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JEE • MATHEMATICS • LIMITS

Limits practice · Mixed difficulty

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391 Easy
391 of 930
Application +4 | -1
Evaluate lim⁡x→1f(x)\lim \limits_{{x \rightarrow 1}} f(x), where, 2x2≤f(x)≤x(x2+1),∀x2 x^2 \leq f(x) \leq x\left(x^2+1\right), \forall x that are near to 1 but not equal to 1.
View answer and explanation
Correct option: A
2
Option A (2) is correct because by the Sandwich theorem, the limit of f(x) as x approaches 1 is squeezed between two functions that both approach 2.
392 Medium
392 of 930
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.
393 Easy
393 of 930
Application +4 | -1
lim⁡x→0(1−ex)sin⁡xx2+x3\lim\limits_{x \to 0} \frac{{\left( {1 - {e^x}} \right)\sin x}}{{{x^2} + {x^3}}} is equal to
View answer and explanation
Correct option: A
−1- 1
Option A (−1-1) is correct because the limit can be evaluated by using Taylor series expansions for exe^x and sin⁡x\sin x or by applying L'Hopital's rule.
394 Medium
394 of 930
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.
395 Easy
395 of 930
Application +4 | -1
The value of lim⁡x→0{ax+bx+cx3}1/x\lim \limits_{{x \rightarrow 0}}\left\{\frac{a^x+b^x+c^x}{3}\right\}^{1 / x}, is
View answer and explanation
Correct option: B
(abc)1/3(a b c)^{1 / 3}
Option B ((abc)1/3(a b c)^{1 / 3}) is correct because the limit can be evaluated using the exponential form of limits and the standard limit lim⁡x→0ax−1x=ln⁡a\lim_{x \rightarrow 0} \frac{{a^x - 1}}{{x}} = \ln a.
396 Medium
396 of 930
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.
397 Easy
397 of 930
Application +4 | -1
lim⁡x→−1x3−2x−1x5−2x−1=\lim \limits_{{x \rightarrow-1}} \frac{x^3-2 x-1}{x^5-2 x-1}=
View answer and explanation
Correct option: B
13\frac{1}{3}
Option B (13\frac{1}{3}) is correct because by applying L'Hopital's rule or factoring, the limit evaluates to 13\frac{1}{3}.
398 Easy
398 of 930
Analyzing +4 | -1
lim⁡x→∞(3x2+2x+1x2+x+2)6x+13x+2\lim\limits_{x \to \infty } {\left( {\frac{{3{x^2} + 2x + 1}}{{{x^2} + x + 2}}} \right)^{\frac{{6x + 1}}{{3x + 2}}}} is equal to
View answer and explanation
Correct option: C
9
Option C (9) is correct because the limit can be evaluated by finding the limits of the base and exponent separately and then combining them.
399 Hard
399 of 930
Analyzing +4 | -1
For each x∈Rx \in R, let [x][x] be the greatest integer less than or equal to xx. Then, lim⁡x→0−x([x]+∣x∣)sin⁡[x]∣x∣\lim\limits_{x \to {0^ - }} \frac{{x([x] + \left| x \right|)\sin [x]}}{{\left| x \right|}} is equal to
View answer and explanation
Correct option: C
- sin 1
Option C (- sin 1) is correct because it is the limit of the given expression as x approaches 0 from the left, obtained by substituting the greatest integer function and absolute value.
400 Medium
400 of 930
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.
401 Easy
401 of 930
Analyzing +4 | -1
Among
(S 1): lim⁡n→∞1n2(2+4+6+…+2n)=1\lim \limits_{{n \rightarrow \infty}} \frac{1}{n^{2}}(2+4+6+\ldots+2 n)=1
(S2): lim⁡n→∞1n16(115+215+315+…+n15)=116\lim \limits_{{n \rightarrow \infty}} \frac{1}{n^{16}}\left(1^{15}+2^{15}+3^{15}+\ldots+n^{15}\right)=\frac{1}{16}
View answer and explanation
Correct option: A
Both (S1) and (S2) are true
Option A (Both (S1) and (S2) are true) is correct because both limits are evaluated correctly using summation formulas and limit properties.
402 Medium
402 of 930
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}.
403 Easy
403 of 930
Application +4 | -1
The value of lim⁡x→0eax−ebxx\lim \limits_{{x \rightarrow 0}} \frac{e^{a x}-e^{b x}}{x}, is
View answer and explanation
Correct option: B
a−ba-b
Option B (a−ba-b) is correct because the limit can be evaluated using the standard limit lim⁡x→0ex−1x=1\lim_{x \to 0} \frac{e^x - 1}{x} = 1.
404 Medium
404 of 930
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.
405 Easy
405 of 930
Analyzing +4 | -1
If lim⁡x→0{(a−n)nx−tan⁡x}sin⁡nxx2=0\lim \limits_{{x \rightarrow 0}} \frac{\{(a-n) n x-\tan x\} \sin n x}{x^2}=0, where nn is aa non-zero real number, then aa is equal to
View answer and explanation
Correct option: D
n+1nn+\frac{1}{n}
Option D (n+1nn+\frac{1}{n}) is correct because by applying L'Hopital's rule or simplifying the limit expression, we find that a must equal n+1nn+\frac{1}{n} for the limit to be 0.

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406 Medium
406 of 930
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.
407 Medium
407 of 930
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.
408 Easy
408 of 930
Analyzing +4 | -1
If lim⁡x→0(1+a3)+8e1/x1+(1−b3)e1/x=2\lim \limits_{{x \rightarrow 0}} \frac{\left(1+a^3\right)+8 e^{1 / x}}{1+\left(1-b^3\right) e^{1 / x}}=2, then
View answer and explanation
Correct option: A
a=1,b=(−3)1/3a=1, b=(-3)^{1 / 3}
Option A (a=1,b=(−3)1/3a=1, b=(-3)^{1 / 3}) is correct because by comparing coefficients after taking the limit as x approaches 0, we find that a=1a=1 and b=(−3)1/3b=(-3)^{1/3}.
409 Easy
409 of 930
Analyzing +4 | -1
If f(1)=g(1)=2f\left( 1 \right) = g\left( 1 \right) = 2, then lim⁡x→1f(1)g(x)−f(x)g(1)−f(1)+g(1)f(x)−g(x)\lim\limits_{x \to 1} \frac{{f\left( 1 \right)g\left( x \right) - f\left( x \right)g\left( 1 \right) - f\left( 1 \right) + g\left( 1 \right)}}{{f\left( x \right) - g\left( x \right)}} is equal to
View answer and explanation
Correct option: D
−2- 2
Option D (-2) is correct because simplifying the expression and applying the given conditions leads to the answer -2.
410 Easy
410 of 930
Application +4 | -1
If f(1)=2f\left( 1 \right) = 2 and f′(1)=1f'\left( 1 \right) = 1, then value of lim⁡x→1  2x−f(x)x−1\lim\limits_{x \to 1} \,\,\frac{{2x - f\left( x \right)}}{{x - 1}} is
View answer and explanation
Correct option: C
1
Option C (1) is correct because by applying L'Hopital's rule and using the given values of f(1) and f'(1), the limit evaluates to 1.
411 Easy
411 of 930
Application +4 | -1
The value of lim⁡h→0(a+h)2sin⁡(a+h)−a2sin⁡ah\lim\limits_{h \to 0} \frac{{{{\left( {a + h} \right)}^2}\sin \left( {a + h} \right) - {a^2}\sin a}}{h}, is
View answer and explanation
Correct option: A
2asin⁡a+a2cos⁡a2a\sin a + {a^2}\cos a
Option A (2asin⁡a+a2cos⁡a2a\sin a + {a^2}\cos a) is correct because it is the derivative of the function x2sin⁡xx^2\sin x evaluated at x=ax=a.
412 Medium
412 of 930
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}.
413 Easy
413 of 930
Application +4 | -1
lim⁡x→0∫0xtsin⁡(10t)dtx\lim\limits_{x \to 0} \frac{{\int_0^x {t\sin (10t)dt} }}{x} is equal to
View answer and explanation
Correct option: A
0
Option B (A−6IA - 6I) is correct because it is the result of using the characteristic equation and Cayley-Hamilton theorem to find 10A−110A^{-1}.
414 Easy
414 of 930
Application +4 | -1
If lim⁡x→1x2−ax+bx−1=5\lim _{x \rightarrow 1} \frac{x^2-a x+b}{x-1}=5; then a+ba+b is equal to:
View answer and explanation
Correct option: C
-7
Option C (-7) is correct because it is the sum of a and b that satisfies the given limit condition after applying L'Hopital's Rule.
415 Easy
415 of 930
Analyzing +4 | -1
Suppose that aa and b(b≠a)b(b \neq a) are real positive numbers, then the value of lim⁡t→0(bt+1−at+1b−a)1/t\lim \limits_{{t \rightarrow 0}}\left(\frac{b^{t+1}-a^{t+1}}{b-a}\right)^{1 / t} is equal to
View answer and explanation
Correct option: D
(bbaa)1b−a\left(\frac{b^b}{a^a}\right)^{\frac{1}{b-a}}
Option D ((bbaa)1b−a\left(\frac{b^b}{a^a}\right)^{\frac{1}{b-a}}) is correct because the limit is of the form 1∞1^{\infty}, and the expression is simplified using exponential and logarithmic properties.
416 Hard
416 of 930
Application +4 | -1
The value of Lim⁡n→∞(1n+m+1n+2m+1n+3m+….+1n+nm)\underset{n \rightarrow \infty}{\operatorname{Lim}}\left(\frac{1}{n+m}+\frac{1}{n+2 m}+\frac{1}{n+3 m}+\ldots .+\frac{1}{n+n m}\right) is equal to
[Note : where m>0\mathrm{m}>0.]
View answer and explanation
Correct option: C
ln⁡(1+m)m\frac{\ln (1+m)}{m}
Option C (ln⁡(1+m)m\frac{\ln (1+m)}{m}) is correct because it correctly evaluates the limit of the Riemann sum as a definite integral.
417 Easy
417 of 930
Application +4 | -1
If f(x)=3x2−xf(x)=3 x^2-x, then lim⁡h→0f(x+h)−f(x)h=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=
View answer and explanation
Correct option: D
6x−16 x-1
Option D (6x−16 x-1) is correct because it is the derivative of the function f(x)=3x2−xf(x)=3x^2-x calculated using the limit definition.
418 Medium
418 of 930
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.
419 Medium
419 of 930
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.
420 Easy
420 of 930
Analyzing +4 | -1
Let f(x)f(x) be a polynomial satisfying (f(α))2+(f′(α))2=0(f(\alpha))^2+\left(f^{\prime}(\alpha)\right)^2=0. Then, lim⁡x→αf(x)f′(x)[f′(x)f(x)]\lim \limits_{{x \rightarrow \alpha}} \frac{f(x)}{f^{\prime}(x)}\left[\frac{f^{\prime}(x)}{f(x)}\right] is equal to (Here [.] denotes the greatest integer function)
View answer and explanation
Correct option: B
1
Option B (1) is correct because after applying the given condition and simplifying the limit expression, it converges to 1.
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