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JEE • MATHEMATICS • LIMITS

Limits practice · Mixed difficulty

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631 Easy
631 of 930
Application +4 | -1
If lim⁡x→∞(1+ax−4x2)2x=e3\lim _{x \rightarrow \infty}\left(1+\frac{a}{x}-\frac{4}{x^2}\right)^{2 x}=e^3, then 'a' is equal to :
View answer and explanation
Correct option: B
32\frac{3}{2}
Option B (3/2) is correct because the limit can be evaluated using the exponential form of limits, leading to a = 3/2.
632 Medium
632 of 930
Analyzing +4 | -1
The value of lim⁡n→∞(cos⁡xn)n\lim\limits_{n \to \infty } {\left( {\cos \frac{x}{n}} \right)^n}, is
View answer and explanation
Correct option: C
1
Option C (1) is correct because the limit can be evaluated by rewriting the expression using the exponential function and L'Hopital's rule or series expansion.
633 Medium
633 of 930
Application +4 | -1
The value of lim⁡h→0f(x+h)−f(x)h\lim \limits_{{h \rightarrow 0}} \frac{f(x+h)-f(x)}{h}, where f(x)=sin⁡(x+1)f(x)=\sin (x+1), is
View answer and explanation
Correct option: B
cos⁡(x+1)\cos (x+1)
Option B (cos⁡(x+1)\cos (x+1)) is correct because it is the derivative of sin(x+1) with respect to x.
634 Easy
634 of 930
Application +4 | -1
If f(x)=lim⁡n→∞[(1+x)(1+x2)(1+x4)…nf(x)=\lim \limits_{{n \rightarrow \infty}}\left[(1+x)\left(1+x^2\right)\left(1+x^4\right) \ldots n\right. terms], then f(12)=……∣x∣<1f\left(\frac{1}{2}\right)=\ldots \ldots \quad|x| < 1
View answer and explanation
Correct option: B
2
Option B (2) is correct because by simplifying the infinite product using the difference of squares, we get a simple expression for f(x) and can evaluate f(1/2).
635 Easy
635 of 930
Analyzing +4 | -1
lim⁡x→1(11−x−31−x3)\lim _{x \rightarrow 1}\left(\frac{1}{1-x}-\frac{3}{1-x^3}\right) is equal to
View answer and explanation
Correct option: D
-1
Option D (-1) is correct because after combining the fractions and factoring, the limit can be evaluated using L'Hopital's rule or by direct substitution.
636 Medium
636 of 930
Analyzing +4 | -1
lim⁡x→0(tan⁡(π4+x))1/x\lim _{x \rightarrow 0}\left(\tan \left(\frac{\pi}{4}+x\right)\right)^{1 / x} is equal to:
View answer and explanation
Correct option: D
e2e^2
Option D (e2e^2) is correct because the limit can be rewritten in the form elim⁡x→0f(x)e^{\lim_{x \to 0} f(x)} and the exponent evaluates to 2.
637 Easy
637 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, where nn is a positive integer, then n=n=
View answer and explanation
Correct option: B
5
Option B (5) is correct because solving the limit equation lim⁡x→2xn−2nx−2=80\lim \limits_{{x \rightarrow 2}} \frac{x^{n}-2^{n}}{x-2}=80 yields n=5n=5.
638 Hard
638 of 930
Analyzing +4 | -1
If lim⁡x→1x2−ax+bx−1=5\lim\limits_{x \to 1} \frac{{{x^2} - ax + b}}{{x - 1}} = 5, then a+b is equal to
View answer and explanation
Correct option: C
-7
Option C (-7) is correct because by applying L'Hôpital's rule or factoring and simplifying, we find that a+b = -7.
639 Medium
639 of 930
Analyzing +4 | -1
The value of lim⁡x→0([11xsin⁡x]+[21sin⁡xx])\lim \limits_{{x \rightarrow 0}}\left(\left[\frac{11 x}{\sin x}\right]+\left[\frac{21 \sin x}{x}\right]\right), where [x][x] is the greatest integer less than or equal to xx is
View answer and explanation
Correct option: B
31
Option B (31) is correct because as x approaches 0, the greatest integer functions evaluate to 11 and 20 respectively, summing to 31.
640 Easy
640 of 930
Analyzing +4 | -1
The value of lim⁡x→01+sin⁡x3−1−sin⁡x3x\lim\limits_{x \to 0} \frac{{\sqrt[3]{{1 + \sin x}} - \sqrt[3]{{1 - \sin x}}}}{x}, is
View answer and explanation
Correct option: A
23\frac{2}{3}
Option A (23\frac{2}{3}) is correct because the limit is evaluated using series expansion or L'Hopital's rule.
641 Easy
641 of 930
Application +4 | -1
If value of lim⁡x→02+x−2x\lim \limits_{{x \rightarrow 0}} \frac{\sqrt{2+x}-\sqrt{2}}{x} is equal to 1a2\frac{1}{a \sqrt{2}} then &#39;aa&#39; equals
View answer and explanation
Correct option: A
2
Option A is correct because the limit evaluates to 1/(2√2), implying that 'a' equals 2.
642 Easy
642 of 930
Analyzing +4 | -1
The value of lim⁡x→0xsin⁡(sin⁡x)−sin⁡2xx6\lim \limits_{{x \rightarrow 0}} \frac{x \sin (\sin x)-\sin ^2 x}{x^6} equals
View answer and explanation
Correct option: C
118\frac{1}{18}
The correct answer is C (118\frac{1}{18}) because after expanding the trigonometric functions using their Taylor series and simplifying, the limit equals 118\frac{1}{18}.
643 Easy
643 of 930
Understanding +4 | -1
If f(x)=x2sin⁡1xf(x)=x^2 \sin \frac{1}{x}, where x≠0x \neq 0, then the value of the function f{f} at x=0{x}=0, so that the function is continuous at x=0x=0, is
View answer and explanation
Correct option: A
0
Option A is correct because the limit of the function as x approaches 0 is 0, making the function continuous at x=0 if f(0)=0.
644 Easy
644 of 930
Analyzing +4 | -1
lim⁡x→1+x2−1+x−1x2−1\lim \limits_{x \rightarrow 1^{+}} \frac{\sqrt{x^2-1}+\sqrt{x-1}}{\sqrt{x^2-1}} is equal to
View answer and explanation
Correct option: D
1+121+\frac{1}{\sqrt{2}}
Option D (1+121+\frac{1}{\sqrt{2}}) is correct because after simplifying the limit expression by factoring and cancelling, the limit can be directly evaluated.
645 Easy
645 of 930
Application +4 | -1
Let P=lim⁡x→0+(1+tan⁡2x)12xP=\lim \limits_{x \rightarrow 0^{+}}\left(1+\tan ^2 \sqrt{x}\right)^{\frac{1}{2 x}}. Then, log⁡eP\log _e P, is equal to
View answer and explanation
Correct option: B
12\frac{1}{2}
Option B (12\frac{{1}}{{2}}) is correct because the limit can be evaluated using the exponential form of limits and the standard limit lim⁡x→0tan⁡xx=1\lim_{x \rightarrow 0} \frac{{\tan x}}{{x}} = 1.

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646 Easy
646 of 930
Analyzing +4 | -1
The value of lim⁡x→0(2x3(sin⁡−1x−tan⁡−1x))2/x2\lim \limits_{{x \rightarrow 0}}\left(\frac{2}{x^3}\left(\sin ^{-1} x-\tan ^{-1} x\right)\right)^{2 / x^2} equals
View answer and explanation
Correct option: D
1e\frac{1}{\sqrt{e}}
The correct answer is D (1e\frac{{1}}{{\sqrt{{e}}}}) because by using Taylor series expansions for sin⁡−1x\sin^{-1}x and tan⁡−1x\tan^{-1}x, the limit can be evaluated to e−1/2e^{-1/2}.
647 Easy
647 of 930
Analyzing +4 | -1
lim⁡x→012x−3x−4x+12cos⁡x+7−3=\lim _{x \rightarrow 0} \frac{12^x-3^x-4^x+1}{\sqrt{2 \cos x+7}-3}=
View answer and explanation
Correct option: D
−6log⁡e3⋅log⁡e4-6 \log _e 3 \cdot \log _e 4
Option D (−6log⁡e3⋅log⁡e4-6 \log _e 3 \cdot \log _e 4) is correct because the limit can be evaluated by factoring the numerator and using standard limit formulas.
648 Medium
648 of 930
Analyzing +4 | -1
The value of lim⁡x→0  ∫txtdt  xtan⁡(x+π)\lim\limits_{x \to 0} \,\,\frac{{\int_t^x {t\mathrm{ }dt} \,\,}}{{x\tan \left( {x + \pi } \right)}} is equal to
View answer and explanation
Correct option: C
12\frac{1}{2}
Option C (12\frac{1}{2}) is correct because, by applying L'Hopital's rule or direct integration and simplification, the limit evaluates to 1/2.
649 Easy
649 of 930
Application +4 | -1
lim⁡θ→π2  π2−θcot⁡θ\lim\limits_{\theta \to \frac{\pi }{2}} \,\,\frac{{\frac{\pi }{2} - \theta }}{{\cot \theta }} is equal to
View answer and explanation
Correct option: C
1
Option C (1) is correct because by applying L'Hopital's rule or converting cot to sin/cos, the limit evaluates to 1.
650 Easy
650 of 930
Analyzing +4 | -1
The value of lim⁡x→252−x\lim\limits_{x \to 2} \frac{5}{{\sqrt 2 - \sqrt 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 are not equal, indicating the limit does not exist.
651 Easy
651 of 930
Application +4 | -1
The value of lim⁡x→0eax−ebxx\lim\limits_{x \to 0} \frac{{{e^{ax}} - {e^{bx}}}}{x} is equal to
View answer and explanation
Correct option: B
a−ba - b
Option B (a−ba - b) is correct because it is the result of applying L'Hopital's rule or using series expansion to evaluate the limit.
652 Medium
652 of 930
Analyzing +4 | -1
Suppose xx is real. Define
f(x)f(x) = x2+1\frac{x}{2}+1
f(2)(x)=f(f(x))f(3)(x)=f(f(2)(x))\begin{aligned} & f^{(2)}(x)=f(f(x)) \\ & f^{(3)}(x)=f\left(f^{(2)}(x)\right) \end{aligned}
and in general
fn+1(x)=f(f(n)(x))f^{n+1}(x)=f\left(f^{(n)}(x)\right)
for n≥1n \geq 1 where
f(1)(x)=f(x)=x2+1f^{(1)}(x)=f(x)=\frac{x}{2}+1
Then lim⁡n→−∞f(n)(x)\lim \limits_{{n \rightarrow-\infty}} f^{(n)}(x) is equal to
View answer and explanation
Correct option: B
2
Option B (2) is correct because the limit of the recursively defined function f(n)(x)f^{(n)}(x) as n approaches negative infinity converges to 2.
653 Medium
653 of 930
Analyzing +4 | -1
Let x0=0,x1=1x_0=0, x_1=1 and xn+1=xn+1+xn2x_{n+1}=x_n+\sqrt{1+x_n^2} for n≥1n \geq 1.
Then
lim⁡n→∞(xn2n−1)=\lim \limits_{{n \rightarrow \infty}}\left(\frac{x_n}{2^{n-1}}\right)=
View answer and explanation
Correct option: D
4π\frac{4}{\pi}
Option D is correct. By using the recurrence relation and trigonometric identities, the limit can be evaluated to 4/π.
654 Easy
654 of 930
Application +4 | -1
If f:R→Rf:R \to R is defined by f(x)={x−2x2−3x+2,ifx∈R−{1,2}2,ifx=11, ifx=2f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{x - 2}}{{{x^2} - 3x + 2}}}&{,\mathrm{if}}&{x \in R - \left\{ {1,2} \right\}}\\ 2&{,\mathrm{if}}&{x = 1}\\ 1&{,\,\mathrm{if}}&{x = 2} \end{array}} \right. Then lim⁡x→2f(x)−f(2)x−2\lim\limits_{x \to 2} \frac{{f\left( x \right) - f\left( 2 \right)}}{{x - 2}} is equal to
View answer and explanation
Correct option: B
−1- 1
Option B (-1) is correct because the limit of the given expression, after simplification and factorization, evaluates to -1.
655 Easy
655 of 930
Application +4 | -1
lim⁡x→1xm−1xn−1\lim\limits_{x \to 1} \frac{{{x^m} - 1}}{{{x^n} - 1}} is equal to
View answer and explanation
Correct option: B
mn\frac{m}{n}
Option B (mn\frac{m}{n}) is correct because by applying L'Hopital's rule, the limit evaluates to m/n.
656 Easy
656 of 930
Application +4 | -1
The value of lim⁡x→0∫0xcos⁡tdtx\lim \limits_{{x \rightarrow 0}} \frac{\int\limits_{0}^x \cos t d t}{x} is
View answer and explanation
Correct option: D
None of these
The correct answer is None of these, as the limit evaluates to 1 using L'Hopital's rule.
657 Easy
657 of 930
Application +4 | -1
lim⁡x→2+{[x]33−[x3]3}\lim\limits _{x \rightarrow 2^{+}}\left\{\frac{[x]^3}{3}-\left[\frac{x}{3}\right]^3\right\} is equal to
View answer and explanation
Correct option: C
83\frac{8}{3}
Option C (83\frac{8}{3}) is correct because it is the result of evaluating the limit as x approaches 2 from the right, considering the greatest integer function.
658 Easy
658 of 930
Application +4 | -1
lim⁡x→∞(x2+5x+3x2+x+2)x\lim \limits_{{x \rightarrow \infty}}\left(\frac{x^2+5 x+3}{x^2+x+2}\right)^x
View answer and explanation
Correct option: A
e4e^4
Option A (e4e^4) is correct because the limit is evaluated using the standard limit formula and algebraic manipulation.
659 Medium
659 of 930
Application +4 | -1
If f:R→Rf: R \rightarrow R is given by f(x)=x+1f(x)=x+1, then the value of
lim⁡n→∞1n[f(0)+f(5n)+f(10n)+…+f(5(n−1)n)], is : \lim \limits_{{n \rightarrow \infty}} \frac{1}{n}\left[f(0)+f\left(\frac{5}{n}\right)+f\left(\frac{10}{n}\right)+\ldots+f\left(\frac{5(n-1)}{n}\right)\right], \text { is : }
View answer and explanation
Correct option: D
72\frac{7}{2}
Option D (72\frac{{7}}{{2}}) is correct because the limit represents a Riemann sum which can be evaluated as a definite integral.
660 Medium
660 of 930
Analyzing +4 | -1
lim⁡n→∞(1+1n2)8n3πsin⁡(π2n)\lim \limits_{{n \rightarrow \infty}}\left(1+\frac{1}{n^2}\right)^{\frac{8 n^3}{\pi} \sin \left(\frac{\pi}{2 n}\right)} is equal to
View answer and explanation
Correct option: B
e4e^4
Option B (e4e^4) is correct because by manipulating the limit into the form of ee, the expression simplifies to e4e^4.
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