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JEE • MATHEMATICS • LIMITS

Limits practice · Medium

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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 241-267 of 267 questions
241 Medium
241 of 267
Analyzing +4 | -1
If a,ba, b and cc are real numbers, then the value of lim⁡t→0ln⁡(1t∫0t(1+asin⁡bx)c/xdx)\lim \limits_{{t \rightarrow 0}} \ln \left(\frac{1}{t} \int_0^t(1+a \sin b x)^{c / x} d x\right) equals
View answer and explanation
Correct option: A
abca b c
Option A (abca b c) is correct because the limit is evaluated using L'Hopital's rule and the exponential limit definition.
242 Medium
242 of 267
Analyzing +4 | -1
ff is a real-valued differentiable and " aa " is a real constant. Further f′f^{\prime} is continuous. If

F(x)=∣f(x+a)f(x+2a)f(x+3a)f(a)f(2a)f(3a)f′(a)f′(2a)f′(3a)∣F(x)=\left|\begin{array}{ccc} f(x+a) & f(x+2 a) & f(x+3 a) \\ f(a) & f(2 a) & f(3 a) \\ f^{\prime}(a) & f^{\prime}(2 a) & f^{\prime}(3 a) \end{array}\right|

then lim⁡x→0F(x)/x\lim \limits_{{x \rightarrow 0}} F(x) / x is
View answer and explanation
Correct option: D
0
Option D (0) is correct because the limit evaluates to 0 after applying L'Hôpital's rule and using the properties of determinants.
243 Medium
243 of 267
Analyzing +4 | -1
lim⁡x→−π∣x+π∣sin⁡x\lim \limits_{{x \rightarrow-\pi}} \frac{|x+\pi|}{\sin x} is equal to
View answer and explanation
Correct option: D
Does not exist
Option D (Does not exist) is correct because the left-hand limit and right-hand limit are different, so the limit does not exist.
244 Medium
244 of 267
Application +4 | -1
The value of lim⁡x→72−x−3x2−49\lim\limits_{x \to 7} \frac{{2 - \sqrt {x - 3} }}{{{x^2} - 49}} is
View answer and explanation
Correct option: D
−1/56- 1/56
Option D (−156\frac{-1}{56}) is correct because it is the result of applying L'Hopital's rule to the limit after recognizing it as an indeterminate form.
245 Medium
245 of 267
Analyzing +4 | -1
The value of lim⁡x→1{x3+2x2+x+1x2+2x+3}1−cos⁡(x−1)(x−1)2\lim\limits_{x \to 1} {\left\{ {\frac{{{x^3} + 2{x^2} + x + 1}}{{{x^2} + 2x + 3}}} \right\}^{\frac{{1 - \cos \left( {x - 1} \right)}}{{{{\left( {x - 1} \right)}^2}}}}}, is
View answer and explanation
Correct option: D
None of these
The correct answer is D (None of these) because the limit evaluates to 56\sqrt{\frac{5}{6}} after simplification and using trigonometric limit properties.
246 Medium
246 of 267
Application +4 | -1
The value of lim⁡x→0(cos⁡x)cot⁡2x\lim\limits_{x \to 0} {\left( {\cos x} \right)^{{{\cot }^2}x}} is
View answer and explanation
Correct option: B
e−12{e^{\frac{{ - 1}}{2}}}
Option B (e−12{e^{\frac{{ - 1}}{2}}}) is correct because it is the result of evaluating the limit using L'Hopital's rule or standard limit approximations.
247 Medium
247 of 267
Analyzing +4 | -1
lim⁡x→0(1−cos⁡2x)22xtan⁡x−xtan⁡2x\lim _{x \rightarrow 0} \frac{(1-\cos 2 x)^2}{2 x \tan x-x \tan 2 x} is :
View answer and explanation
Correct option: C
-2
Option C (-2) is correct because after using trigonometric expansions and simplifying, the limit evaluates to -2.
248 Medium
248 of 267
Analyzing +4 | -1
lim⁡n→∞∏r=3nr3−8r3+8\lim \limits_{{n \rightarrow \infty}} \prod\limits_{{r=3}}^n \frac{r^3-8}{r^3+8} is equal to
View answer and explanation
Correct option: A
27\frac{2}{7}
Option A (27\frac{2}{7}) is correct because after factoring and simplifying the product, the limit equals 27\frac{2}{7}.
249 Medium
249 of 267
Application +4 | -1
lim⁡x→0(1x+2x+3x+⋯+nxn)a/x\lim \limits_{{x \rightarrow 0}}\left(\frac{1^x+2^x+3^x+\cdots+n^x}{n}\right)^{a / x} is equal to
View answer and explanation
Correct option: C
(n‾)a/n(\underline{n})^{a / n}
Option C ((n‾\underline{n})a/n^{a / n}) is correct because the limit is evaluated using the exponential limit formula and properties of logarithms, resulting in (n‾\underline{n})a/n^{a / n}.
250 Medium
250 of 267
Application +4 | -1
lim⁡x→π4∫2sec⁡2xf(t)dtx2−π216\lim\limits_{x \to \frac{\pi }{4}} \frac{{\int \nolimits_2^{{{\sec }^2}x} f\left( t \right)dt}}{{{x^2} - \frac{{{\pi ^2}}}{{16}}}} equals
View answer and explanation
Correct option: A
8πf(2)\frac{8}{\pi }f\left( 2 \right)
Option A (8πf(2)\frac{8}{\pi}f(2)) is correct because it results from applying L'Hopital's rule and evaluating the derivative of sec⁡2x\sec^2 x at x=π4x = \frac{\pi}{4}.
251 Medium
251 of 267
Analyzing +4 | -1
If lim⁡x→0sin⁡(αx−ln⁡(1+βx))e3x−3x−1=1\lim \limits_{{x \rightarrow 0}} \frac{\sin (\alpha x-\ln (1+\beta x))}{e^{3 x}-3 x-1}=1, then α+2β\alpha+2 \beta is equal to
View answer and explanation
Correct option: C
9
Option C (9) is correct because by using L'Hopital's rule and series expansion, we find that α=β\alpha = \beta and β2=9\beta^2 = 9, leading to α+2β=9\alpha + 2\beta = 9.
252 Medium
252 of 267
Understanding +4 | -1
The equivalent definition of the function given by f(x)={2x,x≥00,x<0f(x)=\left\{\begin{array}{ll}2 x, & x \geq 0 \\ 0, & x<0\end{array}\right. is
View answer and explanation
Correct option: C
f(x)=∣x∣+xf(x)=|x|+x
Option C (f(x)=∣x∣+xf(x)=|x|+x) is correct because when x≥0x \geq 0, ∣x∣=x|x| = x, so f(x)=x+x=2xf(x) = x + x = 2x, and when x<0x < 0, ∣x∣=−x|x| = -x, so f(x)=−x+x=0f(x) = -x + x = 0, matching the given definition.
253 Medium
253 of 267
Application +4 | -1
The value of lim⁡x→2x4−4x2+32x−8\lim \limits_{x \rightarrow \sqrt{2}} \frac{x^4-4}{x^2+3 \sqrt{2} x-8} is
View answer and explanation
Correct option: C
85\frac{8}{5}
Option C (8/5) is correct because after factoring and simplifying the expression, the limit as x approaches sqrt(2) is 8/5.
254 Medium
254 of 267
Analyzing +4 | -1
lim⁡x→0xcos⁡x−log⁡(1+x)x2\lim\limits_{x \to 0} \frac{{x\cos x - \log \left( {1 + x} \right)}}{{{x^2}}} equals
View answer and explanation
Correct option: A
12\frac{1}{2}
Option A (12\frac{1}{2}) is correct because using Taylor series expansions for cos(x) and log(1+x) allows us to evaluate the limit.
255 Medium
255 of 267
Application +4 | -1
lim⁡x→0(xx)\lim \limits_{{x \rightarrow 0}}\left(x^x\right) is
View answer and explanation
Correct option: B
1
Option B (1) is correct because by applying L'Hopital's rule to the logarithmic form of the limit, we find the limit equals 1.

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256 Medium
256 of 267
Analyzing +4 | -1
The value of lim⁡x→01x3∫0xtln⁡(1+t)t4+4dt\lim _{x \rightarrow 0} \frac{1}{x^3} \int_0^x \frac{t \ln (1+t)}{t^4+4} d t
View answer and explanation
Correct option: B
112\frac{1}{12}
Option B (112\frac{1}{12}) is correct because applying L'Hopital's rule simplifies the limit to 112\frac{1}{12}.
257 Medium
257 of 267
Application +4 | -1
Suppose f(x)={a+bx,x<14,x=1b−ax,x>1{f}({x})= \begin{cases}a+b x, & x < 1 \\ 4, & x=1 \\ b-a x, & x>1\end{cases} and if lim⁡x→1f(x)=f(1)\lim \limits_{{x \rightarrow 1}} f(x)=f(1) then the value of a+ba+b is
View answer and explanation
Correct option: A
4
Option A is correct because the limit condition implies that a + b = 4.
258 Medium
258 of 267
Analyzing +4 | -1
The value of lim⁡x→π/2{1sec⁡2x+2sec⁡2x+3sec⁡2x+….+nsec⁡2x}cos⁡2x,\lim \limits_{{x \rightarrow \pi / 2}}\left\{1^{\sec ^2 x}+2^{\sec ^2 x}+3^{\sec ^2 x}+\ldots .+n^{\sec ^2 x}\right\}^{\cos ^2 x}, is
View answer and explanation
Correct option: B
nn
Option B (nn) is correct because, as xx approaches π/2\pi/2, the expression simplifies to nn after applying limit properties and L'Hôpital's rule implicitly.
259 Medium
259 of 267
Analyzing +4 | -1
lim⁡λ→0(∫01(1+x)λdx)1/λ\lim _{\lambda \rightarrow 0}\left(\int_0^1(1+x)^\lambda d x\right)^{1 / \lambda} is equal to
View answer and explanation
Correct option: B
4e\frac{4}{e}
Option B (4/e) is correct because, after evaluating the integral and applying L'Hopital's rule, the limit is found to be 4/e.
260 Medium
260 of 267
Understanding +4 | -1
The right hand limit of the function
f(x)={∣x−4∣x−4,x≠40,x=4f(x) = \left\{ {\begin{array}{llllllllllllllllllll} {\frac{{|x - 4|}}{{x - 4}},}&{x \ne 4}\\ {0,}&{x = 4} \end{array}} \right. at x=4x=4, is
View answer and explanation
Correct option: A
1
Option A (1) is correct because the right-hand limit is calculated by considering values of x approaching 4 from the right, where the absolute value simplifies to the expression itself.
261 Medium
261 of 267
Analyzing +4 | -1
lim⁡x→2(∑n=19xn(n+1)x2+2(2n+1)x+4)\lim _{x \rightarrow 2}\left(\sum_{n=1}^9 \frac{x}{n(n+1) x^2+2(2 n+1) x+4}\right) is equal to
View answer and explanation
Correct option: A
944\frac{9}{44}
Option A (944\frac{9}{44}) is correct because it is the result of evaluating the limit of the summation by using partial fraction decomposition.
262 Medium
262 of 267
Understanding +4 | -1
Let f(x)=cos⁡[x]f(x)=\cos [x] where [x][x] is the integral part of xx. Then
View answer and explanation
Correct option: B
lim⁡x→0+0f(x)=1\lim _{x \rightarrow 0+0} f(x)=1
Option B is correct because as x approaches 0 from the right, [x] = 0, so cos([x]) = cos(0) = 1.
263 Medium
263 of 267
Analyzing +4 | -1
lim⁡x→−1π−cos⁡−1xx+1\lim\limits_{x \to - 1} \frac{{\sqrt \pi - \sqrt {{{\cos }^{ - 1}}x} }}{{\sqrt {x + 1} }} is given by
View answer and explanation
Correct option: B
12π\frac{1}{{\sqrt {2\pi } }}
Option B (12π\frac{1}{{\sqrt {2\pi } }}) is correct because by substituting cos⁡−1x=y{\cos ^{ - 1}}x = y and applying trigonometric identities and limit properties, the expression simplifies to 12π\frac{1}{{\sqrt {2\pi } }}.
264 Medium
264 of 267
Analyzing +4 | -1
If f(x)=xe[x]+∣x∣−2[x]+∣x∣f(x)=x \frac{e^{[x]+|x|}-2}{[x]+|x|}, 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, so the limit does not exist.
265 Medium
265 of 267
Application +4 | -1
lim⁡x→0log⁡(1+x3)sin⁡3x\lim \limits_{{x \rightarrow 0}} \frac{\log \left(1+x^3\right)}{\sin ^3 x} is
View answer and explanation
Correct option: D
1
Option D (1) is correct because by using the small angle approximations or L'Hopital's rule, the limit simplifies to 1.
266 Medium
266 of 267
Analyzing +4 | -1
lim⁡x→∞log⁡[x]x\lim \limits_{{x \rightarrow \infty}} \frac{\log [x]}{x}, where [x][x] denotes the greatest integer less than or equal to xx, is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the limit is evaluated using L'Hospital's Rule after applying the greatest integer function.
267 Medium
267 of 267
Analyzing +4 | -1
If f(x)f(x) is a polynomial satisfying f(x)⋅f(1x)=f(x)+f(1x)f(x)\cdot f\left(\frac{1}{x}\right)=f(x)+f\left(\frac{1}{x}\right) and f(2)>1f(2) > 1, then lim⁡x→1f(x)\lim \limits_{{x \rightarrow 1}} f(x) is
View answer and explanation
Correct option: A
2
Option A (2) is correct because the polynomial satisfying the given condition is f(x)=xn+1f(x) = x^n + 1, and thus lim⁡x→1f(x)=2\lim_{x \rightarrow 1} f(x) = 2.
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