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JEE • MATHEMATICS • LIMITS

Limits practice · Easy

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271 Easy
271 of 647
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.
272 Easy
272 of 647
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.
273 Easy
273 of 647
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.
274 Easy
274 of 647
Understanding +4 | -1
Let f(x)=[sin⁡xx],x≠0f(x)=\left[\frac{\sin x}{x}\right], x \neq 0, where [.] denotes the greatest integer function then lim⁡x→0f(x)\lim \limits_{{x \rightarrow 0}} f(x)
View answer and explanation
Correct option: C
Is equal to 0
Option C (Is equal to 0) is correct because as x approaches 0, sin⁡xx\frac{\sin x}{x} approaches 1, but remains less than 1, so the greatest integer function of it is 0.
275 Easy
275 of 647
Analyzing +4 | -1
The value of lim⁡x→0{11sin⁡2x+21sin⁡2x+31sin⁡2x+….+n1sin⁡2x}sin⁡2x,\lim \limits_{{x \rightarrow 0}}\left\{1^{\frac{1}{\sin ^2 x}+} 2^{\frac{1}{\sin ^2 x}}+3^{\frac{1}{\sin ^2 x}}+\ldots .+n^{\frac{1}{\sin ^2 x}}\right\}^{\sin ^2 x}, is
View answer and explanation
Correct option: C
nn
Option C (n) is correct because, as x approaches 0, the expression simplifies to n.
276 Easy
276 of 647
Application +4 | -1
lim⁡x→04x−13x−1\lim \limits_{{x \rightarrow 0}} \frac{4^x-1}{3^x-1} equals
View answer and explanation
Correct option: D
log⁡34\log _3 4
Option D (log⁡34\log _3 4) is correct because the limit is evaluated using the standard limit lim⁡x→0ax−1x=log⁡a\lim_{x \rightarrow 0} \frac{a^x - 1}{x} = \log a.
277 Easy
277 of 647
Analyzing +4 | -1
lim⁡x→π482−(cos⁡x+sin⁡x)72−2sin⁡2x\lim\limits_{x \to \frac{\pi }{4}} \frac{{8\sqrt 2 - {{\left( {\cos x + \sin x} \right)}^7}}}{{\sqrt 2 - \sqrt 2 \sin 2x}} is equal to
View answer and explanation
Correct option: A
14
Option A (14) is correct because applying L'Hopital's rule twice simplifies the limit to a determinate form.
278 Easy
278 of 647
Application +4 | -1
If f(9)=9,f′(9)=4f(9)=9, f^{\prime}(9)=4, then lim⁡x→9f(x)−3x−3=\lim \limits_{{x \rightarrow 9}} \frac{\sqrt{f(x)-3}}{\sqrt{x}-3}=
View answer and explanation
Correct option: B
4
Option B (4) is correct because applying L'Hopital's rule and using the given values of f(9)f(9) and f′(9)f'(9) yields the limit as 4.
279 Easy
279 of 647
Application +4 | -1
lim⁡x→π6  sin⁡2xsin⁡x\lim\limits_{x \to \frac{\pi }{6}} \,\,\frac{{\sin 2x}}{{\sin x}} is equal to
View answer and explanation
Correct option: A
3\sqrt 3
Option A (3\sqrt 3) is correct because by simplifying the expression using trigonometric identities, the limit evaluates to the square root of 3.
280 Easy
280 of 647
Application +4 | -1
If lim⁡x→∞(1+ax+bx2)2x=e2\lim \limits_{{x \rightarrow \infty}}\left(1+\frac{a}{x}+\frac{b}{x^2}\right)^{2 x}=e^2, then
View answer and explanation
Correct option: C
a=1,b∈Ra=1, b \in R
Option C (a=1,b∈Ra=1, b \in R) is correct because by evaluating the limit, we find that a=1a=1, while bb can be any real number.
281 Easy
281 of 647
Application +4 | -1
The value of lim⁡x→11+log⁡x−x1−2x+x2\lim \limits_{{x \rightarrow 1}} \frac{1+\log x-x}{1-2 x+x^2} is
View answer and explanation
Correct option: B
−(1/2)-(1 / 2)
Option B (−(1/2)-(1 / 2)) is correct because L'Hôpital's rule is applied to resolve the indeterminate form of the limit.
282 Easy
282 of 647
Application +4 | -1
Let f : R→Rf\,:\,R \to R be a function such that f(2)=4f\left( 2 \right) = 4 and f′(2)=1f'\left( 2 \right) = 1. Then, the value of lim⁡x→2x2f(2)−4f(x)x−2\lim\limits_{x \to 2} \frac{{{x^2}f\left( 2 \right) - 4f\left( x \right)}}{{x - 2}} is equal to
View answer and explanation
Correct option: D
12
Option D (12) is correct because applying L'Hopital's rule to the limit and using the given values of f(2) and f'(2), we can find the value of the limit.
283 Easy
283 of 647
Application +4 | -1
lim⁡x→12x2+x−33x3−3x2+2x−2\lim\limits_{x \to 1} \frac{{2{x^2} + x - 3}}{{3{x^3} - 3{x^2} + 2x - 2}} is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because it is the result of factoring the numerator and denominator, canceling the common factor, and then evaluating the limit.
284 Easy
284 of 647
Application +4 | -1
lim⁡x→0e5x−e4xx\lim\limits_{x \to 0} \frac{{{e^{5x}} - {e^{4x}}}}{x} is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because by using the series expansion of e^x or L'Hopital's rule, the limit evaluates to 1.
285 Easy
285 of 647
Application +4 | -1
lim⁡x→π3sin⁡(π3−x)2cos⁡x−1\lim \limits_{x \rightarrow \frac{\pi}{3}} \frac{\sin \left(\frac{\pi}{3}-x\right)}{2 \cos x-1} is equal to:
View answer and explanation
Correct option: B
13\frac{1}{\sqrt{3}}
Option B (1/sqrt(3)) is correct because after applying L'Hopital's rule or trigonometric identities, the limit evaluates to 1/sqrt(3).

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286 Easy
286 of 647
Analyzing +4 | -1
The value of the constant α\alpha and β\beta such that lim⁡x→∞(x2+1x+1−αx−β)=0\lim\limits_{x \to \infty } \left( {\frac{{{x^2} + 1}}{{x + 1}} - \alpha x - \beta } \right) = 0 are respectively
View answer and explanation
Correct option: C
(1,−1)\left( {1, - 1} \right)
Option C ((1, -1)) is correct because by equating coefficients after rationalizing, we find α=1\alpha = 1 and β=−1\beta = -1.
287 Easy
287 of 647
Analyzing +4 | -1
If lim⁡x→0αex+βe−x+γsin⁡xxsin⁡2x=23\lim _{x \rightarrow 0} \frac{\alpha e^x+\beta e^{-x}+\gamma \sin x}{x \sin ^2 x}=\frac{2}{3}, where α,β,γ∈R\alpha, \beta, \gamma \in \mathbf{R}, then which of the following is NOT correct?
View answer and explanation
Correct option: C
αβ2+βγ2+γα2+3=0\alpha \beta^2+\beta \gamma^2+\gamma \alpha^2+3=0
Option C (αβ2+βγ2+γα2+3=0\alpha \beta^2+\beta \gamma^2+\gamma \alpha^2+3=0) is NOT correct because after finding the values of α\alpha, β\beta, and γ\gamma, substituting them into the expression does not result in 0.
288 Easy
288 of 647
Analyzing +4 | -1
If lim⁡x→aax−xaxx−aa=−1\lim \limits_{{x \rightarrow a}} \frac{a^x-x^a}{x^x-a^a}=-1, then the value of aa is
View answer and explanation
Correct option: A
1
Option A (1) is correct because after applying L'Hopital's rule and simplifying, the equation holds true only when a = 1.
289 Easy
289 of 647
Application +4 | -1
The value of lim⁡x→∞(3x−43x+2)x+13\lim\limits_{x \to \infty } {\left( {\frac{{3x - 4}}{{3x + 2}}} \right)^{\frac{{x + 1}}{3}}} is equal to
View answer and explanation
Correct option: B
e−2/3{e^{ - 2/3}}
Option B (e−2/3{e^{ - 2/3}}) is correct because by rewriting the expression and taking the limit as x approaches infinity, the result is e−2/3{e^{ - 2/3}}.
290 Easy
290 of 647
Analyzing +4 | -1
The value of lim⁡h→0ln⁡(1+2h)−2ln(1+h)h2\lim\limits_{h \to 0} \frac{{\ln \left( {1 + 2h} \right) - 2l\mathrm{n}\left( {1 + h} \right)}}{{{h^2}}} is
View answer and explanation
Correct option: B
−1- 1
Option B (−1-1) is correct because using logarithm properties and limit evaluation, the expression simplifies to −1-1.
291 Easy
291 of 647
Application +4 | -1
lim⁡x→π/6sin⁡(x−π/6)3/2−cos⁡x\lim \limits_{{x \rightarrow \pi / 6}} \frac{\sin (x-\pi / 6)}{\sqrt{3} / 2-\cos x} is equal to
View answer and explanation
Correct option: A
2
Option A (2) is correct because applying L'Hôpital's rule, lim⁡x→π/6sin⁡(x−π/6)3/2−cos⁡x=lim⁡x→π/6cos⁡(x−π/6)sin⁡x=cos⁡(0)sin⁡(π/6)=11/2=2\lim_{x \rightarrow \pi / 6} \frac{\sin (x-\pi / 6)}{\sqrt{3} / 2-\cos x} = \lim_{x \rightarrow \pi / 6} \frac{\cos (x-\pi / 6)}{\sin x} = \frac{\cos(0)}{\sin(\pi/6)} = \frac{1}{1/2} = 2.
292 Easy
292 of 647
Analyzing +4 | -1
If lim⁡x→∞{x2+1x+1−ax−b}=0\lim \limits_{{x \rightarrow \infty}}\left\{\frac{x^2+1}{x+1}-a x-b\right\}=0, then
View answer and explanation
Correct option: C
a=1,b=−1a=1, b=-1
Option C (a=1,b=−1a=1, b=-1) is correct because for the limit to be 0, the coefficients of x2x^2 and xx in the numerator must be 0.
293 Easy
293 of 647
Analyzing +4 | -1
The value of lim⁡n→∞cos⁡(x2)cos⁡(x4)cos⁡(x8)…cos⁡(x2n)\lim\limits_{n \to \infty } \cos \left( {\frac{x}{2}} \right)\cos \left( {\frac{x}{4}} \right)\cos \left( {\frac{x}{8}} \right) \ldots \cos \left( {\frac{x}{{{2^n}}}} \right), is
View answer and explanation
Correct option: B
sin⁡xx\frac{{\sin x}}{x}
Option B (sin⁡xx\frac{{\sin x}}{x}) is correct because using the trigonometric identity and limit properties, the expression simplifies to the given form.
294 Easy
294 of 647
Application +4 | -1
lim⁡x→∞(x+5x−1)x\lim \limits_{{x \rightarrow \infty}}\left(\frac{x+5}{x-1}\right)^x is equal to
View answer and explanation
Correct option: B
e6e^6
Option B (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.
295 Easy
295 of 647
Analyzing +4 | -1
lim⁡x→0(1−cos⁡2x)(3+cos⁡x)xtan⁡4x\lim \limits_{{x \rightarrow 0}} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x} is equal to
View answer and explanation
Correct option: D
2
Option D (2) is correct because the limit is evaluated using trigonometric identities and standard limit results, resulting in 2.
296 Easy
296 of 647
Analyzing +4 | -1
lim⁡x→0sin⁡(πcos⁡2x)x2\lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^2 x\right)}{x^2} is equal to:
View answer and explanation
Correct option: B
π\pi
Option B (π\pi) is correct because the limit can be simplified using trigonometric identities and standard limits.
297 Easy
297 of 647
Application +4 | -1
lim⁡x→0[2x−11+x−1]\lim\limits_{x \to 0} \left[ {\frac{{{2^x} - 1}}{{\sqrt {1 + x} - 1}}} \right] is equal to
View answer and explanation
Correct option: C
log⁡e4{\log _e}4
Option C (log⁡e4\log_e 4) is correct because by applying L'Hopital's rule, the limit evaluates to ln(4).
298 Easy
298 of 647
Analyzing +4 | -1
lim⁡n→∞{(nn+1)α+sin⁡1n}n\lim \limits_{{n \rightarrow \infty}}\left\{\left(\frac{n}{n+1}\right)^\alpha+\sin \frac{1}{n}\right\}^n, when α∈Q\alpha \in Q is equal to
View answer and explanation
Correct option: C
e1−αe^{1-\alpha}
Option C (e1−αe^{1-\alpha}) is correct because the limit can be evaluated by using the binomial expansion and the limit definition of e.
299 Easy
299 of 647
Application +4 | -1
lim⁡x→0ex2−cos⁡xsin⁡2x\lim _{x \rightarrow 0} \frac{e^{x^2}-\cos x}{\sin ^2 x} is equal to:
View answer and explanation
Correct option: C
32\frac{3}{2}
Option C (32\frac{3}{2}) is correct because it is the value of the limit after applying L'Hopital's Rule twice.
300 Easy
300 of 647
Analyzing +4 | -1
lim⁡x→0sin⁡xn(sin⁡x)m,  (m<n)\lim\limits_{x \to 0} \frac{{\sin {x^n}}}{{{{\left( {\sin x} \right)}^m}}},\;(m < n) is equal to
View answer and explanation
Correct option: B
0
Option B (0) is correct because, as x approaches 0, the expression simplifies to x(n-m), and since m < n, the limit is 0.
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