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JEE • MATHEMATICS • LIMITS

Limits practice · Mixed difficulty

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271 Medium
271 of 930
Application +4 | -1
If lim⁡x→−12f(x)−22x2−1=3π\lim \limits_{x \rightarrow \frac{-1}{\sqrt{2}}} \frac{f(x)-2}{2 x^2-1}=3 \pi, then lim⁡x→−12ef(x)\lim \limits_{x \rightarrow \frac{-1}{\sqrt{2}}} e^{f(x)} will be
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the limit of f(x) as x approaches -1/sqrt(2) is 2, so the limit of e^(f(x)) is e2.
272 Easy
272 of 930
Analyzing +4 | -1
The value of lim⁡x→−∞(x2−x+1+x)\lim \limits_{{x \rightarrow-\infty}}\left(\sqrt{x^2-x+1}+x\right), is
View answer and explanation
Correct option: B
12\frac{1}{2}
Option B (12\frac{1}{2}) is correct because the limit is evaluated by rationalizing the expression and simplifying.
273 Easy
273 of 930
Analyzing +4 | -1
lim⁡n→∞2−n(n2+5n+6)(n+4)(n+5)\lim\limits_{n \to \infty } \frac{{{2^{ - n}}\left( {{n^2} + 5n + 6} \right)}}{{\left( {n + 4} \right)\left( {n + 5} \right)}} is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because as n approaches infinity, the exponential term 2(-n) dominates, causing the entire expression to approach 0.
274 Easy
274 of 930
Application +4 | -1
lim⁡x→0x2sin⁡πx\lim\limits_{x \to 0} {x^2}\sin \frac{\pi }{x} is equal to
View answer and explanation
Correct option: B
0
Option B (0) is correct because the sine function is bounded between -1 and 1, and the x2x^2 term approaches 0 as x approaches 0, making the entire expression approach 0.
275 Easy
275 of 930
Analyzing +4 | -1
If lim⁡x→0sin⁡(sin⁡x)−sin⁡xax3+bx5+c=−112\lim \limits_{{x \rightarrow 0}} \frac{\sin (\sin x)-\sin x}{a x^3+b x^5+c}=-\frac{1}{12}, then
View answer and explanation
Correct option: A
a=2,b∈R,c=0a=2, b \in R, c=0
Option A (a=2,b∈R,c=0a=2, b \in R, c=0) is correct because the limit exists and equals -1/12 only when a=2 and c=0, with b being any real number.
276 Easy
276 of 930
Application +4 | -1
The value of lim⁡x→0(1+5x21+3x2)1/x2\lim\limits_{x \to 0} {\left( {\frac{{1 + 5{x^2}}}{{1 + 3{x^2}}}} \right)^{1/{x^2}}}, is
View answer and explanation
Correct option: A
e2{e^2}
Option A (e2{e^2}) is correct because it is the result of evaluating the limit using the exponential limit formula.
277 Easy
277 of 930
Application +4 | -1
lim⁡x→∞(x2+4x−3x2−2x+5)x\lim \limits_{{x \rightarrow \infty}}\left(\frac{x^2+4 x-3}{x^2-2 x+5}\right)^x is equal to
View answer and explanation
Correct option: A
e6e^6
Option A (e6e^6) is correct because the limit can be evaluated by rewriting the expression and using the standard limit lim⁡x→∞(1+ax)x=ea\lim_{x \rightarrow \infty} (1 + \frac{{a}}{{x}})^x = e^a.
278 Medium
278 of 930
Application +4 | -1
lim⁡x→0(ln⁡(a+x)−ln⁡ax)+klim⁡x→eln⁡x−1x−e=1\lim \limits_{{x \rightarrow 0}}\left(\frac{\ln (a+x)-\ln a}{x}\right)+k \lim \limits_{{x \rightarrow e}} \frac{\ln x-1}{x-e}=1 then
View answer and explanation
Correct option: A
k=e(1−1a)k=e\left(1-\frac{1}{a}\right)
Option A (k=e(1−1a)k=e\left(1-\frac{{1}}{{a}}\right)) is correct because by applying the definition of derivative and simplifying, we can find the value of k.
279 Easy
279 of 930
Analyzing +4 | -1
lim⁡x→∞x2+1−x3+13x4+14−x4+15\lim\limits_{x \to \infty } \frac{{\sqrt {{x^2} + 1} - \sqrt[3]{{{x^3} + 1}}}}{{\sqrt[4]{{{x^4} + 1}} - \sqrt[5]{{{x^4} + 1}}}} equals
View answer and explanation
Correct option: B
0
Option B (0) is correct because by dividing the numerator and denominator by the highest power of x and simplifying, the limit evaluates to 0.
280 Easy
280 of 930
Application +4 | -1
If f(1)=1f(1)=1 and f′(1)=4f^{\prime}(1)=4, then the value of lim⁡x→1f(x)−1x−1\lim _{x \rightarrow 1} \frac{\sqrt{f(x)}-1}{\sqrt{x}-1} is
View answer and explanation
Correct option: B
4
Option B (4) is correct because by applying L'Hôpital's rule, the limit simplifies to f′(1)1/(21)=41/2=4\frac{{f'(1)}}{{1 / (2 \sqrt{{1}})}} = \frac{{4}}{{1/2}} = 4.
281 Easy
281 of 930
Analyzing +4 | -1
lim⁡x→02log⁡(1+x)−log⁡(1+2x)x2\lim \limits_{{x \rightarrow 0}} \frac{2 \log (1+x)-\log (1+2 x)}{x^2} is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because after applying L'Hopital's rule or using series expansion, the limit evaluates to 1.
282 Easy
282 of 930
Analyzing +4 | -1
If ∣cos⁡−11n∣<π2\left|\cos ^{-1} \frac{1}{n}\right| < \frac{\pi}{2}, then lim⁡n→∞{(n+1)2πcos⁡−11n−n}\lim \limits_{{n \rightarrow \infty}}\left\{(n+1) \frac{2}{\pi} \cos ^{-1} \frac{1}{n}-n\right\} is equal to
View answer and explanation
Correct option: B
π−2π\frac{\pi-2}{\pi}
Option B (π−2π\frac{\pi-2}{\pi}) is correct because after applying trigonometric identities and evaluating the limit, the expression simplifies to π−2π\frac{\pi-2}{\pi}.
283 Medium
283 of 930
Application +4 | -1
ff is a differentiable function such that f(xy)=f(x)f(y),x≠0,y≠0,f(y)≠0f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}, x \neq 0, y \neq 0, f(y) \neq 0. If f′(1)=2f^{\prime}(1)=2, then f′(x)=……f^{\prime}(x)=\ldots \ldots
View answer and explanation
Correct option: D
2f(x)x\frac{2 f(x)}{x}
Option C (π3\frac{\pi}{3}) is correct because by using the sine rule and differentiating, we find the angle to be π3\frac{\pi}{3}.
284 Easy
284 of 930
Application +4 | -1
lim⁡x→0log⁡(1+x3)sin⁡3x\lim\limits_{x \to 0} \frac{{\log \left( {1 + {x^3}} \right)}}{{{{\sin }^3}x}} is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because by using the series expansion of log(1+x) and sin(x), the limit simplifies to 1.
285 Medium
285 of 930
Application +4 | -1
Let f(x)={e−1x2sin⁡1xx≠0, then f′(0)=0x=0f(x)=\left\{\begin{array}{cl}e^{-\frac{1}{x^2}} \sin \frac{1}{x} & x \neq 0, \text { then } f^{\prime}(0)= \\ 0 & x=0\end{array}\right.
View answer and explanation
Correct option: C
0
Option C (0) is correct because the derivative at x=0 is found using the limit definition and the squeeze theorem.

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286 Easy
286 of 930
Application +4 | -1
lim⁡x→π/32sin⁡(x−π/3)1−2cos⁡x\lim \limits_{{x \rightarrow \pi / 3}} \frac{2 \sin (x-\pi / 3)}{1-2 \cos x} is
View answer and explanation
Correct option: B
2/32 / \sqrt{3}
Option B (2/32 / \sqrt{3}) is correct because L'Hopital's rule is applied to evaluate the limit of the given expression as x approaches π/3\pi/3.
287 Medium
287 of 930
Analyzing +4 | -1
lim⁡x→∞(3x+1+3x−1)6+(3x+1−3x−1)6(x+x2−1)6+(x−x2−1)6x3\lim \limits_{{x \rightarrow \infty}} \frac{(\sqrt{3 x+1}+\sqrt{3 x-1})^{6}+(\sqrt{3 x+1}-\sqrt{3 x-1})^{6}}{\left(x+\sqrt{x^{2}-1}\right)^{6}+\left(x-\sqrt{x^{2}-1}\right)^{6}} x^{3}
View answer and explanation
Correct option: C
Is equal to 272\frac{27}{2}
Option C (Is equal to 272\frac{27}{2}) is correct because by applying binomial expansion and simplifying the limit, we get 272\frac{27}{2}.
288 Medium
288 of 930
Application +4 | -1
lim⁡x→∞(1+ax+bx2)2x=e2\lim _{x \rightarrow \infty}\left(1+\frac{a}{x}+\frac{b}{x^2}\right)^{2 x}=\mathrm{e}^2, then the values of a& b\mathrm{a} \& \mathrm{~b} are -
View answer and explanation
Correct option: B
a=1,b∈Ra=1, b \in R
Option B (a=1,b∈Ra=1, b \in R) is correct because the limit can be evaluated using L'Hopital's rule or by comparing coefficients, leading to a=1 while b can be any real number.
289 Easy
289 of 930
Analyzing +4 | -1
The value of lim⁡x→∞(x3+x23−x3−x23)\lim \limits_{{x \rightarrow \infty}}\left(\sqrt[3]{x^3+x^2}-\sqrt[3]{x^3-x^2}\right), is
View answer and explanation
Correct option: B
23\frac{2}{3}
Option B (2/3) is correct because the limit is evaluated by rationalizing the expression and simplifying.
290 Easy
290 of 930
Analyzing +4 | -1
The value of lim⁡x→0sin⁡(sin⁡x)−tan⁡(sin⁡x)sin⁡3(sin⁡x)\lim \limits_{{x \rightarrow 0}} \frac{\sin (\sin x)-\tan (\sin x)}{\sin ^3(\sin x)}, is
View answer and explanation
Correct option: A
−12-\frac{1}{2}
Option A (-1/2) is correct because it is obtained by applying the limit properties and using the known limit lim⁡x→0tan⁡x−sin⁡xsin⁡3x=12\lim \limits_{{x \rightarrow 0}} \frac{{\tan x-\sin x}}{{\sin ^3 x}}=\frac{{1}}{{2}}
291 Easy
291 of 930
Application +4 | -1
lim⁡n→∞(2n+5n)1/n\lim \limits_{{n \rightarrow \infty}}\left(2^n+5^n\right)^{1 / n} is equal to
View answer and explanation
Correct option: B
5
Option B, 5, is correct because it is the result of factoring out the largest term and then taking the limit.
292 Medium
292 of 930
Application +4 | -1
Which of the following is/are true?

I. The derivative of x2−2x^2-2 at x=10x=10 is 18.

II. The derivative of 99x99 x at x=100x=100 is 99.

III. The derivative of xx at x=1x=1 is 1.
View answer and explanation
Correct option: C
II and III are true
Option C is correct because statements II and III are true. The derivative of 99x is 99, and the derivative of x is 1, while the derivative of x²-2 at x=10 is 20, not 18.
293 Easy
293 of 930
Analyzing +4 | -1
lim⁡x→1sec⁡π2xlog⁡x\lim \limits_{{x \rightarrow 1}} \sec \frac{\pi}{2 x} \log x equals
View answer and explanation
Correct option: C
2π\frac{2}{\pi}
Option C (2π\frac{2}{\pi}) is correct because after applying L'Hôpital's rule, the limit evaluates to 2π\frac{2}{\pi}.
294 Easy
294 of 930
Analyzing +4 | -1
The value of lim⁡n→∞{13+221+391+…+nn4+n2+1}\lim\limits_{n \to \infty } \left\{ {\frac{1}{3} + \frac{2}{{21}} + \frac{3}{{91}} + \ldots + \frac{n}{{{n^4} + {n^2} + 1}}} \right\}, is
View answer and explanation
Correct option: B
12\frac{1}{2}
Option B (12\frac{1}{2}) is correct because the limit of the given series as n approaches infinity is 1/2, obtained by telescoping the sum.
295 Medium
295 of 930
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 (1/(2(23/3) * z)) is correct because after simplifying the expression and taking the limit, the result is 1/(2(23/3) * z).
296 Easy
296 of 930
Application +4 | -1
lim⁡x→∞xsin⁡(2x)\lim\limits_{x \to \infty } x\sin \left( {\frac{2}{x}} \right) is equal to
View answer and explanation
Correct option: C
2
Option C (2) is correct because by substituting u=2xu = \frac{2}{x}, the limit transforms into a standard limit: lim⁡u→0sin⁡uu=1\lim_{u \to 0} \frac{\sin u}{u} = 1.
297 Easy
297 of 930
Application +4 | -1
lim⁡x→1x8−2x+1x4−2x+1\lim \limits_{{x \rightarrow 1}} \frac{x^8-2 x+1}{x^4-2 x+1} equals
View answer and explanation
Correct option: A
3
Option A (3) is correct because it is the result of factoring and simplifying the limit expression as x approaches 1.
298 Easy
298 of 930
Application +4 | -1
Let f(x)={x+2,x≤−1cx2,x>−1{f}({x})= \begin{cases}{x}+2, & {x} \leq-1 \\ {cx}^2, & {x}>-1\end{cases}
If lim⁡x→−1f(x)\lim \limits_{{x \rightarrow-1}} f(x) exists, then cc is equal to
View answer and explanation
Correct option: A
1
Option A is correct because for the limit to exist at x = -1, the value of c must be 1.
299 Easy
299 of 930
Analyzing +4 | -1
If f(x)={sin⁡[x][x], for [x]≠00, for [x]=0f(x)=\left\{\begin{array}{cc}\frac{\sin [x]}{[x]}, & \text { for }[x] \neq 0 \\ 0, & \text { for }[x]=0\end{array}\right., where [x][x] denotes the greatest integer less than or equal to xx, then lim⁡x→0f(x)\lim \limits_{{x \rightarrow 0}} f(x) equals
View answer and explanation
Correct option: D
doesn't exist
Option D (doesn't exist) is correct because the left-hand limit and the right-hand limit at x=0 are not equal, indicating that the limit does not exist.
300 Easy
300 of 930
Application +4 | -1
The value of lim⁡x→0(1+sin⁡x)2cot⁡x\lim \limits_{{x \rightarrow 0}}(1+\sin x)^{2 \cot 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 trigonometric identities.
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