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JEE • MATHEMATICS • LIMITS

Limits practice · Mixed difficulty

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Showing 421-450 of 930 questions

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421 Easy
421 of 930
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.
422 Easy
422 of 930
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.
423 Easy
423 of 930
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.
424 Easy
424 of 930
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.
425 Easy
425 of 930
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.
426 Hard
426 of 930
Application +4 | -1
lim⁡n→∞∑r=1n1nern\lim _{n \rightarrow \infty} \sum\limits_{{r=1}}^{n} \frac{1}{n} e^{\frac{r}{n}}
View answer and explanation
Correct option: B
e−1e-1
Option B (e-1) is correct because the limit of the Riemann sum can be evaluated as a definite integral, which simplifies to e-1.
427 Medium
427 of 930
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.
428 Easy
428 of 930
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.
429 Easy
429 of 930
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.
430 Easy
430 of 930
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.
431 Easy
431 of 930
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.
432 Easy
432 of 930
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.
433 Easy
433 of 930
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.
434 Easy
434 of 930
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).
435 Easy
435 of 930
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.

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436 Medium
436 of 930
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.
437 Easy
437 of 930
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.
438 Easy
438 of 930
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.
439 Easy
439 of 930
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}}.
440 Easy
440 of 930
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.
441 Hard
441 of 930
Analyzing +4 | -1
lim⁡x→−∞x5tan⁡(1πx2)+3∣x∣2+7∣x∣3+7∣x∣+8\lim\limits_{{x \rightarrow -\infty}}\frac{x^5 \tan \left(\frac{1}{\pi x^2}\right)+3|x|^2+7}{|x|^3+7|x|+8} is equal to 
View answer and explanation
Correct option: A
−1π-\frac{1}{\pi}
Option A (-1/π) is correct because after dividing by |x|^3 and using the limit tan(θ)/θ as θ approaches 0, the expression simplifies to -1/π.
442 Medium
442 of 930
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}.
443 Easy
443 of 930
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.
444 Medium
444 of 930
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.
445 Medium
445 of 930
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.
446 Easy
446 of 930
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.
447 Easy
447 of 930
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.
448 Easy
448 of 930
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.
449 Easy
449 of 930
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.
450 Easy
450 of 930
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.
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