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JEE • MATHEMATICS • LIMITS

Limits practice · Easy

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211 Easy
211 of 647
Application +4 | -1
lim⁡x→π2(sec⁡x−tan⁡x)=\lim\limits_{x \to \frac{\pi }{2}} \left( {\sec x - \tan x} \right) =
View answer and explanation
Correct option: B
0
Option B (0) is correct because the limit of (sec x - tan x) as x approaches pi/2 is 0 after rewriting in terms of sine and cosine and simplifying.
212 Easy
212 of 647
Analyzing +4 | -1
lim⁡x→+∞20+2x+3x1/32+4x−3+(8x−4)1/3=\lim _{x \rightarrow+\infty} \frac{20+2 \sqrt{x}+3 x^{1 / 3}}{2+\sqrt{4 x-3}+(8 x-4)^{1 / 3}}=
View answer and explanation
Correct option: A
1
Option A (1) is correct because after dividing both numerator and denominator by √x, the limit as x approaches infinity is 1.
213 Easy
213 of 647
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.
214 Easy
214 of 647
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.
215 Easy
215 of 647
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}}.
216 Easy
216 of 647
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.
217 Easy
217 of 647
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.
218 Easy
218 of 647
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.
219 Easy
219 of 647
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.
220 Easy
220 of 647
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.
221 Easy
221 of 647
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.
222 Easy
222 of 647
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.
223 Easy
223 of 647
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.
224 Easy
224 of 647
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.
225 Easy
225 of 647
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.

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226 Easy
226 of 647
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.
227 Easy
227 of 647
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}.
228 Easy
228 of 647
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.
229 Easy
229 of 647
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.
230 Easy
230 of 647
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.
231 Easy
231 of 647
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.
232 Easy
232 of 647
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.
233 Easy
233 of 647
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.
234 Easy
234 of 647
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}.
235 Easy
235 of 647
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.
236 Easy
236 of 647
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.
237 Easy
237 of 647
Understanding +4 | -1
The value of lim⁡x→−∞[tan⁡−1x]\lim \limits_{{x \rightarrow-\infty}}\left[\tan ^{-1} x\right] is
View answer and explanation
Correct option: A
−2-2
Option A (-2) is correct because as x approaches negative infinity, tan⁡−1x\tan^{-1}x approaches -π2\frac{{\pi}}{{2}}, and the greatest integer function of a value approaching -π2\frac{{\pi}}{{2}} is -2, since -π2≈−1.57\frac{{\pi}}{{2}} \approx -1.57.
238 Easy
238 of 647
Application +4 | -1
lim⁡x→+∞[2x−1tan⁡(32x)]=\lim _{x \rightarrow+\infty}\left[2^{x-1} \tan \left(\frac{3}{2^x}\right)\right]=
View answer and explanation
Correct option: B
32\frac{3}{2}
Option B (32\frac{3}{2}) is correct because, after using the limit property lim⁡x→0tan⁡xx=1\lim_{x \to 0} \frac{\tan x}{x} = 1, the limit of the expression as x approaches infinity is 3/2.
239 Easy
239 of 647
Analyzing +4 | -1
The value of lim⁡n→∞xnxn+1\lim\limits_{n \to \infty } \frac{{{x^n}}}{{{x^n} + 1}}, where x<−1x < - 1 is
View answer and explanation
Correct option: C
1
Option C (1) is correct because, by dividing both the numerator and denominator by x^n, the limit simplifies to 1 as n approaches infinity, given that x < -1.
240 Easy
240 of 647
Application +4 | -1
The magnitude of the two forces forming a couple is 36 N36 {~N} each and the arm of the couple is 4 m4 {~m}. The magnitude of each force of an equivalent couple whose arm 9 m9 {~m}, is (in Newtons)
View answer and explanation
Correct option: C
16
Option C (16) is correct because the moment of a couple remains constant, so the force must adjust inversely with the arm length.
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