Skip to main content
Back to LIMITS

Your Progress

0 / 930 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • MATHEMATICS • LIMITS

Limits practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

Take a chapter quiz →
Showing 751-780 of 930 questions

Preparing practice and saved progress…

751 Easy
751 of 930
Analyzing +4 | -1
lim⁡x→axa−axxx−aa\lim \limits_{{x \rightarrow a}} \frac{x^a-a^x}{x^x-a^a} is equal to
View answer and explanation
Correct option: B
log⁡e(e/a)log⁡e(ae)\frac{\log _e(e / a)}{\log _e(a e)}
Option B (log⁡e(e/a)log⁡e(ae)\frac{\log _e(e / a)}{\log _e(a e)}) is correct because applying L'Hopital's rule and simplifying leads to this expression.
752 Easy
752 of 930
Application +4 | -1
If lim⁡x→∞(x2+1x+1−ax−b)=0\lim _{x \rightarrow \infty}\left(\frac{x^2+1}{x+1}-a x-b\right)=0 then (a,b)(a, b) value is
View answer and explanation
Correct option: B
(1,−1)(1,-1)
Option B ((1,-1)) is correct because, for the limit to be 0, the coefficients of x2x^2 and xx in the numerator must be zero, leading to a=1a=1 and b=−1b=-1.
753 Easy
753 of 930
Application +4 | -1
The value of lim⁡x→π4(1+[x])1ln⁡(tan⁡x)\lim \limits_{x \rightarrow \frac{\pi}{4}}(1+[x])^{\frac{1}{\ln (\tan x)}} is:
(where [.] denotes greatest integer function).
View answer and explanation
Correct option: B
1
Option B (1) is correct because as x approaches π4\frac{\pi}{4}, the greatest integer function of x is 1, and any constant raised to a finite power is the constant itself.
754 Easy
754 of 930
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.
755 Medium
755 of 930
Application +4 | -1
Let f:R→Rf: \mathbb{R} \rightarrow \mathbb{R} be such that f(1)=3f(1)=3 and f′(1)=6f^{\prime}(1)=6. Then

lim⁡x→0(f(1+x)f(1))1/x=\lim \limits_{{x \rightarrow 0}}\left(\frac{f(1+x)}{f(1)}\right)^{1 / x}=
View answer and explanation
Correct option: A
e2e^2
Option A (e2e^2) is correct because by taking the natural logarithm and applying L'Hôpital's rule, the limit simplifies to e2e^2.
756 Easy
756 of 930
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.
757 Easy
757 of 930
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.
758 Easy
758 of 930
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}}.
759 Medium
759 of 930
Analyzing +4 | -1
lim⁡x→0(ex+e−x−2x2)1/x2\lim\limits_{x \to 0} {\left( {\frac{{{e^x} + {e^{ - x}} - 2}}{{{x^2}}}} \right)^{1/{x^2}}} is equal to
View answer and explanation
Correct option: D
e1/12{e^{1/12}}
The correct answer is D (e1/12{e^{1/12}}) because it is the result of evaluating the limit using series expansion and properties of exponential functions.
760 Easy
760 of 930
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}.
761 Easy
761 of 930
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.
762 Medium
762 of 930
Understanding +4 | -1
The value of lim⁡x→∞x1/x\lim\limits_{x \to \infty } {x^{1/x}} equals
View answer and explanation
Correct option: B
1
Option C (7) is correct because the minimum value of the absolute value function ∣3−x∣|3 - x| is 0, which occurs when x=3x = 3. Therefore, the minimum value of f(x)f(x) is 0+7=70 + 7 = 7.
763 Medium
763 of 930
Analyzing +4 | -1
If xn+1=1+xn2x_{n+1}=\sqrt{\frac{1+x_n}{2}} and ∣x0∣<1,n≥0\left|x_0\right|<1, n \geq 0, then lim⁡n→∞(1−x02x1x2x3…xn+1)\lim \limits_{{n \rightarrow \infty}}\left(\frac{\sqrt{1-x_0^2}}{x_1 x_2 x_3 \ldots x_{n+1}}\right) is equal to
View answer and explanation
Correct option: D
cos⁡−1(x0)\cos ^{-1}\left(x_0\right)
Option D, cos⁡−1(x0)\cos^{-1}(x_0), is correct because by substituting x0=cos⁡θx_0 = \cos \theta and using trigonometric identities, the limit simplifies to θ\theta, which is cos⁡−1(x0)\cos^{-1}(x_0).
764 Easy
764 of 930
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.
765 Easy
765 of 930
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.

Explore Other Chapters in MATHEMATICS

View All →
SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs
766 Easy
766 of 930
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.
767 Easy
767 of 930
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}.
768 Easy
768 of 930
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.
769 Easy
769 of 930
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}.
770 Easy
770 of 930
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.
771 Medium
771 of 930
Application +4 | -1
If a1=1a_1=1 and an+1=4+3an3+2an,n≥1a_{n+1}=\frac{4+3 a_n}{3+2 a_n}, n \geq 1 and if ana_n has a limit ll as n→∞n \rightarrow \infty, then
View answer and explanation
Correct option: B
l=2l=\sqrt{2}
Option B (l=2l=\sqrt{2}) is correct because the limit ll satisfies the equation l=4+3l3+2ll = \frac{4+3l}{3+2l}, and solving for ll yields l=2l=\sqrt{2} (since l>0l>0).
772 Easy
772 of 930
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.
773 Easy
773 of 930
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.
774 Easy
774 of 930
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}.
775 Easy
775 of 930
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.
776 Easy
776 of 930
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.
777 Easy
777 of 930
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 &#39;nn&#39; is
View answer and explanation
Correct option: A
5
Option A is correct because the value of n is 5.
778 Easy
778 of 930
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}.
779 Medium
779 of 930
Application +4 | -1
lim⁡x→0x7[1x3]=……\lim \limits_{{x \rightarrow 0}} x^7\left[\frac{1}{x^3}\right]=\ldots \ldots
View answer and explanation
Correct option: A
0
Option A (0) is correct because as x approaches 0, the expression simplifies to a limit that evaluates to 0 using the sandwich theorem.
780 Easy
780 of 930
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.
Revision notes and key takeaways

Review LIMITS in your textbook or class notes, then use this set to check your understanding.

Showing 751-780 of 930 questions

Continue Your Practice

📚 More Sets in This Chapter

Try different difficulty levels to master this topic

→ ENGINEERING MATHEMATICS - LIMITS MCQs (Easy) 647 MCQs → ENGINEERING MATHEMATICS - LIMITS MCQs (Hard) 16 MCQs → ENGINEERING MATHEMATICS - LIMITS MCQs (Medium) 267 MCQs
🎯 Next Chapters to Practice

Continue building your subject mastery

→ SEQUENCE OF SERIES 1764 MCQs → DEFINITE INTEGRATION 1443 MCQs → MATRICES 1363 MCQs → INDEFINITE INTEGRATION 1179 MCQs

Your practice

0 / 930 answered · 930 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Practice settings

Type at least 2 characters or jump directly:
🩺 NEET MCQs 🔬 JEE MCQs 🎯 JEE College Predictor Study guides 📊 Dashboard
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.

Continue with Free Account →