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JEE • MATHEMATICS • LIMITS

Limits practice · Easy

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511 Easy
511 of 647
Application +4 | -1
lim⁡x→π4tan⁡x−1x−π4\lim\limits_{x \to \frac{\pi }{4}} \frac{{\tan x - 1}}{{x - \frac{\pi }{4}}} is equal to
View answer and explanation
Correct option: C
2
Option C (2) is correct because by applying L'Hopital's rule, the limit can be evaluated to 2.
512 Easy
512 of 647
Analyzing +4 | -1
lim⁡x→01−cos⁡2x2x\lim\limits_{x \to 0} \frac{{\sqrt {1 - \cos 2x} }}{{\sqrt 2 x}} is
View answer and explanation
Correct option: D
Does not exist
Option D (Does not exist) is correct because the left-hand limit and right-hand limit of the function as x approaches 0 are not equal.
513 Easy
513 of 647
Analyzing +4 | -1
Let f:R→Rf:R \to R be appositive increasing function with lim⁡x→∞f(3x)f(x)=1\lim\limits_{x \to \infty } \frac{{f\left( {3x} \right)}}{{f\left( x \right)}} = 1. Then, lim⁡x→∞f(2x)f(x)\lim\limits_{x \to \infty } \frac{{f\left( {2x} \right)}}{{f\left( x \right)}} is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because, by the Sandwich theorem, the limit of f(2x)/f(x) as x approaches infinity is squeezed between 1 and the limit of f(3x)/f(x), which is given as 1.
514 Easy
514 of 647
Application +4 | -1
lim⁡x→a3x−a−x+ax−a=\lim \limits_{{x \rightarrow a}} \frac{\sqrt{3 x-a}-\sqrt{x+a}}{x-a}=
View answer and explanation
Correct option: B
1/2a1 / \sqrt{2 a}
Option B (12a\frac{1}{\sqrt{2a}}) is correct because rationalizing the numerator and simplifying the expression, the limit evaluates to 12a\frac{1}{\sqrt{2a}}.
515 Easy
515 of 647
Analyzing +4 | -1
lim⁡x→0{log⁡e(1+x)x2+x−1x}\lim \limits_{{x \rightarrow 0}}\left\{\frac{\log _e(1+x)}{x^2}+\frac{x-1}{x}\right\} is equal to
View answer and explanation
Correct option: A
12\frac{1}{2}
Option A (12\frac{1}{2}) is correct because by using the series expansion of log(1+x), the limit evaluates to 12\frac{1}{2}.
516 Easy
516 of 647
Analyzing +4 | -1
lim⁡x→∞{121−x3+31+x2+521−x3+71+x2+…}\lim \limits_{{x \rightarrow \infty}}\left\{\frac{1^2}{1-x^3}+\frac{3}{1+x^2}+\frac{5^2}{1-x^3}+\frac{7}{1+x^2}+\ldots\right\} equals
View answer and explanation
Correct option: B
−103-\frac{10}{3}
Option B (-10/3) is correct because after simplifying the expression and taking the limit as x approaches infinity, the result is -10/3.
517 Easy
517 of 647
Application +4 | -1
lim⁡x→0(tan⁡(π4+x))1/x\lim\limits_{x \to 0} {\left( {\tan \left( {\frac{\pi }{4} + x} \right)} \right)^{1/x}} is equal to
View answer and explanation
Correct option: D
e2{e^2}
Option D (e2{e^2}) is correct because the limit is of the form 1^infinity, and using the exponential limit formula, the answer is e2.
518 Easy
518 of 647
Analyzing +4 | -1
Let f′′(x)f''\left( x \right) be continuous at x=0x = 0 and f′′(0)=4f''\left( 0 \right) = 4. Then lim⁡x→02f(x)−3f(2x)+f(4x)x2\lim\limits_{x \to 0} \frac{{2f\left( x \right) - 3f\left( {2x} \right) + f\left( {4x} \right)}}{{{x^2}}} is equal to
View answer and explanation
Correct option: C
12
Option C (12) is correct because applying L'Hopital's rule twice and using the given condition f′′(0)=4f''(0) = 4 results in the limit being equal to 12.
519 Easy
519 of 647
Analyzing +4 | -1
For x>0,lim⁡x→0{(sin⁡x)1/x+(1x)sin⁡x}x > 0, \lim \limits_{{x \rightarrow 0}}\left\{(\sin x)^{1 / x}+\left(\frac{1}{x}\right)^{\sin x}\right\}, is
View answer and explanation
Correct option: C
1
Option C (1) is correct because as x approaches 0, (sin⁡x)1/x(\sin x)^{1/x} approaches 0 and (1/x)sin⁡x(1/x)^{\sin x} approaches 1, so the sum approaches 1.
520 Easy
520 of 647
Application +4 | -1
lim⁡x→01−cos⁡mx1−cos⁡nx=\lim _{x \rightarrow 0} \frac{1-\cos m x}{1-\cos n x}=
View answer and explanation
Correct option: A
m2n2\frac{m^2}{n^2}
Option A (m2n2\frac{m^2}{n^2}) is correct because using the identity 1−cos⁡x=2sin⁡2(x/2)1 - \cos x = 2 \sin^2 (x/2) and the limit lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1, the limit evaluates to m2n2\frac{m^2}{n^2}.
521 Easy
521 of 647
Application +4 | -1
lim⁡x→∞exx3\lim \limits_{{x \rightarrow \infty}} \frac{e^x}{x^3} is
View answer and explanation
Correct option: D
∞\infty
Option D (∞\infty) is correct because by applying L'Hôpital's rule repeatedly, the limit simplifies to ∞\infty.
522 Easy
522 of 647
Application +4 | -1
The value of lim⁡x→0ex−(1+x)x2\lim\limits_{x \to 0} \frac{{{e^x} - \left( {1 + x} \right)}}{{{x^2}}}, is
View answer and explanation
Correct option: B
12\frac{1}{2}
Option B (12\frac{1}{2}) is correct because applying L'Hopital's rule twice to the limit results in 12\frac{1}{2}.
523 Easy
523 of 647
Application +4 | -1
lim⁡x→0  ∫0x2  sin⁡  t  dtx3\lim\limits_{x \to 0} \,\,\frac{{\int \nolimits_0^{{x^2}} \,\,\sin \,\,\sqrt t \,\,dt}}{{{x^3}}} is equal to
View answer and explanation
Correct option: C
2/32/3
Option C (2/32/3) is correct because applying L'Hospital's rule yields the limit as 2/32/3.
524 Easy
524 of 647
Application +4 | -1
lim⁡x→0log⁡(sin⁡7x+cos⁡7x)sin⁡3x\lim _{x \rightarrow 0} \frac{\log (\sin 7 x+\cos 7 x)}{\sin 3 x} equals :
View answer and explanation
Correct option: B
73\frac{7}{3}
Option B (73\frac{7}{3}) is correct because by using L'Hopital's rule and trigonometric identities, the limit evaluates to 7/3.
525 Easy
525 of 647
Application +4 | -1
If f(5)=7f\left( 5 \right) = 7 and f′(5)=7f'\left( 5 \right) = 7, then lim⁡x→5xf(5)−5f(x)x−5\lim\limits_{x \to 5} \frac{{xf\left( 5 \right) - 5f\left( x \right)}}{{x - 5}} is given by
View answer and explanation
Correct option: D
−28- 28
Option D (-28) is correct because by applying L'Hopital's rule, the limit can be evaluated to -28.

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526 Easy
526 of 647
Analyzing +4 | -1
Let the rr th term, trt_r of a series is given by tr=r1+r2+r4t_r=\frac{r}{1+r^2+r^4}. The value of lim⁡n→∞∑r=1ntr\lim \limits_{{n \rightarrow \infty}} \sum\limits_{{r=1}}^n t_r is
View answer and explanation
Correct option: B
12\frac{1}{2}
Option B (12\frac{1}{2}) is correct because the given series telescopes, and the limit of the sum as n approaches infinity is 12\frac{1}{2}.
527 Easy
527 of 647
Analyzing +4 | -1
If {x}\{x\} denotes the fractional part of xx, then lim⁡x→0{x}tan⁡{x}=\lim \limits_{{x \rightarrow 0}} \frac{\{x\}}{\tan \{x\}}=
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the left-hand limit and right-hand limit are not equal, hence the limit does not exist.
528 Easy
528 of 647
Analyzing +4 | -1
The value of lim⁡n→∞n2+1+nn3+n−4n4\lim\limits_{n \to \infty } \frac{{\sqrt {{n^2} + 1} + \sqrt n }}{{\sqrt[4]{{{n^3} + n - }}\sqrt[4]{n}}}, is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the limit approaches infinity after simplification.
529 Easy
529 of 647
Application +4 | -1
If lim⁡x→2xn−2nx−2=80\lim \limits_{{x \rightarrow 2}} \frac{x^n-2^n}{x-2}=80 and n∈Nn \in N, then the value of 'nn' is
View answer and explanation
Correct option: A
5
Option A is correct because the value of n is 5.
530 Easy
530 of 647
Application +4 | -1
lim⁡n→∞12+22+32+…+n2n3=\lim \limits_{{n \rightarrow \infty}} \frac{1^2+2^2+3^2+\ldots+n^2}{n^3}=
View answer and explanation
Correct option: B
13\frac{1}{3}
Option B (13\frac{1}{3}) is correct because the limit of the given expression as n approaches infinity is equal to 13\frac{1}{3}.
531 Easy
531 of 647
Application +4 | -1
lim⁡x→0(cosec⁡x)x\lim \limits_{{x \rightarrow 0}}(\operatorname{cosec} x)^x is
View answer and explanation
Correct option: A
1
Option A (1) is correct because by taking the natural logarithm and applying L'Hôpital's rule, the limit simplifies to 1.
532 Easy
532 of 647
Analyzing +4 | -1
For each t∈Rt \in R, let [t] denote the greatest integer less than or equal to t. Then,
lim⁡x→1+(1−∣x∣+sin⁡∣1−x∣)sin⁡(π2[1−x])∣1−x∣[1−x]\lim \limits_ {x \rightarrow 1^{+}} \frac{(1-|x|+\sin |1-x|) \sin \left(\frac{\pi}{2}[1-x]\right)}{|1-x|[1-x]}
View answer and explanation
Correct option: D
Equals 0
Option D (Equals 0) is correct because after simplifying the expression and evaluating the limit as x approaches 1 from the right, the result is 0.
533 Easy
533 of 647
Understanding +4 | -1
The value of lim⁡x→0[sin⁡xx]\lim \limits_{{x \rightarrow 0}}\left[\frac{\sin x}{x}\right] is
View answer and explanation
Correct option: A
0
Option A (0) is correct because as x approaches 0, sin⁡xx\frac{{\sin x}}{{x}} approaches 1 from below, and the greatest integer function of a value approaching 1 from below is 0.
534 Easy
534 of 647
Application +4 | -1
If lim⁡x→∞(x+cx−c)x=4\lim \limits_{{x \rightarrow \infty}}\left(\frac{x+c}{x-c}\right)^x=4, then the value of cc is
View answer and explanation
Correct option: B
log⁡e2\log _e 2
Option B (log⁡e2\log _e 2) 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.
535 Easy
535 of 647
Analyzing +4 | -1
lim⁡x→01+sin⁡2x3−1−2tan⁡x4sin⁡x+tan⁡2x\lim \limits_{{x \rightarrow 0}} \frac{\sqrt[3]{1+\sin ^2 x}-\sqrt[4]{1-2 \tan x}}{\sin x+\tan ^2 x} is equal to
View answer and explanation
Correct option: C
12\frac{1}{2}
Option C (12\frac{{1}}{{2}}) is correct because using approximations for small x, the limit simplifies to 12\frac{{1}}{{2}}.
536 Easy
536 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: A
log⁡34\log _3 4
Option A (log⁡34\log_3 4) is correct because the limit can be evaluated using the standard limit lim⁡x→0ax−1x=ln⁡a\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a.
537 Easy
537 of 647
Analyzing +4 | -1
The value of lim⁡x→0(4x−1)3sin⁡x24log⁡(1+3x)\lim \limits_{{x \rightarrow 0}} \frac{\left(4^x-1\right)^3}{\sin \frac{x^2}{4} \log (1+3 x)}, is
View answer and explanation
Correct option: B
43(ln⁡4)3\frac{4}{3}(\ln 4)^3
Option B (43(ln⁡4)3\frac{{4}}{{3}}(\ln 4)^3) is correct because the limit is evaluated using standard limit results, resulting in 43(ln⁡4)3\frac{{4}}{{3}}(\ln 4)^3.
538 Easy
538 of 647
Application +4 | -1
The value of lim⁡x→1log⁡xsin⁡πx\lim \limits_{{x \rightarrow 1}} \frac{\log x}{\sin \pi x}, is
View answer and explanation
Correct option: D
−1π-\frac{1}{\pi}
Option D (−1π-\frac{1}{\pi}) is correct because the limit can be evaluated using the standard limit lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1 and lim⁡x→1ln⁡xx−1=1\lim_{x \to 1} \frac{\ln x}{x-1} = 1.
539 Easy
539 of 647
Analyzing +4 | -1
If lim⁡x→0log⁡(x+a)−log⁡ax+klim⁡x→elog⁡x−1x−e=1\lim\limits_{x \to 0} \frac{{\log \left( {x + a} \right) - \log a}}{x} + k\lim\limits_{x \to e} \frac{{\log x - 1}}{{x - e}} = 1, then the value of kk is
View answer and explanation
Correct option: C
e(1−1a)e\left( {1 - \frac{1}{a}} \right)
Option C (e(1−1a)e(1 - \frac{1}{a})) is correct because it is obtained by applying L'Hopital's rule to both limits and solving for k.
540 Easy
540 of 647
Analyzing +4 | -1
The value of lim⁡h→02{3sin⁡(π6+h)−cos⁡(π6+h)}h(3sinh⁡−cosh⁡)\lim \limits_{{h \rightarrow 0}} \frac{2\left\{\sqrt{3} \sin \left(\frac{\pi}{6}+h\right)-\cos \left(\frac{\pi}{6}+h\right)\right\}}{h(\sqrt{3} \sinh -\cosh )} is
View answer and explanation
Correct option: A
−4-4
The correct answer is A (-4) because by applying trigonometric identities and limit properties, the expression simplifies to -4.
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