3}{2}
Option B (3/2) is correct because the limit can be evaluated by rewriting the expression and applying L'Hopital's rule or using a substitution.
332 Easy
332 of 930
Application +4 | -1
Let f(x)=x2−1,0<x<2f(x)=x^2-1,0 < x < 2 and 2x+3,2≤x<32 x+3,2 \leq x < 3. The quadratic equation whose roots are lim⁡x→2−0f(x)\lim \limits_{{x \rightarrow 2-0}} f(x) and lim⁡x→2+0f(x)\lim \limits_{{x \rightarrow 2+0}} f(x), is
View answer and explanation
Correct option: B
x2−10x+21=0x^2-10 x+21=0
Option B (x2−10x+21=0x^2-10x+21=0) is correct because the roots of the quadratic equation are the left-hand limit and right-hand limit of the function at x=2, which are 3 and 7 respectively.
333 Easy
333 of 930
Analyzing +4 | -1
lim⁡x→a(a+2x)13−(3x)13(3a+x)13−(4x)13(a≠0)\lim _{x \rightarrow a} \frac{(a+2 x)^{\frac{1}{3}}-(3 x)^{\frac{1}{3}}}{(3 a+x)^{\frac{1}{3}}-(4 x)^{\frac{1}{3}}}(a \neq 0) is equal to:
View answer and explanation
Correct option: B
(23)(29)13\left(\frac{2}{3}\right)\left(\frac{2}{9}\right)^{\frac{1}{3}}
Option B ((23)(29)13\left(\frac{2}{3}\right)\left(\frac{2}{9}\right)^{\frac{1}{3}}) is correct because after applying L'Hopital's rule, the limit evaluates to (23)(29)13\left(\frac{2}{3}\right)\left(\frac{2}{9}\right)^{\frac{1}{3}}.
334 Easy
334 of 930
Application +4 | -1
If lim⁡x→0log⁡(3+x)−log⁡(3−x)x=k\lim \limits_{{x \rightarrow 0}} \frac{\log (3+x)-\log (3-x)}{x}=k, then k=k=
View answer and explanation
Correct option: C
23\frac{2}{3}
Option C (23\frac{2}{3}) is correct because applying L'Hopital's rule to the limit results in (1/(3+x) + 1/(3-x)), which evaluates to 2/3 as x approaches 0.
335 Easy
335 of 930
Analyzing +4 | -1
lim⁡x→0ax+a−x−2x2\lim\limits_{x \to 0} \frac{{{a^x} + {a^{ - x}} - 2}}{{{x^2}}} is equal to
View answer and explanation
Correct option: A
(log⁡a)2{\left( {\log a} \right)^2}
Option C (a=1a = 1 and b=−2b = - 2) is correct because by simplifying the expression and taking the limit as x approaches infinity, we find that a = 1 and b = -2.
336 Easy
336 of 930
Application +4 | -1
Let f(x)={1−tan⁡x4x−πx≠π4λx=π4;x∈[0,π2)f(x)=\left\{\begin{array}{cl}\frac{1-\tan x}{4 x-\pi} & x \neq \frac{\pi}{4} \\ \lambda & x=\frac{\pi}{4}\end{array} ; x \in\left[0, \frac{\pi}{2}\right)\right..
If f(x)f(x) is continuous in [0,π2)\left[0, \frac{\pi}{2}\right) then λ\lambda is equal to :
View answer and explanation
Correct option: C
−12-\frac{1}{2}
Option C (-12\frac{1}{2}) is correct because it is the limit of the function as x approaches π4\frac{\pi}{4}, ensuring continuity.
337 Easy
337 of 930
Application +4 | -1
If a,ba, b are chosen from {1,2,3,4,5\{1,2,3,4,5, 6,7}6,7\} randomly with replacement. The probability that lim⁡x→0(ax+bx2)2/x=7\lim \limits_{{x \rightarrow 0}}\left(\frac{a^x+b^x}{2}\right)^{2 / x}=7 is
View answer and explanation
Correct option: B
2/492 / 49
Option B (2/49) is correct because the probability is calculated by finding the pairs (a, b) such that ab = 7 and dividing by the total number of possible pairs.
338 Easy
338 of 930
Analyzing +4 | -1
P(x)P(x) is a polynomial such that P(x)+P(2x)=5x2−18P(x)+P(2 x)=5 x^2-18. Then
lim⁡x→3(P(x)x−3)=\lim \limits_{{x \rightarrow 3}}\left(\frac{P(x)}{x-3}\right)=
View answer and explanation
Correct option: A
6
Option A (6) is correct because by finding the polynomial P(x) and applying L'Hopital's rule, we can evaluate the limit.
339 Easy
339 of 930
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: A
2
Option A (2) is correct because the limit can be evaluated by using trigonometric identities and standard limits.
340 Easy
340 of 930
Analyzing +4 | -1
The value of lim⁡n→∞∑k=1nn−Kn2cos⁡(4Kn)\lim \limits_{{n \rightarrow \infty}} \sum\limits_{{k=1}}^n \frac{n-K}{n^2} \cos \left(\frac{4 K}{n}\right) equals
View answer and explanation
Correct option: D
116(1−cos⁡4)\frac{1}{16}(1-\cos 4)
The correct answer is D (116(1−cos⁡4)\frac{{1}}{{16}}(1-\cos 4)) because the limit of the Riemann sum can be expressed as a definite integral, which can then be evaluated using integration by parts.
341 Medium
341 of 930
Analyzing +4 | -1
lim⁡x→−1(1+x)(1−x2)(1+x3)(1−x4)…(1−x4n)[(1+x)(1−x2)(1+x3)(1−x4)…(1−x2n)]2\lim\limits_{x \to - 1} \frac{{\left( {1 + x} \right)\left( {1 - {x^2}} \right)\left( {1 + {x^3}} \right)\left( {1 - {x^4}} \right) \ldots \left( {1 - {x^{4n}}} \right)}}{{{{\left[ {\left( {1 + x} \right)\left( {1 - {x^2}} \right)\left( {1 + {x^3}} \right)\left( {1 - {x^4}} \right) \ldots \left( {1 - {x^{2n}}} \right)} \right]}^2}}} is equal to
View answer and explanation
Correct option: A
4nC2n^{4n}{C_{2n}}
Option A (4nC2n^{4n}{C_{2n}}) is correct because after simplifying the expression and applying L'Hopital's rule repeatedly, the limit evaluates to the binomial coefficient 4nC2n^{4n}{C_{2n}}.
342 Easy
342 of 930
Application +4 | -1
For x∈R,Lim⁡x→∞(x−3x+2)x=x \in R, \operatorname{Lim}_{x \rightarrow \infty}\left(\frac{x-3}{x+2}\right)^x=
View answer and explanation
Correct option: C
e−5e^{-5}
Option C is correct because it is the result of applying the limit definition of e and simplifying the expression.
343 Medium
343 of 930
Application +4 | -1
Let α\alpha and β\beta be the distinct roots of ax2+bx+c=0a{x^2} + bx + c = 0, then lim⁡x→α1−cos⁡(ax2+bx+c)(x−α)2\lim\limits_{x \to \alpha } \frac{{1 - \cos \left( {a{x^2} + bx + c} \right)}}{{{{\left( {x - \alpha } \right)}^2}}} is equal to
View answer and explanation
Correct option: D
a22(α−β)2\frac{{{a^2}}}{2}{\left( {\alpha - \beta } \right)^2}
Option D (a22(α−β)2\frac{{{a^2}}}{2}{\left( {\alpha - \beta } \right)^2}) is correct because by using trigonometric identities and the fact that alpha and beta are roots, the limit evaluates to the given expression.
344 Medium
344 of 930
Application +4 | -1
lim⁡x→01+x+x2−exx2\lim \limits_{{x \rightarrow 0}} \frac{1+x+x^2-e^x}{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 the series expansion of e^x, the limit evaluates to 1/2.
345 Easy
345 of 930
Application +4 | -1
lim⁡x→0(3x2+27x2+2)1/x2\lim\limits_{x \to 0} {\left( {\frac{{3{x^2} + 2}}{{7{x^2} + 2}}} \right)^{1/{x^2}}} is equal to
View answer and explanation
Correct option: C
1e2\frac{1}{{{e^2}}}
Option C (1e2\frac{1}{{{e^2}}}) is correct because it is the result of evaluating the limit using the standard limit formula for 1∞1^\infty form.

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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
346 Medium
346 of 930
Application +4 | -1
lim⁡x→∞(1−4x−1)3x−1\lim\limits_{x \to \infty } {\left( {1 - \frac{4}{{x - 1}}} \right)^{3x - 1}} is equal to
View answer and explanation
Correct option: B
e−12{e^{ - 12}}
Option B (e−12{e^{ - 12}}) is correct because the limit of the given expression as x approaches infinity equals e−12{e^{ - 12}} after applying limit rules.
347 Easy
347 of 930
Application +4 | -1
lim⁡x→01+x+x2−exx2\lim \limits_{{x \rightarrow 0}} \frac{1+x+x^2-e^x}{x^2} is equal to
View answer and explanation
Correct option: C
1/21/2
Option C, 1/21/2, is correct because it is the result of expanding exe^x as a Taylor series and then taking the limit.
348 Medium
348 of 930
Application +4 | -1
If f′(a)=2f'\left( a \right) = 2 and f(a)=4f\left( a \right) = 4, then lim⁡x→axf(a)−af(x)x−a\lim\limits_{x \to a} \frac{{xf\left( a \right) - af\left( x \right)}}{{x - a}} equals
View answer and explanation
Correct option: B
4−2a4 - 2a
Option B (4 - 2a) is correct because it is the result of applying L'Hopital's rule or algebraic manipulation to the limit expression.
349 Easy
349 of 930
Application +4 | -1
If f:R→Rf:R \to R is defined by f(x)=[x−3]+[x−4]f\left( x \right) = \left[ {x - 3} \right] + \left[ {x - 4} \right] for x∈Rx \in R, then lim⁡x→3−f(x)\lim\limits_{x \to {3^ - }} f\left( x \right) is equal to
View answer and explanation
Correct option: C
0
Option C (0) is correct because the limit evaluates to 0 after substituting x=3−hx = 3 - h and simplifying the greatest integer function and absolute value.
350 Easy
350 of 930
Application +4 | -1
lim⁡x→01+x−exx2\lim \limits_{{x \rightarrow 0}} \frac{1+x-e^x}{x^2} is
View answer and explanation
Correct option: B
−12-\frac{1}{2}
The correct answer is B (-12\frac{1}{2}) because applying L'Hôpital's rule twice to the limit yields the value -12\frac{1}{2}.
351 Easy
351 of 930
Application +4 | -1
lim⁡x→−2sin⁡−1(x+2)x2+2x\lim \limits_{{x \rightarrow-2}} \frac{\sin ^{-1}(x+2)}{x^2+2 x} is equal to
View answer and explanation
Correct option: A
−12-\frac{1}{2}
Option A (-1/2) is correct because after substitution and applying the limit lim⁡t→0sin⁡−1tt=1\lim_{t \rightarrow 0} \frac{\sin^{-1} t}{t} = 1, the limit evaluates to -1/2.
352 Easy
352 of 930
Analyzing +4 | -1
The value of lim⁡x→0ex+log⁡(1+x)−(1−x)−2x2\lim\limits_{x \to 0} \frac{{{e^x} + \log \left( {1 + x} \right) - {{\left( {1 - x} \right)}^{ - 2}}}}{{{x^2}}} is equal to
View answer and explanation
Correct option: B
−3- 3
Option B (−3-3) is correct because it is obtained by applying L'Hopital's rule twice and then evaluating the limit.
353 Easy
353 of 930
Analyzing +4 | -1
Let lim⁡x→1xa−ax+a−1(x−1)2=f(a)\lim \limits_{{x \rightarrow 1}} \frac{x^a-a x+a-1}{(x-1)^2}=f(a). The value of f(101)f(101) equals
View answer and explanation
Correct option: A
5050
Option A (5050) is correct because the limit is evaluated using L'Hopital's rule or Taylor series expansion.
354 Easy
354 of 930
Analyzing +4 | -1
lim⁡x→a−{∣x∣3a−[xa]3},a>0\lim \limits_ {x \rightarrow a^{-}}\left\{\frac{|x|^3}{a}-\left[\frac{x}{a}\right]^3\right\}, a > 0, where [x][x] denotes the greatest integer less than or equal to xx, is
View answer and explanation
Correct option: C
a2a^2
Option C (a2a^2) is correct because as x approaches a from the left, the greatest integer function evaluates to 0, simplifying the limit.
355 Medium
355 of 930
Application +4 | -1
Let x1=1 and xn+1=4+3xn3+2xn for n≥1x_1=1 \quad \text { and } \quad x_{n+1}=\frac{4+3 x_n}{3+2 x_n} \text { for } n \geq 1 If lim⁡n→∞xn\lim _{n \rightarrow \infty} x_n exists finitely, then the limit is equal to
View answer and explanation
Correct option: A
2\sqrt{2}
Option A (2\sqrt{2}) is correct because the limit, if it exists, must satisfy the recursive relation, leading to the equation l=4+3l3+2ll = \frac{4+3l}{3+2l}, which solves to l=2l = \sqrt{2}.
356 Easy
356 of 930
Analyzing +4 | -1
The value of lim⁡x→∞5x+1−7x+15x−7x\lim\limits_{x \to \infty } \frac{{{5^{x + 1}} - {7^{x + 1}}}}{{{5^x} - {7^x}}}, is
View answer and explanation
Correct option: C
7
Option C (7) is correct because, as x approaches infinity, the term (5/7)^x approaches 0, simplifying the expression to 7.
357 Easy
357 of 930
Analyzing +4 | -1
lim⁡x→∞x+sin⁡xx−cos⁡x\lim _{x \rightarrow \infty} \sqrt{\frac{x+\sin x}{x-\cos x}} is equal to
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⁡xx\frac{{\cos x}}{{x}} approach 0, leaving 11=1\sqrt{{\frac{{1}}{{1}}}} = 1.
358 Easy
358 of 930
Analyzing +4 | -1
The value of lim⁡x→0(sin⁡xx)1x2\lim \limits_{{x \rightarrow 0}}\left(\frac{\sin x}{x}\right)^{\frac{1}{x^2}}, is
View answer and explanation
Correct option: C
e−1/6e^{-1 / 6}
Option C (e−1/6e^{-1 / 6}) is correct because using the series expansion of sin(x) and properties of limits, the expression simplifies to e−1/6e^{-1/6}.
359 Easy
359 of 930
Analyzing +4 | -1
lim⁡x→0log⁡xn−[x][x],n∈N,([x]\lim _{x \rightarrow 0} \frac{\log x^n-[x]}{[x]}, n \in N,([x] denotes greatest integer less than or equal to xx )
View answer and explanation
Correct option: D
does not exist
Option D (does not exist) is correct because the limit from the left and right sides are not equal, thus the limit does not exist.
360 Easy
360 of 930
Analyzing +4 | -1
The value of lim⁡x→0ex−esin⁡x2(x−sin⁡x)\lim \limits_{{x \rightarrow 0}} \frac{e^x-e^{\sin x}}{2(x-\sin x)}, is
View answer and explanation
Correct option: B
12\frac{1}{2}
Option B (1/2) is correct because the limit is evaluated using standard limit results and L'Hopital's rule, resulting in 1/2.
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