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JEE • MATHEMATICS • LIMITS

Limits practice · Medium

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Showing 211-240 of 267 questions

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211 Medium
211 of 267
Analyzing +4 | -1
The value of lim⁡x→0cos⁡(sin⁡2x)−cos⁡(x2)x6\lim \limits_{{x \rightarrow 0}} \frac{\cos \left(\sin ^2 x\right)-\cos \left(x^2\right)}{x^6} is
View answer and explanation
Correct option: C
1/31 / 3
Option C (13\frac{1}{3}) is correct because using L'Hopital's rule or Taylor series expansion, the limit evaluates to 1/3.
212 Medium
212 of 267
Analyzing +4 | -1
lim⁡n→∞(sin⁡n1+cos⁡n1)n\lim \limits_{{n \rightarrow \infty}}\left(\sin ^n 1+\cos ^n 1\right)^n is equal to
View answer and explanation
Correct option: D
sin⁡1\sin 1
The correct answer is D (sin⁡1\sin 1) because as n approaches infinity, cos⁡n(1)\cos^n(1) approaches 0 faster than sin⁡n(1)\sin^n(1), making the limit approximately sin⁡(1)\sin(1).
213 Medium
213 of 267
Application +4 | -1
lim⁡x→0(cos⁡x)cot⁡2x\lim \limits_{{x \rightarrow 0}}(\cos x)^{\cot ^2 x} is
View answer and explanation
Correct option: D
1e\frac{1}{\sqrt{e}}
Option D (1e\frac{{1}}{{\sqrt{{e}}}}) is correct because by applying L'Hôpital's rule to the logarithmic form of the limit, we find the limit to be e−12=1ee^{-\frac{{1}}{{2}}} = \frac{{1}}{{\sqrt{{e}}}}.
214 Medium
214 of 267
Analyzing +4 | -1
lim⁡x→+∞(2xx10)\lim \limits_{{x \rightarrow+\infty}}\left(\frac{2^x}{x^{10}}\right) is equal to
View answer and explanation
Correct option: D
+∞+\infty
Option D (+∞+\infty) is correct because exponential functions grow faster than polynomial functions as x approaches infinity. Applying L'Hôpital's rule repeatedly confirms this.
215 Medium
215 of 267
Application +4 | -1
The value of lim⁡x→0{sin⁡x−x+x36x5}\lim\limits_{x \to 0} \left\{ {\frac{{\sin x - x + \frac{{{x^3}}}{6}}}{{{x^5}}}} \right\} is
View answer and explanation
Correct option: D
1120\frac{1}{{120}}
Option D (1120\frac{1}{{120}}) is correct because it is the result of evaluating the limit using the Taylor series expansion of sin(x).
216 Medium
216 of 267
Analyzing +4 | -1
The value of lim⁡x→∞2(x)1/2+3(x)1/3+4(x)1/4+…+n(x)1/n(2x−3)1/2+(2x−3)1/3+…+(2x−3)1/n\lim \limits_{{x \rightarrow \infty}} \frac{2(x)^{1 / 2}+3(x)^{1 / 3}+4(x)^{1 / 4}+\ldots+n(x)^{1 / n}}{(2 x-3)^{1 / 2}+(2 x-3)^{1 / 3}+\ldots+(2 x-3)^{1 / n}} is
View answer and explanation
Correct option: A
2\sqrt{2}
Option A (2\sqrt{2}) is correct. By dividing numerator and denominator by x1/2x^{1/2} and taking the limit as x approaches infinity, the expression simplifies to 2\sqrt{2}.
217 Medium
217 of 267
Analyzing +4 | -1
The value of lim⁡x→∞x{tan⁡−1(x+1x+2)−tan⁡−1(xx+2)}\lim\limits_{x \to \infty } x\left\{ {{{\tan }^{ - 1}}\left( {\frac{{x + 1}}{{x + 2}}} \right) - {{\tan }^{ - 1}}\left( {\frac{x}{{x + 2}}} \right)} \right\} is
View answer and explanation
Correct option: C
1/21/2
Option C (1/21/2) is correct because using the arctangent subtraction formula and then the small angle approximation for tangent yields the correct limit.
218 Medium
218 of 267
Application +4 | -1
Let f(2)=4f\left( 2 \right) = 4 and f′(2)=4f'\left( 2 \right) = 4. Then, lim⁡x→2xf(2)−2f(x)x−2\lim\limits_{x \to 2} \frac{{xf\left( 2 \right) - 2f\left( x \right)}}{{x - 2}} is given by
View answer and explanation
Correct option: C
−4- 4
Option C (−4-4) is correct because it is obtained by applying L'Hopital's rule or by rewriting the limit in terms of the derivative of f(x).
219 Medium
219 of 267
Analyzing +4 | -1
lim⁡x→0(cosec  x)1log⁡ex\lim\limits_{x \to 0} {\left( {{\text{cosec}}\,\,x} \right)^{\frac{1}{{{{\log }_e}x}}}} equals
View answer and explanation
Correct option: C
1e\frac{1}{e}
Option C (1e\frac{1}{e}) is correct because by applying L'Hopital's rule to the logarithm of the expression, we find the limit is -1, thus the original limit is 1e\frac{1}{e}.
220 Medium
220 of 267
Application +4 | -1
lim⁡α→π4sin⁡α−cos⁡αα−π4=……\lim \limits_ {\alpha \rightarrow \frac{\pi}{4}} \frac{\sin \alpha-\cos \alpha}{\alpha-\frac{\pi}{4}}=\ldots \ldots
View answer and explanation
Correct option: A
2\sqrt{2}
Option A (2\sqrt{2}) is correct because the limit can be evaluated using L'Hopital's rule, resulting in cos(α\alpha) + sin(α\alpha), which evaluates to 2\sqrt{2} at α\alpha = π\pi/4.
221 Medium
221 of 267
Analyzing +4 | -1
Let f(x)=sin⁡x/xf(x)=\sin x / x for x≠0x \neq 0 and [.] denote the greatest integer function. Then lim⁡x→0[f(x)]\lim _{x \rightarrow 0}[f(x)] is
View answer and explanation
Correct option: A
0
Option A (0) is correct because for x near 0, but not equal to 0, sin⁡xx\frac{{\sin x}}{{x}} is between 0 and 1, so its greatest integer function is 0.
222 Medium
222 of 267
Application +4 | -1
Let f:R→Rf:R \to R be a differentiable function having f(2)=6,  f′(2)=(148)f\left( 2 \right) = 6,\;f'\left( 2 \right) = \left( {\frac{1}{{48}}} \right). Then, lim⁡x→2∫6f(x)4  t3  dtx−2\lim\limits_{x \to 2} \frac{{\int \nolimits_6^{f\left( x \right)} 4\;{t^3}\;dt}}{{x - 2}} is equals
View answer and explanation
Correct option: A
18
Option A (18) is correct because it is the result of applying L'Hopital's rule and the Fundamental Theorem of Calculus to the given limit.
223 Medium
223 of 267
Analyzing +4 | -1
If f(x+y)=f(x)+f(y)f(x+y)=f(x)+f(y) for all x,y∈Rx, y \in R and f(1)=1f(1)=1, then lim⁡x→02f(tan⁡x)−2f(sin⁡x)x2f(sin⁡x)\lim \limits_{{x \rightarrow 0}} \frac{2^{f(\tan x)}-2^{f(\sin x)}}{x^2 f(\sin x)} equals
View answer and explanation
Correct option: C
log⁡2\log \sqrt{2}
Option C (log⁡2\log \sqrt{2}) is correct because after simplifying the expression using the given functional equation and limit properties, the limit evaluates to 12log⁡2=log⁡2\frac{1}{2} \log 2 = \log \sqrt{2}.
224 Medium
224 of 267
Analyzing +4 | -1
If a.lim⁡x→1x1/1−x+b=e−1(a≥1,b≥0)a.\lim \limits_{{x \rightarrow 1}} x^{1 / 1-x}+b=e^{-1}(a \geq 1, b \geq 0), then
View answer and explanation
Correct option: D
a=1,b=0a=1, b=0
Option D (a=1,b=0a=1, b=0) is correct because by evaluating the limit and comparing with the given equation, we find a=1 and b=0.
225 Medium
225 of 267
Analyzing +4 | -1
lim⁡x→0e1/x−1e1/x+1\lim\limits _{x \rightarrow 0} \frac{e^{1 / x}-1}{e^{1 / x}+1}, 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 as x approaches 0.

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226 Medium
226 of 267
Analyzing +4 | -1
The value of lim⁡x→1(log⁡22x)log⁡x5\lim\limits_{x \to 1} {\left( {{{\log }_2}2x} \right)^{{{\log }_x}5}} is
View answer and explanation
Correct option: B
elog⁡25{e^{{{\log }_2}5}}
Option B (elog⁡25{e^{{{\log }_2}5}}) is correct because after applying logarithmic properties and simplifying, the limit evaluates to elog⁡25{e^{{{\log }_2}5}}.
227 Medium
227 of 267
Analyzing +4 | -1
The value of lim⁡x→0cos⁡(sin⁡x)−cos⁡xx4\lim\limits_{x \to 0} \frac{{\cos \left( {\sin x} \right) - \cos x}}{{{x^4}}} is equal to
View answer and explanation
Correct option: B
1/61/6
Option B (1/61/6) is correct because the limit is evaluated using L'Hopital's rule or series expansion of trigonometric functions.
228 Medium
228 of 267
Application +4 | -1
Let f:R→Rf: \mathbb{R} \rightarrow \mathbb{R} be such that f(1)=3f(1)=3 and f′(1)=6f^{\prime}(1)=6. Then

lim⁡x→0(f(1+x)f(1))1/x=\lim \limits_{{x \rightarrow 0}}\left(\frac{f(1+x)}{f(1)}\right)^{1 / x}=
View answer and explanation
Correct option: A
e2e^2
Option A (e2e^2) is correct because by taking the natural logarithm and applying L'Hôpital's rule, the limit simplifies to e2e^2.
229 Medium
229 of 267
Analyzing +4 | -1
lim⁡x→0(ex+e−x−2x2)1/x2\lim\limits_{x \to 0} {\left( {\frac{{{e^x} + {e^{ - x}} - 2}}{{{x^2}}}} \right)^{1/{x^2}}} is equal to
View answer and explanation
Correct option: D
e1/12{e^{1/12}}
The correct answer is D (e1/12{e^{1/12}}) because it is the result of evaluating the limit using series expansion and properties of exponential functions.
230 Medium
230 of 267
Understanding +4 | -1
The value of lim⁡x→∞x1/x\lim\limits_{x \to \infty } {x^{1/x}} equals
View answer and explanation
Correct option: B
1
Option C (7) is correct because the minimum value of the absolute value function ∣3−x∣|3 - x| is 0, which occurs when x=3x = 3. Therefore, the minimum value of f(x)f(x) is 0+7=70 + 7 = 7.
231 Medium
231 of 267
Analyzing +4 | -1
If xn+1=1+xn2x_{n+1}=\sqrt{\frac{1+x_n}{2}} and ∣x0∣<1,n≥0\left|x_0\right|<1, n \geq 0, then lim⁡n→∞(1−x02x1x2x3…xn+1)\lim \limits_{{n \rightarrow \infty}}\left(\frac{\sqrt{1-x_0^2}}{x_1 x_2 x_3 \ldots x_{n+1}}\right) is equal to
View answer and explanation
Correct option: D
cos⁡−1(x0)\cos ^{-1}\left(x_0\right)
Option D, cos⁡−1(x0)\cos^{-1}(x_0), is correct because by substituting x0=cos⁡θx_0 = \cos \theta and using trigonometric identities, the limit simplifies to θ\theta, which is cos⁡−1(x0)\cos^{-1}(x_0).
232 Medium
232 of 267
Application +4 | -1
If a1=1a_1=1 and an+1=4+3an3+2an,n≥1a_{n+1}=\frac{4+3 a_n}{3+2 a_n}, n \geq 1 and if ana_n has a limit ll as n→∞n \rightarrow \infty, then
View answer and explanation
Correct option: B
l=2l=\sqrt{2}
Option B (l=2l=\sqrt{2}) is correct because the limit ll satisfies the equation l=4+3l3+2ll = \frac{4+3l}{3+2l}, and solving for ll yields l=2l=\sqrt{2} (since l>0l>0).
233 Medium
233 of 267
Application +4 | -1
lim⁡x→0x7[1x3]=……\lim \limits_{{x \rightarrow 0}} x^7\left[\frac{1}{x^3}\right]=\ldots \ldots
View answer and explanation
Correct option: A
0
Option A (0) is correct because as x approaches 0, the expression simplifies to a limit that evaluates to 0 using the sandwich theorem.
234 Medium
234 of 267
Application +4 | -1
lim⁡x→∞3x[x4]=pq\lim \limits_{{x \rightarrow \infty}} \frac{3}{x}\left[\frac{x}{4}\right]=\frac{p}{q} (where [.] denotes greatest integer function), then p+qp+q (where p,qp, q are relative prime) is:
View answer and explanation
Correct option: B
7
Option B (7) is correct because the limit simplifies to 3/4, and the sum of the numerator and denominator is 7.
235 Medium
235 of 267
Application +4 | -1
The derivative of x2/3x^{2 / 3} by using first principle is
View answer and explanation
Correct option: A
23x−1/3\frac{2}{3} x^{-1 / 3}
Option A (23x−1/3\frac{2}{3} x^{-1 / 3}) is correct because it is the derivative of x2/3x^{2/3} obtained using the first principle of differentiation.
236 Medium
236 of 267
Analyzing +4 | -1
lim⁡x→0xz2−(z−x)23(8xz−4x23+8xz3)4\lim \limits_{{x \rightarrow 0}} \frac{x \sqrt[3]{z^2-(z-x)^2}}{\left(\sqrt[3]{8 x z-4 x^2}+\sqrt[3]{8 x z}\right)^4} is equal to
View answer and explanation
Correct option: B
1223/3⋅z\frac{1}{2^{23 / 3} \cdot z}
Option B (1223/3⋅z\frac{1}{2^{23 / 3} \cdot z}) is correct because it is the result of simplifying the limit expression by factoring and canceling terms.
237 Medium
237 of 267
Understanding +4 | -1
lim⁡x→0xtan⁡−12x\lim\limits_{x \to 0} \frac{x}{{{{\tan }^{ - 1}}2x}} is equal to
View answer and explanation
Correct option: B
12\frac{1}{2}
Option B (12\frac{1}{2}) is correct because applying limit properties and the standard limit formula for inverse tangent yields the result.
238 Medium
238 of 267
Analyzing +4 | -1
The value of lim⁡x→0(sin⁡xx)sin⁡xx−sin⁡x\lim\limits_{x \to 0} {\left( {\frac{{\sin x}}{x}} \right)^{\frac{{\sin x}}{{x - \sin x}}}}, is
View answer and explanation
Correct option: A
e−1{e^{ - 1}}
Option A (e−1{e^{ - 1}}) is correct because using the exponential limit formula and simplifying the expression leads to the correct result.
239 Medium
239 of 267
Analyzing +4 | -1
If ana_n and bnb_n are positive integers and an+2bna_n+\sqrt{2} b_n =(2+2)n=(2+\sqrt{2})^n, then
lim⁡n→−(anbn)=\lim \limits_{{n \rightarrow-}}\left(\frac{a_n}{b_n}\right)=
View answer and explanation
Correct option: B
2\sqrt{2}
Option B (2\sqrt{2}) is correct because after finding expressions for ana_n and bnb_n and taking the limit of their ratio, the result is 2\sqrt{2}.
240 Medium
240 of 267
Analyzing +4 | -1
The value of lim⁡x→1(x2−1)sin⁡2(πx)x4−2x3+2x−1\lim\limits _{x \to 1} \frac{\left(x^2-1\right) \sin ^2(\pi x)}{x^4-2 x^3+2 x-1} is equal to
View answer and explanation
Correct option: D
π2\pi^2
Option D (π2\pi^2) is correct because after simplification and applying L'Hopital's rule or trigonometric limits, the limit evaluates to π2\pi^2.
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