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JEE • MATHEMATICS • LIMITS

Limits practice · Mixed difficulty

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511 Medium
511 of 930
Analyzing +4 | -1
The value of lim⁡x→π/422−(cos⁡x+sin⁡x)31−sin⁡2x\lim\limits_{x \to \pi /4} \frac{{2\sqrt 2 - {{\left( {\cos x + \sin x} \right)}^3}}}{{1 - \sin 2x}}, is
View answer and explanation
Correct option: A
32\frac{3}{{\sqrt 2 }}
Option A (32\frac{3}{{\sqrt 2 }}) is correct because it is obtained by using a substitution and then evaluating the limit.
512 Medium
512 of 930
Analyzing +4 | -1
Evaluate lim⁡x→1x2−xx−1\lim \limits_{{x \rightarrow 1}} \frac{x^2-\sqrt{x}}{\sqrt{x}-1}.
View answer and explanation
Correct option: B
3
Option B (3) is correct because the limit is evaluated by factoring and simplifying the expression.
513 Easy
513 of 930
Analyzing +4 | -1
Let x=2x=2 be a root of the equation x2+px+q=0x^{2}+p x+q=0 and
f(x)−{1−cos⁡(x2−4px+q2+8q+16)(x−2p)4,x≠2p0,,x=2pf(x)-\left\{\begin{array}{cc} \frac{1-\cos \left(x^{2}-4 p x+q^{2}+8 q+16\right)}{(x-2 p)^{4}}, & x \neq 2 p \\ 0, & , x=2 p \end{array}\right.

Then lim⁡x→2p+[f(x)]\lim\limits_{x \to 2{p^ + }} [f(x)],
where [.] denotes greatest integer function, is
View answer and explanation
Correct option: C
0
Option C (0) is correct because the limit of the function as x approaches 2p from the right is 1/2, and the greatest integer function of 1/2 is 0.
514 Easy
514 of 930
Analyzing +4 | -1
lim⁡x→0(27+x)13−39−(27+x)23\lim _{x \rightarrow 0} \frac{(27+x)^{\frac{1}{3}}-3}{9-(27+x)^{\frac{2}{3}}} equals.
View answer and explanation
Correct option: C
−16-\frac{1}{6}
Option C (−16\frac{-1}{6}) is correct because after applying L'Hopital's rule, the limit evaluates to −16\frac{-1}{6}.
515 Medium
515 of 930
Analyzing +4 | -1
The value of lim⁡n→∞n−n2[(n+1)(n+12)(n+122)\lim \limits_{{n \rightarrow \infty}} n^{-n^2}\left[(n+1)\left(n+\frac{1}{2}\right)\left(n+\frac{1}{2^2}\right)\right. …(n+12n−1)]n\left.\ldots\left(n+\frac{1}{2^{n-1}}\right)\right]^n is
View answer and explanation
Correct option: B
e2e^2
Option B (e2e^2) is correct because the given limit can be simplified to an infinite product of exponentials, which converges to e2e^2 after summing the exponents.
516 Easy
516 of 930
Application +4 | -1
If lim⁡x→−ax9+a9x+a=9\lim \limits_{{x \rightarrow-a}} \frac{x^9+a^9}{x+a}=9, then the value of aa is
View answer and explanation
Correct option: B
−1-1
Option B (-1) is correct because the limit can be evaluated using the formula for the limit of (x^n + a^n)/(x+a) as x approaches -a.
517 Medium
517 of 930
Analyzing +4 | -1
The value of lim⁡x→2{[x3−4xx3−8)−1−(x+2xx−2−2x−2)−1}\lim\limits_{x \to 2} \left\{ {{{\left[ {\frac{{{x^3} - 4x}}{{{x^3} - 8}}} \right)}^{ - 1}} - {{\left( {\frac{{x + \sqrt {2x} }}{{x - 2}} - \frac{{\sqrt 2 }}{{\sqrt x - \sqrt 2 }}} \right)}^{ - 1}}} \right\} is
View answer and explanation
Correct option: A
12\frac{1}{2}
Option A (12\frac{1}{2}) is correct because after simplification and applying the limit, the expression evaluates to 12\frac{1}{2}.
518 Easy
518 of 930
Analyzing +4 | -1
lim⁡x→1[log⁡e{ex(x−1)−ex(1−x)}−log⁡e{4x(x−1)}]\lim \limits_{{x \rightarrow 1}}\left[\log _e\left\{e^{x(x-1)}-e^{x(1-x)}\right\}-\log _e\{4 x(x-1)\}\right] is equal to
View answer and explanation
Correct option: D
−log⁡e2-\log _e 2
The correct answer is D (−log⁡e2-\log _e 2) because after applying logarithmic properties and limits, the expression simplifies to −log⁡e2-\log _e 2.
519 Easy
519 of 930
Application +4 | -1
The value of lim⁡x→0(cos⁡x)cot⁡x\lim \limits_{{x \rightarrow 0}}(\cos x)^{\cot x}, is
View answer and explanation
Correct option: C
1
Option C (1) is correct because the limit can be evaluated by converting the cosine function to sine, then using the exponential form and trigonometric identities to simplify to 1.
520 Easy
520 of 930
Analyzing +4 | -1
lim⁡x→π2[1−tan⁡(x2)][1−sin⁡x][1+tan⁡(x2)][π−2x]3 is \lim _{x \rightarrow \frac{\pi}{2}} \frac{\left[1-\tan \left(\frac{x}{2}\right)\right][1-\sin x]}{\left[1+\tan \left(\frac{x}{2}\right)\right][\pi-2 x]^3} \text { is }
View answer and explanation
Correct option: D
132\frac{1}{32}
Option D (132\frac{1}{32}) is correct because after applying L'Hopital's rule and trigonometric identities, the limit evaluates to 132\frac{1}{32}.
521 Easy
521 of 930
Application +4 | -1
lim⁡x→02x−1(1+x)1/2−1=\lim \limits_{{x \rightarrow 0}} \frac{2^{x}-1}{(1+x)^{1 / 2}-1}=
View answer and explanation
Correct option: B
log⁡4\log 4
Option B (log⁡4\log 4) is correct because applying L'Hopital's rule yields 2log⁡22 \log 2, which simplifies to log⁡4\log 4.
522 Easy
522 of 930
Application +4 | -1
lim⁡x→1(1−x)tan⁡πx2=\lim _{x \rightarrow 1}(1-x) \tan \frac{\pi x}{2}=
View answer and explanation
Correct option: C
2π\frac{2}{\pi}
Option C (2π\frac{2}{\pi}) is correct because by substituting 1−x=θ1-x = \theta and using L'Hopital's rule or trigonometric identities, the limit evaluates to 2π\frac{2}{\pi}.
523 Easy
523 of 930
Analyzing +4 | -1
The value of lim⁡x→0a2−ax+x2−a2+ax+x2a+x−a−x\lim \limits_{{x \rightarrow 0}} \frac{\sqrt{a^2-a x+x^2}-\sqrt{a^2+a x+x^2}}{\sqrt{a+x}-\sqrt{a-x}}, is
View answer and explanation
Correct option: D
−a-\sqrt{a}
Option D (−a-\sqrt{a}) is correct because after rationalizing and simplifying the limit expression, the limit can be directly evaluated.
524 Easy
524 of 930
Analyzing +4 | -1
If lim⁡x→π/442−(cos⁡x+sin⁡x)51−sin⁡2x=\lim \limits_{{x \rightarrow \pi / 4}} \frac{4 \sqrt{2}-(\cos x+\sin x)^5}{1-\sin 2 x}=
View answer and explanation
Correct option: A
525 \sqrt{2}
Option A (525 \sqrt{2}) is correct because the limit is evaluated by using trigonometric identities and substitution to simplify the expression.
525 Easy
525 of 930
Analyzing +4 | -1
If lim⁡x→∞[x3+1x2+1−(ax+b)]=2\lim\limits_{x \to \infty } \left[ {\frac{{{x^3} + 1}}{{{x^2} + 1}} - \left( {ax + b} \right)} \right] = 2, then
View answer and explanation
Correct option: C
a=1a = 1 and b=−2b = - 2
Option C (a=1a = 1 and b=−2b = - 2) is correct because these values ensure that the limit exists and equals 2 when x approaches infinity.

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526 Easy
526 of 930
Analyzing +4 | -1
lim⁡x→−1x+1(17+x)1/4−2=\lim _{x \rightarrow-1} \frac{x+1}{(17+x)^{1 / 4}-2}=
View answer and explanation
Correct option: C
32
Option C (32) is correct because by substituting y=(17+x)1/4y = (17+x)^{1/4}, the limit is transformed and evaluated using the formula for the limit of xn−anx−a\frac{x^n - a^n}{x-a}.
527 Easy
527 of 930
Application +4 | -1
lim⁡x→a−(∣x∣3a−[xa]3)(a>0)\lim \limits_ {x \rightarrow a^{-}}\left(\frac{|x|^{3}}{a}-\left[\frac{x}{a}\right]^{3}\right)(a > 0), where [∗][*] denotes the greatest integer less than or equal to xx, is
View answer and explanation
Correct option: C
a2a^{2}
Option C (a2a^{2}) is correct because it is the limit obtained by substituting the value of x approaching a from the left into the given expression, considering the greatest integer function.
528 Easy
528 of 930
Understanding +4 | -1
The derivative of an odd function is always
View answer and explanation
Correct option: A
An even function
Option A (An even function) is correct because the derivative of an odd function is always an even function, which can be shown by differentiating the property f(-x) = -f(x).
529 Medium
529 of 930
Understanding +4 | -1
Let f(x)=[x]=f\left( x \right) = \left[ x \right]= Greatest integer less than or equal to xx and kk be an integer. Then, which one of the following is not correct?
View answer and explanation
Correct option: C
lim⁡x→kf(x)\lim\limits_{x \to k} f\left( x \right) exists
Option C (lim⁡x→kf(x)\lim\limits_{x \to k} f\left( x \right) exists) is correct because the limit does not exist as the left-hand limit and right-hand limit are not equal.
530 Easy
530 of 930
Analyzing +4 | -1
For each t∈Rt \in R let [t][t] be 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 \to 1 + } \frac{{(1 - \left| x \right| + \sin \left| {1 - x} \right|\sin \left( {\frac{\pi }{2}[1 - x]} \right)}}{{\left| {1 - x} \right|[1 - x]}}
View answer and explanation
Correct option: A
equals 0
Option A (equals 0) is correct because after substituting x = 1 + h and simplifying, the limit evaluates to 0.
531 Easy
531 of 930
Understanding +4 | -1
lim⁡x→0(Sin⁡−1xx)=\lim _{x \rightarrow 0}\left(\frac{\operatorname{Sin}^{-1} x}{x}\right)=
View answer and explanation
Correct option: B
1
Option B (1) is correct because the limit of (arcsin(x))/x as x approaches 0 is a standard limit equal to 1.
532 Easy
532 of 930
Analyzing +4 | -1
lim⁡x→0xlog⁡e(sin⁡x)\lim\limits_{x \to 0} x{\log _e}\left( {\sin x} \right) is equal to
View answer and explanation
Correct option: B
log⁡e1{\log _e}1
Option B (log⁡e1{\log _e}1) is correct because the limit evaluates to 0, and log⁡e1=0{\log _e}1 = 0.
533 Easy
533 of 930
Application +4 | -1
lim⁡x→+∞x(x2+1−x)=\lim _{x \rightarrow+\infty} x\left(\sqrt{x^2+1}-x\right)=
View answer and explanation
Correct option: A
12\frac{1}{2}
Option A (12\frac{1}{2}) is correct because after rationalizing and simplifying, the limit of the expression as x approaches infinity is 1/2.
534 Easy
534 of 930
Application +4 | -1
The value of lim⁡x→0(1−cos⁡2x)x2\lim\limits_{x \to 0} \frac{{\left( {1 - \cos 2x} \right)}}{{{x^2}}} is
View answer and explanation
Correct option: D
2
Option D (2) is correct because applying the trigonometric identity and the standard limit yields a value of 2.
535 Easy
535 of 930
Application +4 | -1
Let f(x)f(x) be polynomial of degree 4 with roots 1,2,3,41,2,3,4 and leading coefficient 1 and g(x)g(x) be the polynomial of degree 4 with roots 1,12,131, \frac{1}{2}, \frac{1}{3} and 14\frac{1}{4} with leading coefficient 1. Then, lim⁡x→1f(x)g(x)\lim \limits_{{x \rightarrow 1}} \frac{f(x)}{g(x)} equals
View answer and explanation
Correct option: B
−24-24
Option B (-24) is correct because the limit can be evaluated by factoring out (x-1) from both polynomials and then substituting x=1.
536 Medium
536 of 930
Application +4 | -1
lim⁡x→1tan⁡(x2−1)x−1\lim\limits_{x \to 1} \frac{{\tan \left( {{x^2} - 1} \right)}}{{x - 1}} is equal to
View answer and explanation
Correct option: A
2
Option A (2) is correct because by applying L'Hopital's rule, the limit can be evaluated to 2.
537 Easy
537 of 930
Understanding +4 | -1
The value of lim⁡x→1[sin⁡−1x]\lim \limits_{{x \rightarrow 1}}\left[\sin ^{-1} x\right] is
View answer and explanation
Correct option: B
1
Option B (1) is correct because as x approaches 1, sin⁡−1x\sin^{-1}x approaches π2\frac{{\pi}}{{2}}, and the greatest integer function of a value approaching π2\frac{{\pi}}{{2}} from below is 1, since π2≈1.57\frac{{\pi}}{{2}} \approx 1.57.
538 Medium
538 of 930
Analyzing +4 | -1
If lim⁡x→01−cos⁡xeax−bx−1=2\lim \limits_{{x \rightarrow 0}} \frac{1-\cos x}{e^{a x}-b x-1}=2 then the value of ab\frac{a}{b} is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because applying L'Hopital's rule twice and using the given limit, we find that a must equal b, hence a/b = 1.
539 Easy
539 of 930
Analyzing +4 | -1
If ff is a real-valued function satisfying the relation

f(x)+2f(1x)=3xf(x)+2 f\left(\frac{1}{x}\right)=3 x

for all real x≠0x \neq 0, then lim⁡x→0(sin⁡x)f(x)\lim \limits_{{x \rightarrow 0}}(\sin x) f(x) is equal to
View answer and explanation
Correct option: B
2
Option B (2) is correct because by substituting 1/x into the original equation, we can solve for f(x) and then evaluate the limit.
540 Medium
540 of 930
Analyzing +4 | -1
The value of lim⁡x→01+sin⁡x−cos⁡x+log⁡(1−x)x3\lim\limits_{x \to 0} \frac{{1 + \sin x - \cos x + \log \left( {1 - x} \right)}}{{{x^3}}}, is
View answer and explanation
Correct option: B
−12\frac{{ - 1}}{2}
Option B (−12\frac{{ - 1}}{2}) is correct because after expanding the trigonometric and logarithmic functions using their Maclaurin series and simplifying, the limit evaluates to -1/2.
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