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JEE • MATHEMATICS • LIMITS

Limits practice · Mixed difficulty

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781 Medium
781 of 930
Application +4 | -1
lim⁡x→∞3x[x4]=pq\lim \limits_{{x \rightarrow \infty}} \frac{3}{x}\left[\frac{x}{4}\right]=\frac{p}{q} (where [.] denotes greatest integer function), then p+qp+q (where p,qp, q are relative prime) is:
View answer and explanation
Correct option: B
7
Option B (7) is correct because the limit simplifies to 3/4, and the sum of the numerator and denominator is 7.
782 Easy
782 of 930
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.
783 Medium
783 of 930
Application +4 | -1
The derivative of x2/3x^{2 / 3} by using first principle is
View answer and explanation
Correct option: A
23x−1/3\frac{2}{3} x^{-1 / 3}
Option A (23x−1/3\frac{2}{3} x^{-1 / 3}) is correct because it is the derivative of x2/3x^{2/3} obtained using the first principle of differentiation.
784 Medium
784 of 930
Analyzing +4 | -1
lim⁡x→0xz2−(z−x)23(8xz−4x23+8xz3)4\lim \limits_{{x \rightarrow 0}} \frac{x \sqrt[3]{z^2-(z-x)^2}}{\left(\sqrt[3]{8 x z-4 x^2}+\sqrt[3]{8 x z}\right)^4} is equal to
View answer and explanation
Correct option: B
1223/3⋅z\frac{1}{2^{23 / 3} \cdot z}
Option B (1223/3⋅z\frac{1}{2^{23 / 3} \cdot z}) is correct because it is the result of simplifying the limit expression by factoring and canceling terms.
785 Easy
785 of 930
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.
786 Easy
786 of 930
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.
787 Easy
787 of 930
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}}.
788 Easy
788 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: 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.
789 Easy
789 of 930
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.
790 Easy
790 of 930
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.
791 Easy
791 of 930
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.
792 Easy
792 of 930
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.
793 Easy
793 of 930
Analyzing +4 | -1
If lim⁡x→aax−xaxx−aa=−1\lim\limits_{x \to a} \frac{{{a^x} - {x^a}}}{{{x^x} - {a^a}}} = - 1, then aa equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because by applying L'Hospital's rule, the value of 'a' is found to be 1.
794 Easy
794 of 930
Application +4 | -1
The value of lim⁡n→∞1⋅2+2⋅3+3⋅4+…+n(n+1)n3\lim \limits_{{n \rightarrow \infty}} \frac{1 \cdot 2+2 \cdot 3+3 \cdot 4+\ldots+n(n+1)}{n^3} 30, is
View answer and explanation
Correct option: C
13\frac{1}{3}
Option C (13\frac{1}{3}) is correct because the sum in the numerator simplifies to n(n+1)(n+2)3\frac{n(n+1)(n+2)}{3}, and dividing by n3n^3 and taking the limit as nn approaches infinity yields 13\frac{1}{3}.
795 Easy
795 of 930
Understanding +4 | -1
If f(x)={xsin⁡1x,x≠00,x=0f(x)=\left\{\begin{array}{rr}x \sin \frac{1}{x}, & x \neq 0 \\ 0, & x=0\end{array}\right., then lim⁡x→0f(x)=\lim \limits_{{x \rightarrow 0}} f(x)=
View answer and explanation
Correct option: B
0
Option B (0) is correct because the limit of the function as x approaches 0 is 0, as shown by the Squeeze Theorem.

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796 Easy
796 of 930
Analyzing +4 | -1
If α\alpha and β\beta are roots of x2−(1−cos⁡2θ)x+θ=0x^2-(\sqrt{1-\cos 2 \theta}) x+\theta=0, where 0<θ<π20<\theta<\frac{\pi}{2}. Then, lim⁡θ→0+(1α+1β)\lim \limits_{\theta \rightarrow 0^{+}}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right) is
View answer and explanation
Correct option: C
2\sqrt{2}
The correct answer is C (2\sqrt{2}) because by simplifying the expression and using the limit definition, the value can be determined as 2\sqrt{2}.
797 Easy
797 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 a=a=
View answer and explanation
Correct option: A
±1\pm 1
Option C (1/2) is correct because simplifying the expression using the sum of arithmetic series leads to a limit of 1/2 as n approaches infinity.
798 Medium
798 of 930
Understanding +4 | -1
lim⁡x→0xtan⁡−12x\lim\limits_{x \to 0} \frac{x}{{{{\tan }^{ - 1}}2x}} is equal to
View answer and explanation
Correct option: B
12\frac{1}{2}
Option B (12\frac{1}{2}) is correct because applying limit properties and the standard limit formula for inverse tangent yields the result.
799 Easy
799 of 930
Analyzing +4 | -1
If lim⁡x→1asin⁡(x−1)+bcos⁡(x−1)+4x2−1=2\lim \limits_{{x \rightarrow 1}} \frac{a \sin (x-1)+b \cos (x-1)+4}{x^2-1}=2, then (a,b)(a, b) is equal to
View answer and explanation
Correct option: D
(4,−4)(4,-4)
Option B (3x+4y-5=0) is correct because the values of a and b obtained from the limit condition satisfy this equation.
800 Easy
800 of 930
Analyzing +4 | -1
If lim⁡x→∞{x2+1x+1−(ax+b)}→∞\lim \limits_{{x \rightarrow \infty}}\left\{\frac{x^2+1}{x+1}-(a x+b)\right\} \rightarrow \infty, then
View answer and explanation
Correct option: C
a∈(−∞,1)a \in(-\infty, 1)
Option C (a∈(−∞,1)a \in(-\infty, 1)) is correct because for the limit to be infinite, the degree of numerator must be greater than the degree of the denominator, implying 1−a>01-a > 0.
801 Easy
801 of 930
Application +4 | -1
lim⁡x→0(1+6x21+3x2)1/x2=\lim _{x \rightarrow 0}\left(\frac{1+6 x^2}{1+3 x^2}\right)^{1 / x^2}=
View answer and explanation
Correct option: C
e3e^3
Option C (e3e^3) is correct because it is obtained by applying the limit formula lim⁡x→0(1+ax)1/x=ea\lim_{x \rightarrow 0} (1 + ax)^{1/x} = e^a to both the numerator and denominator.
802 Medium
802 of 930
Analyzing +4 | -1
The value of lim⁡x→0(sin⁡xx)sin⁡xx−sin⁡x\lim\limits_{x \to 0} {\left( {\frac{{\sin x}}{x}} \right)^{\frac{{\sin x}}{{x - \sin x}}}}, is
View answer and explanation
Correct option: A
e−1{e^{ - 1}}
Option A (e−1{e^{ - 1}}) is correct because using the exponential limit formula and simplifying the expression leads to the correct result.
803 Easy
803 of 930
Application +4 | -1
The value of lim⁡x→π2(sin⁡x)tan⁡x\lim\limits_{x \to \frac{\pi }{2}} {\left( {\sin x} \right)^{\tan x}} is
View answer and explanation
Correct option: A
1
Option A (1) is correct because by taking the logarithm and applying L'Hopital's rule, the limit simplifies to 1.
804 Easy
804 of 930
Application +4 | -1
If f(1)=1,f′(1)=2f(1)=1, f^{\prime}(1)=2 then lim⁡x→1f(x)−1x−1\lim \limits_{{x \rightarrow 1}} \frac{\sqrt{f(x)}-1}{\sqrt{x}-1} is
View answer and explanation
Correct option: A
2
Option A (2) is correct because the limit can be evaluated using L'Hopital's rule or by algebraic manipulation and the given derivative.
805 Easy
805 of 930
Analyzing +4 | -1
If lim⁡x→∞{x2+1x+1−ax−b}=2\lim \limits_{{x \rightarrow \infty}}\left\{\frac{x^2+1}{x+1}-a x-b\right\}=2, then
View answer and explanation
Correct option: D
a=1,b=−3a=1, b=-3
Option D (a=1,b=−3a=1, b=-3) is correct because for the limit to be 2, the coefficients of x2x^2 must be 0 and the ratio of the coefficients of xx must be 2.
806 Easy
806 of 930
Application +4 | -1
The value of lim⁡x→∞xcos⁡(π4x)sin⁡(π4x)\lim\limits_{x \to \infty } x\cos \left( {\frac{\pi }{{4x}}} \right)\sin \left( {\frac{\pi }{{4x}}} \right), is
View answer and explanation
Correct option: B
π4\frac{\pi }{4}
Option B (π4\frac{\pi }{4}) is correct because by using the identity 2sin⁡xcos⁡x=sin⁡2x2\sin x \cos x = \sin 2x and the limit lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{{\sin x}}{x} = 1, the given limit simplifies to π4\frac{\pi }{4}.
807 Easy
807 of 930
Application +4 | -1
lim⁡x→a(a+2x)1/3−(3x)1/3(3a+x)1/3−(4x)1/3,(a≠0)\lim \limits_{{x \rightarrow a}} \frac{(a+2 x)^{1 / 3}-(3 x)^{1 / 3}}{(3 a+x)^{1 / 3}-(4 x)^{1 / 3}},(a \neq 0) is equal to
View answer and explanation
Correct option: D
23(29)1/3\frac{2}{3}\left(\frac{2}{9}\right)^{1 / 3}
Option D, 23(29)1/3\frac{2}{3}(\frac{2}{9})^{1 / 3}, is correct because it is the result of applying L'Hopital's rule to the limit and simplifying the expression.
808 Medium
808 of 930
Analyzing +4 | -1
If ana_n and bnb_n are positive integers and an+2bna_n+\sqrt{2} b_n =(2+2)n=(2+\sqrt{2})^n, then
lim⁡n→−(anbn)=\lim \limits_{{n \rightarrow-}}\left(\frac{a_n}{b_n}\right)=
View answer and explanation
Correct option: B
2\sqrt{2}
Option B (2\sqrt{2}) is correct because after finding expressions for ana_n and bnb_n and taking the limit of their ratio, the result is 2\sqrt{2}.
809 Medium
809 of 930
Analyzing +4 | -1
The value of lim⁡x→1(x2−1)sin⁡2(πx)x4−2x3+2x−1\lim\limits _{x \to 1} \frac{\left(x^2-1\right) \sin ^2(\pi x)}{x^4-2 x^3+2 x-1} is equal to
View answer and explanation
Correct option: D
π2\pi^2
Option D (π2\pi^2) is correct because after simplification and applying L'Hopital's rule or trigonometric limits, the limit evaluates to π2\pi^2.
810 Easy
810 of 930
Application +4 | -1
If lim⁡x→1x4−1x−1=lim⁡x→kx3−k3x2−k2\lim \limits_{{x \rightarrow 1}} \frac{x^4-1}{x-1}=\lim \limits_{{x \rightarrow k}} \frac{x^3-k^3}{x^2-k^2}, then k=k=
View answer and explanation
Correct option: B
83\frac{8}{3}
Option B (1/2) is correct because rationalizing the expression and simplifying leads to a limit of 1/2 as x approaches infinity.
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