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JEE • MATHEMATICS • LIMITS

Limits practice · Medium

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121 Medium
121 of 267
Analyzing +4 | -1
lim⁡x→∞log⁡xn−[x][x]\lim \limits_{{x \rightarrow \infty}} \frac{\log x^n-[x]}{[x]}, where n∈Nn \in N and [.] denotes the greatest integer function, is
View answer and explanation
Correct option: B
−1-1
Option B (-1) is correct because the limit of the given expression as x approaches infinity simplifies to -1.
122 Medium
122 of 267
Application +4 | -1
lim⁡x→∞(x+6x+1)x+4=\lim _{x \rightarrow \infty}\left(\frac{x+6}{x+1}\right)^{x+4}=
View answer and explanation
Correct option: B
e5e^5
Option B (e5e^5) is correct because the limit can be evaluated using the standard limit formula lim⁡x→∞(1+ax)x=ea\lim_{x \rightarrow \infty} (1 + \frac{a}{x})^x = e^a.
123 Medium
123 of 267
Analyzing +4 | -1
The value of lim⁡x→alog⁡(x−a)log⁡(ex−ea)\lim\limits_{x \to a} \frac{{\log \left( {x - a} \right)}}{{\log \left( {{e^x} - {e^a}} \right)}}, is
View answer and explanation
Correct option: A
1
Option A (1) is correct because applying L'Hopital's rule or using the definition of the derivative, the limit evaluates to 1.
124 Medium
124 of 267
Application +4 | -1
lim⁡x→∞(x+6x+1)x+4\lim \limits_{{x \rightarrow \infty}}\left(\frac{x+6}{x+1}\right)^{x+4} is equal to
View answer and explanation
Correct option: B
e5e^5
Option B (e5e^5) is correct because by using the limit property lim⁡x→∞(1+ax)x=ea\lim_{x \to \infty} (1 + \frac{{a}}{{x}})^x = e^a, the limit evaluates to e5e^5.
125 Medium
125 of 267
Analyzing +4 | -1
lim⁡x→a{[(a1/2+x1/2a1/4−x1/4)−1−2(ax)1/4x3/4−a1/4x1/2+a1/2x1/4−a3/4]−1−2log⁡4a}8\lim \limits_{{x \rightarrow a}}\left\{\left[\left(\frac{a^{1 / 2}+x^{1 / 2}}{a^{1 / 4}-x^{1 / 4}}\right)^{-1}\right.\right.\left.-\frac{2(a x)^{1 / 4}}{x^{3 / 4}-a^{1 / 4} x^{1 / 2}+a^{1 / 2} x^{1 / 4}-a^{3 / 4}}\right]^{-1}\left.-\sqrt{2}^{\log_4a}\right\}^8 is equal to
View answer and explanation
Correct option: C
a2a^2
Option C (a2) is correct because after simplifying the expression and substituting x=a, the limit evaluates to a2.
126 Medium
126 of 267
Analyzing +4 | -1
lim⁡x→1cos⁡−1(1−x1−x)\lim\limits_{x \to 1} {\cos ^{ - 1}}\left( {\frac{{1 - \sqrt x }}{{1 - x}}} \right) is equal to
View answer and explanation
Correct option: A
π3\frac{\pi }{3}
Option A (π3\frac{\pi}{3}) is correct because the limit can be evaluated by rationalizing the expression and then applying the inverse cosine function.
127 Medium
127 of 267
Analyzing +4 | -1
The value of
lim⁡n→∞1+2−3+4+5−6+…+(3n−2)+(3n−1)−3n2n4+4n+3−n4+5n+4\lim \limits_{{n \rightarrow \infty}} \frac{1+2-3+4+5-6+\ldots+(3 n-2)+(3 n-1)-3 n}{\sqrt{2 n^{4}+4 n+3}-\sqrt{n^{4}+5 n+4}}
is :
View answer and explanation
Correct option: B
32(2+1)\frac{3}{2}(\sqrt{2}+1)
Option B (32(2+1)\frac{3}{2}(\sqrt{2}+1)) is correct because we simplify the numerator and denominator separately, then rationalize the denominator and take the limit as n approaches infinity.
128 Medium
128 of 267
Analyzing +4 | -1
If lim⁡x→1ax2+bx+c(x−1)2=2,  \lim\limits_{x \to 1} \frac{{a{x^2} + bx + c}}{{{{\left( {x - 1} \right)}^2}}} = 2,\; then (a, b, c) is
View answer and explanation
Correct option: A
(2,−4,2)\left( {2, - 4,2} \right)
Option A ((2, -4, 2)) is correct because for the limit to exist, the numerator must be equivalent to 2(x-1)2, leading to a=2, b=-4, and c=2.
129 Medium
129 of 267
Analyzing +4 | -1
lim⁡x→0(sin⁡x)tan⁡x\lim \limits_{{x \rightarrow 0}}(\sin x)^{\tan x} is
View answer and explanation
Correct option: A
1
Option A (1) is correct because after applying L'Hôpital's rule to the logarithm of the function, the limit evaluates to 0, and thus the original limit is e0 = 1.
130 Medium
130 of 267
Analyzing +4 | -1
lim⁡x→2x−2+x−2x2−4\lim \limits_{{x \rightarrow 2}} \frac{\sqrt{x-2}+\sqrt{x}-\sqrt{2}}{\sqrt{x^2-4}} is equal to
View answer and explanation
Correct option: A
12\frac{1}{2}
Option A (1/2) is correct because after rationalizing and simplifying, the limit evaluates to 1/2.
131 Medium
131 of 267
Analyzing +4 | -1
x→0lim⁡(sin⁡2(πcos⁡4x)x4)=_{x \to 0}^{\lim }\left( {\frac{{{{\sin }^2}\left( {\pi {{\cos }^4}x} \right)}}{{{x^4}}}} \right) =
View answer and explanation
Correct option: C
4π24{\pi ^2}
Option C (4π24{\pi ^2}) is correct because the limit is evaluated using trigonometric identities and L'Hopital's rule or series expansion.
132 Medium
132 of 267
Analyzing +4 | -1
Let f:R→Rf:R \to R be a continuously differentiable function such that f(2)=6f\left( 2 \right) = 6 and f′(2)=148.  If∫6f(x)4t3dt=(x−2)g(x)f'\left( 2 \right) = \frac{1}{{48}}.\,\,\mathrm{If }\int_6^{f\left( x \right)} {4{t^3}dt = \left( {x - 2} \right)g\left( x \right)}, then lim⁡x→2g(x)\lim\limits_{x \to 2} g\left( x \right) is equal to
View answer and explanation
Correct option: A
18
Option A (18) is correct because L'Hopital's rule is applied to evaluate the limit after expressing g(x) as a ratio of integrals.
133 Medium
133 of 267
Application +4 | -1
lim⁡h→0cos⁡2(x+h)−cos⁡2xh\lim \limits_{{h \rightarrow 0}} \frac{\cos ^2(x+h)-\cos ^2 x}{h} is equal to
View answer and explanation
Correct option: B
−sin⁡2x-\sin 2 x
Option B (−sin⁡2x-\sin 2x) is correct because the limit represents the derivative of cos⁡2x\cos^2 x, which is −2cos⁡xsin⁡x=−sin⁡2x-2 \cos x \sin x = -\sin 2x.
134 Medium
134 of 267
Analyzing +4 | -1
The value of lim⁡x→01−cos⁡x21−cos⁡x\lim \limits_{{x \rightarrow 0}} \frac{\sqrt{1-\cos x^2}}{1-\cos x} is
View answer and explanation
Correct option: C
2\sqrt{2}
Option C (2\sqrt{2}) is correct because by using trigonometric identities and standard limits, the expression simplifies to 2\sqrt{2}.
135 Medium
135 of 267
Application +4 | -1
If G(x)=−36−x2G(x)=-\sqrt{36-x^{2}}, then Ltx→1G(x)−G(1)x−1=\underset{x \rightarrow 1}{L t} \frac{G(x)-G(1)}{x-1}=
View answer and explanation
Correct option: A
135\frac{1}{\sqrt{35}}
Option A (1/sqrt(35)) is correct because it is the result of applying L'Hopital's rule to find the limit of the difference quotient, which represents the derivative of G(x) at x=1.

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136 Medium
136 of 267
Application +4 | -1
The value of lim⁡x→elog⁡x−1x−e\lim \limits_{{x \rightarrow e}} \frac{\log x-1}{x-e}, is
View answer and explanation
Correct option: B
1e\frac{1}{e}
Option B (1e\frac{1}{e}) is correct because the limit can be evaluated using the standard limit lim⁡x→0ln⁡(1+x)x=1\lim_{x \to 0} \frac{\ln(1+x)}{x} = 1.
137 Medium
137 of 267
Analyzing +4 | -1
If ϕ(x)=lim⁡n→∞x2nf(x)+g(x)1+x2n\phi(x)=\lim \limits_{{n \rightarrow \infty}} \frac{x^{2 n} f(x)+g(x)}{1+x^{2 n}}, then
View answer and explanation
Correct option: D
ϕ(x)={g(x)  for ∣x∣<1f(x)  for ∣x∣>1f(x)+g(x)2  for ∣x∣=1\phi(x)= \begin{cases}g(x) \,\text { for }|x| < 1 \\ f(x) \,\text { for }|x| > 1 \\ \frac{f(x)+g(x)}{2} \,\text { for }|x|=1\end{cases}
Option D is correct because it considers the limit of the given function for different ranges of x: |x| < 1, |x| > 1, and |x| = 1.
138 Medium
138 of 267
Analyzing +4 | -1
The value of lim⁡x→1sec⁡(π2x)log⁡x\lim \limits_{{x \rightarrow 1}} \sec \left(\frac{\pi}{2^x}\right) \log x is
View answer and explanation
Correct option: C
2πlog⁡2\frac{2}{\pi \log 2}
Option C (2πlog⁡2\frac{2}{\pi \log 2}) is correct because after applying L'Hopital's rule and limit properties, the expression simplifies to 2πlog⁡2\frac{2}{\pi \log 2}.
139 Medium
139 of 267
Application +4 | -1
lim⁡x→+∞3x2−1−x2−12x+3=\lim _{x \rightarrow+\infty} \frac{\sqrt{3 x^2-1}-\sqrt{x^2-1}}{2 x+3}=
View answer and explanation
Correct option: B
3−12\frac{\sqrt{3}-1}{2}
Option B (3−12\frac{\sqrt{3}-1}{2}) is correct because, after dividing by x and simplifying, the limit of the expression as x approaches infinity is (3\sqrt{3}-1)/2.
140 Medium
140 of 267
Analyzing +4 | -1
The limiting value of (cos⁡x)1/sin⁡x(\cos x)^{1 / \sin x} as x→0x \rightarrow 0 is
View answer and explanation
Correct option: A
1
The correct answer is A (1) because by rewriting the expression and applying limit properties, the limit evaluates to 1.
141 Medium
141 of 267
Analyzing +4 | -1
If f(x)=sin⁡(1x)f(x)=\sin \left(\frac{1}{x}\right), then at x=0x=0
View answer and explanation
Correct option: C
Both LHL and RHL do not exist
Option C (Both LHL and RHL do not exist) is correct because the function oscillates between -1 and 1 as x approaches 0 from either side.
142 Medium
142 of 267
Analyzing +4 | -1
lim⁡x→−11∣x∣−{−x}\lim \limits_{{x \rightarrow-1}} \frac{1}{\sqrt{|x|-\{-x\}}} (where {x}\{x\} denotes the fractional part of xx 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 (LHL) approaches infinity, while the right-hand limit (RHL) approaches 1, indicating the limit does not exist.
143 Medium
143 of 267
Analyzing +4 | -1
Let f(x)f(x) be a polynomial of degree 4 having extreme values at x=1x=1 and x=2x=2. If lim⁡x→0(f(x)x2+1)=3\lim _{x \rightarrow 0}\left(\frac{f(x)}{x^2}+1\right)=3 then f(−1)f(-1) is equal to
View answer and explanation
Correct option: D
92\frac{9}{2}
Option D (92\frac{9}{2}) is correct because after finding the polynomial f(x) using the given conditions, f(-1) evaluates to 92\frac{9}{2}.
144 Medium
144 of 267
Application +4 | -1
lim⁡x→axsin⁡a−asin⁡xx−a\lim \limits_{{x \rightarrow a}} \frac{x \sin a-a \sin x}{x-a} is
View answer and explanation
Correct option: D
sin⁡a−acos⁡a\sin a-a \cos a
The correct answer is D (sin⁡a−acos⁡a\sin a - a \cos a) because applying L'Hôpital's rule to the limit yields the value sin⁡a−acos⁡a\sin a - a \cos a.
145 Medium
145 of 267
Analyzing +4 | -1
Let ff be differentiable on some interval (a,+∞)(a,+\infty) and suppose

lim⁡x→∞[f(x)+f′(x)]=l\lim \limits_{{x \rightarrow \infty}}\left[f(x)+f^{\prime}(x)\right]=l

Then lim⁡x→∞f(x)\lim \limits_{{x \rightarrow \infty}} f(x) is equal to
View answer and explanation
Correct option: C
ll
Option C (l) is correct because applying L'Hôpital's rule to the expression f(x)e^x / e^x and using the given limit results in the limit of f(x) being equal to l.
146 Medium
146 of 267
Analyzing +4 | -1
If
lim⁡x→0[1+xlog⁡(1+b2)](1ix)=2bsin⁡2θ,b>0\lim \limits_{{x \rightarrow 0}}\left[1+x \log \left(1+b^2\right)\right]^{(1 i x)}=2 b \sin ^2 \theta, b > 0
and θ∈(−π,π]\theta \in(-\pi, \pi] then the value of θ\theta is
View answer and explanation
Correct option: D
±π2\pm \frac{\pi}{2}
Option D (±π2\pm \frac{\pi}{2}) is correct because by evaluating the limit and solving the resulting equation, the value of θ\theta is found to be ±π2\pm \frac{\pi}{2}.
147 Medium
147 of 267
Understanding +4 | -1
Which of the following is/are true?

I. lim⁡x→0sin⁡xx=1\lim \limits_{{x \rightarrow 0}} \frac{\sin x}{x}=1

II. lim⁡x→01−cos⁡xx=0\lim \limits_{{x \rightarrow 0}} \frac{1-\cos x}{x}=0
View answer and explanation
Correct option: A
Both I and II are true
Option A (Both I and II are true) is correct because both given limits are standard results in calculus.
148 Medium
148 of 267
Application +4 | -1
lim⁡x→∞(x2+x+1−x2+1)=\lim \limits_{{x \rightarrow \infty}}\left(\sqrt{x^2+x+1}-\sqrt{x^2+1}\right)=
View answer and explanation
Correct option: B
12\frac{1}{2}
Option C (1/4) is correct because simplifying the expression using the sum of cubes formula leads to a limit of 1/4 as n approaches infinity.
149 Medium
149 of 267
Analyzing +4 | -1
lim⁡x→0{1+tan⁡x1+sin⁡x}cosecx\lim\limits_{x \to 0} {\left\{ {\frac{{1 + \tan x}}{{1 + \sin x}}} \right\}^{\mathrm{cosec}x}} is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because by rewriting the expression and using the limit definition of e, the limit simplifies to 1.
150 Medium
150 of 267
Analyzing +4 | -1
lim⁡x→04x8(1−cos⁡x22−cos⁡x24+cos⁡x22⋅cos⁡x24)=(1n1)n\lim\limits_{x \to 0} \frac{4}{{{x^8}}}\left( {1 - \cos \frac{{{x^2}}}{2} - \cos \frac{{{x^2}}}{4} + \cos \frac{{{x^2}}}{2} \cdot \cos \frac{{{x^2}}}{4}} \right) = {\left( {\frac{1}{{{n_1}}}} \right)^n} Where n1{n_1} is a Prime number, then value of n1+nn_1+n equals
View answer and explanation
Correct option: A
8
The correct answer is A. By simplifying the limit using trigonometric identities and standard limit results, the value of n1+nn_1 + n is found to be 8.
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