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JEE • MATHEMATICS • LIMITS

Limits practice · Medium

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Showing 31-60 of 267 questions
31 Medium
31 of 267
Understanding +4 | -1
The value of lim⁡x→k−x−[x]\lim\limits _{x \rightarrow k^{-}} x-[x], where kk is an integer, is
View answer and explanation
Correct option: B
1
Option B (1) is correct because as x approaches k from the left, x - [x] approaches 1.
32 Medium
32 of 267
Analyzing +4 | -1
If f(x)={sin⁡x,x≠nπ,n∈Z0,other   wisef\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\sin x,x \ne n\mathrm{ }\pi ,n \in Z}\\ {0,\mathrm{other}\,\,\,\mathrm{wise}} \end{array}} \right. and g(x)={x2+1,x≠0,24,x=05,x=2g\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {{x^2} + 1,x \ne 0,2}\\ {4,x = 0}\\ {5,x = 2} \end{array}} \right., then lim⁡x→0g(f(x))\lim\limits_{x \to 0} g\left( {f\left( x \right)} \right) is
View answer and explanation
Correct option: A
1
Option A (1) is correct because as x approaches 0, sin(x) also approaches 0, and g(0) = 4. However, since x != n*pi, we use the other definition of g(x) which is x2 + 1. Thus, the limit is 02 + 1 = 1.
33 Medium
33 of 267
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}.
34 Medium
34 of 267
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}}.
35 Medium
35 of 267
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.
36 Medium
36 of 267
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.
37 Medium
37 of 267
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.
38 Medium
38 of 267
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.
39 Medium
39 of 267
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.
40 Medium
40 of 267
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.
41 Medium
41 of 267
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.
42 Medium
42 of 267
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.
43 Medium
43 of 267
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.
44 Medium
44 of 267
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.
45 Medium
45 of 267
Analyzing +4 | -1
The value of lim⁡x→0(1+x)1/x−e+12exx2\lim \limits_{{x \rightarrow 0}} \frac{(1+x)^{1 / x}-e+\frac{1}{2} e x}{x^2} is
View answer and explanation
Correct option: A
11e24\frac{11 e}{24}
Option A (11e24\frac{11 e}{24}) is correct because the limit is evaluated using the series expansion of (1+x)1/x(1+x)^{1/x} and simplifying the expression.

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46 Medium
46 of 267
Analyzing +4 | -1
Let L=lim⁡x→0  a−a2−x2−x2/4x4,a>0L = \lim\limits_{x \to 0} \,\,\frac{{a - \sqrt {{a^2} - {x^2}} - {x^2}/4}}{{{x^4}}},a > 0. If LL is finite, then
View answer and explanation
Correct option: A
a=2,L=164a = 2,L = \frac{1}{{64}}
Option A (a=2,L=164a = 2, L = \frac{1}{{64}}) is correct because these values make the limit finite after using the binomial expansion.
47 Medium
47 of 267
Application +4 | -1
The value of lim⁡x→∞(x2−1x2+1)x2\lim \limits_{{x \rightarrow \infty}}\left(\frac{x^2-1}{x^2+1}\right)^{x^2} is
View answer and explanation
Correct option: C
e−2e^{-2}
Option C (e−2e^{-2}) is correct because the limit can be evaluated by rewriting the expression in the form of elim⁡x→∞f(x)e^{\lim_{x \rightarrow \infty} f(x)}.
48 Medium
48 of 267
Application +4 | -1
Let ff is twice differentiable on RR such that f(0)=1,f′(0)=0f(0)=1, f^{\prime}(0)=0 and f′′(0)=−1f^{\prime \prime}(0)=-1, then for a ∈R\in R, lim⁡x→∞(f(ax))x=\lim _{x \rightarrow \infty}\left(f\left(\frac{a}{\sqrt{x}}\right)\right)^x=
View answer and explanation
Correct option: C
ea22e^{\frac{a^2}{2}}
Option C (ea22e^{\frac{a^2}{2}}) is correct because it is the limit calculated using L'Hôpital's rule and the given conditions.
49 Medium
49 of 267
Analyzing +4 | -1
The value of lim⁡x→0(∫0x2sec⁡2tdtxsin⁡x)\lim\limits_{x \to 0} \left( {\frac{{\int \nolimits_0^{{x^2}} {{\sec }^2}t\mathrm{ }dt}}{{x\sin x}}} \right) is
View answer and explanation
Correct option: C
1
Option C (1) is correct because after applying L'Hopital's rule, the limit evaluates to 1.
50 Medium
50 of 267
Analyzing +4 | -1
If α,  β\alpha ,\,\,\beta are the distinct roots of x2+bx+c=0,{x^2} + bx + c = 0, then,
lim⁡x→βe2(x2+bx+c)−1−2(x2+bx+c)(x→β)2\lim\limits_{x \to \beta } \frac{{{e^{2({x^2} + bx + c)}} - 1 - 2({x^2} + bx + c)}}{{{{(x \to \beta )}^2}}} is equal to
View answer and explanation
Correct option: C
2(b2−4c)2\left( {{b^2} - 4c} \right)
Option C (2(b2−4c)2\left( {{{b^2} - 4c}} \right)) is correct because by factoring the quadratic and using L'Hopital's rule, the limit simplifies to 2(β−α)22(\beta - \alpha)^2, which equals 2(b2−4c)2(b^2 - 4c).
51 Medium
51 of 267
Analyzing +4 | -1
lim⁡x→∞[xx2+4−x4+16]=\lim _{x \rightarrow \infty}\left[x \sqrt{x^2+4}-\sqrt{x^4+16}\right]=
View answer and explanation
Correct option: C
2
Option C (2) is correct because after simplifying the expression and taking the limit as x approaches infinity, the result is 2.
52 Medium
52 of 267
Analyzing +4 | -1
The value of lim⁡α→0[cosec−1(sec⁡α)+cot⁡−1(tan⁡α)+cot⁡−1cos⁡(sin⁡−1α)]α\lim\limits_{\alpha \to 0} \frac{{\left[ {\mathrm{cose}{\mathrm{c}^{ - 1}}\left( {\sec \alpha } \right) + {{\cot }^{ - 1}}\left( {\tan \alpha } \right) + {{\cot }^{ - 1}}\cos \left( {{{\sin }^{ - 1}}\alpha } \right)} \right]}}{\alpha } is
View answer and explanation
Correct option: C
−2- 2
Option C (−2-2) is correct because after simplifying the trigonometric inverse functions and applying L'Hopital's rule, the limit evaluates to -2.
53 Medium
53 of 267
Analyzing +4 | -1
If [x][x] denotes the greatest integer less than or equal to xx, then, lim⁡n→∞[12x]+[22x]+[32x]+…..+[n2x]n3\lim \limits_{{n \rightarrow \infty}} \frac{\left[1^2 x\right]+\left[2^2 x\right]+\left[3^2 x\right]+\ldots . .+\left[n^2 x\right]}{n^3} equals
View answer and explanation
Correct option: B
x/3x / 3
Option B (x/3x / 3) is correct because the limit is evaluated using the Squeeze Theorem and the sum of squares formula.
54 Medium
54 of 267
Analyzing +4 | -1
lim⁡x→0tan⁡−1x−sin⁡−1xx3\lim \limits_{{x \rightarrow 0}} \frac{\tan ^{-1} x-\sin ^{-1} x}{x^3} is equal to
View answer and explanation
Correct option: B
−12-\frac{1}{2}
Option B (-12\frac{1}{2}) is correct because the limit is evaluated using L'Hopital's rule or series expansion.
55 Medium
55 of 267
Analyzing +4 | -1
Let f(x)f\left( x \right) be twice differentiable function such that f′′(0)=2f''\left( 0 \right) = 2. Then, lim⁡x→02f(x)−3f(2x)+f(4x)x2\lim\limits_{x \to 0} \frac{{2f\left( x \right) - 3f\left( {2x} \right) + f\left( {4x} \right)}}{{{x^2}}} is
View answer and explanation
Correct option: A
6
Option A (6) is correct because by applying L'Hôpital's Rule twice and using the given condition f′′(0)=2f''(0) = 2, the limit evaluates to 6.
56 Medium
56 of 267
Analyzing +4 | -1
Ltx→1 [[4x2−x−1−1−3x+x21−x3]−1+3(x4−1)x3−x−1]=\underset{x\to 1}{\mathop{Lt}}\,\left[ {{\left[ \frac{4}{{{x}^{2}}-{{x}^{-1}}}-\frac{1-3x+{{x}^{2}}}{1-{{x}^{3}}} \right]}^{-1}}+\frac{3\left( {{x}^{4}}-1 \right)}{{{x}^{3}}-{{x}^{-1}}} \right]=
View answer and explanation
Correct option: B
3
Option B (3) is correct because after simplification and applying the limit, the expression evaluates to 3.
57 Medium
57 of 267
Analyzing +4 | -1
lim⁡x→0((1−cos⁡2(3x)cos⁡3(4x))(sin⁡3(4x)(log⁡e(2x+1))5))\lim \limits_{{x \rightarrow 0}}\left(\left(\frac{1-\cos ^{2}(3 x)}{\cos ^{3}(4 x)}\right)\left(\frac{\sin ^{3}(4 x)}{\left(\log _{e}(2 x+1)\right)^{5}}\right)\right) is equal to
View answer and explanation
Correct option: C
18
Option C (18) is correct because after applying L'Hopital's rule and trigonometric limits, the expression simplifies to 18.
58 Medium
58 of 267
Analyzing +4 | -1
lim⁡x→0(1+x)yx−ex\lim \limits_{{x \rightarrow 0}} \frac{(1+x)^{y x}-e}{x} is
View answer and explanation
Correct option: B
−e2\frac{-e}{2}
Option B (-e/2) is correct because after applying L'Hôpital's rule and simplifying, the limit evaluates to -e/2.
59 Medium
59 of 267
Analyzing +4 | -1
lim⁡x→∞{(xx+1)aa+sin⁡1x}x\lim \limits_{{x \rightarrow \infty}}\left\{\left(\frac{x}{x+1}\right)^a a+\sin \frac{1}{x}\right\}^x is equal to
View answer and explanation
Correct option: B
e1−ae^{1-a}
Option B (e1−ae^{1-a}) is correct because the limit can be evaluated using L'Hopital's rule and properties of exponential functions.
60 Medium
60 of 267
Understanding +4 | -1
If ff is an odd function and if lim⁡x→0f(x)\lim\limits _{x \rightarrow 0} f(x) exists, then lim⁡x→0f(x)\lim\limits _{x \rightarrow 0} f(x) is
View answer and explanation
Correct option: A
0
Option A (0) is correct because for an odd function, if the limit exists as x approaches 0, it must be 0.
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