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JEE • MATHEMATICS • LIMITS

Limits practice · Easy

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571 Easy
571 of 647
Analyzing +4 | -1
The value of lim⁡n→∞{11.3+13.5+15.7+….+1(2n+1)(2n+3)},\lim \limits_{{n \rightarrow \infty}}\left\{\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots .+\frac{1}{(2 n+1)(2 n+3)}\right\}, is
View answer and explanation
Correct option: B
12\frac{1}{2}
Option B (12\frac{1}{2}) is correct because the series telescopes, and the limit as nn approaches infinity is 12\frac{1}{2}.
572 Easy
572 of 647
Analyzing +4 | -1
Let β=lim⁡x→0αx−(e3x−1)αx(e3x−1)\beta = \lim\limits_{x \to 0} \frac{{\alpha x - \left( {{e^{3x}} - 1} \right)}}{{\alpha x\left( {{e^{3x}} - 1} \right)}} . Then the value of α+β\alpha+\beta is
View answer and explanation
Correct option: C
52\frac{5}{2}
Option C (52\frac{5}{2}) is correct because it involves using L'Hopital's rule or series expansion to evaluate the limit and then summing α\alpha and β\beta.
573 Easy
573 of 647
Analyzing +4 | -1
lim⁡n→∞{11−n2+21−n2+31−n2+…+n1−n2}\lim\limits_{n \to \infty } \left\{ {\frac{1}{{1 - {n^2}}} + \frac{2}{{1 - {n^2}}} + \frac{3}{{1 - {n^2}}} + \ldots + \frac{n}{{1 - {n^2}}}} \right\} is equal to
View answer and explanation
Correct option: B
−12\frac{{ - 1}}{2}
The correct answer is B (−12\frac{{ - 1}}{2}) because the limit is evaluated by summing the series and then simplifying the expression.
574 Easy
574 of 647
Analyzing +4 | -1
Let α,β∈R\alpha, \beta \in R be such that lim⁡x→0x2sin⁡βxαx−sin⁡x=1\lim \limits_{{x \rightarrow 0}} \frac{x^2 \sin \beta x}{\alpha x-\sin x}=1. Then, 6(α+β)6(\alpha+\beta) equals
View answer and explanation
Correct option: B
7
Option B (7) is correct because by applying L'Hopital's rule and standard limits, we find α=1\alpha = 1 and β=16\beta = \frac{{1}}{{6}}, thus 6(α+β)=76(\alpha + \beta) = 7.
575 Easy
575 of 647
Analyzing +4 | -1
The integer ' nn ' for which the lim⁡x→0cos⁡2x−cos⁡x−excos⁡x+ex−x32xn\lim \limits_{{x \rightarrow 0}} \frac{\cos ^2 x-\cos x-e^x \cos x+e^x-\frac{x^3}{2}}{x^n} is a finite non-zero number, is
View answer and explanation
Correct option: C
4
Option C (4) is correct because when n=4, the limit evaluates to a finite non-zero number after expanding the trigonometric and exponential functions using their Maclaurin series.
576 Easy
576 of 647
Analyzing +4 | -1
lim⁡x→0x(e1+x2+x4−1x−1)1+x2+x4−1\lim \limits_{{x \rightarrow 0}} \frac{x\left(e^{\frac{\sqrt{1+x^2+x^4}-1}{x}}-1\right)}{\sqrt{1+x^2+x^4}-1}
View answer and explanation
Correct option: B
Is equal to 1
Option B (Is equal to 1) is correct because the limit is evaluated using substitution and standard limit results, resulting in 1.
577 Easy
577 of 647
Analyzing +4 | -1
lim⁡h→0f(2h+2+h2)−f(2)f(h−h2+1)−f(1)\lim\limits_{h \to 0} \frac{{f\left( {2h + 2 + {h^2}} \right) - f\left( 2 \right)}}{{f\left( {h - {h^2} + 1} \right) - f\left( 1 \right)}}, given that f′(2)=f'\left( 2 \right) = and f′(1)=4f'\left( 1 \right) = 4
View answer and explanation
Correct option: D
is equal to 3
Option D (is equal to 3) is correct because applying L'Hopital's rule and substituting the given derivatives yields the correct result.
578 Easy
578 of 647
Analyzing +4 | -1
The value of lim⁡x→∞(x2+2x+32x2+x+5)3x−23x+2\lim\limits_{x \to \infty } {\left( {\frac{{{x^2} + 2x + 3}}{{2{x^2} + x + 5}}} \right)^{\frac{{3x - 2}}{{3x + 2}}}}, is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the limit evaluates to 1/2, which is not among the given options.
579 Easy
579 of 647
Analyzing +4 | -1
The value of lim⁡x→∞(1+1xn)x,n>0\lim \limits_{{x \rightarrow \infty}}\left(1+\frac{1}{x^n}\right)^x, n > 0, is
View answer and explanation
Correct option: B
1, if n>1n > 1
Option B (1, if n>1n > 1) is correct because the limit evaluates to 1 when n > 1, using properties of exponential functions.
580 Easy
580 of 647
Application +4 | -1
The value of lim⁡x→2e3x−6−1sin⁡(2−x)\lim\limits_{x \to 2} \frac{{{e^{3x - 6}} - 1}}{{\sin \left( {2 - x} \right)}} is
View answer and explanation
Correct option: C
−3- 3
Option C (-3) is correct because by applying L'Hopital's rule, the limit can be evaluated to -3.
581 Easy
581 of 647
Application +4 | -1
lim⁡x→∞(1−4x−1)3x−1\lim \limits_{{x \rightarrow \infty}}\left(1-\frac{4}{x-1}\right)^{3 x-1} is ....
View answer and explanation
Correct option: A
e−12e^{-12}
Option A (e−12e^{-12}) is correct because by applying the limit definition of e, the expression converges to e−12e^{-12}.
582 Easy
582 of 647
Analyzing +4 | -1
If f(x)=sin⁡(ex−2−1)log⁡(x−1)f\left( x \right) = \frac{{\sin \left( {{e^{x - 2}} - 1} \right)}}{{\log \left( {x - 1} \right)}}, then lim⁡x→2f(x)\lim\limits_{x \to 2} f\left( x \right) is given by
View answer and explanation
Correct option: D
1
Option D (1) is correct. By using algebraic manipulation and standard limits, the limit can be evaluated to 1.
583 Easy
583 of 647
Analyzing +4 | -1
If f(x)f\left( x \right) is the integral function of the function 2sin⁡x−sin⁡2xx3,x≠0\frac{{2\sin x - \sin 2x}}{{{x^3}}},x \ne 0, then lim⁡→f′(x){\lim _ \to }f'\left( x \right) is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because it is the result of evaluating the limit of the derivative of the integral function.
584 Easy
584 of 647
Application +4 | -1
lim⁡x→1e−x−e−1x−1\lim\limits_{x \to 1} \frac{{{e^{ - x}} - {e^{ - 1}}}}{{x - 1}} is equal to
View answer and explanation
Correct option: B
−1e- \frac{1}{e}
Option B (−1e\frac{-1}{e}) is correct because the limit is the derivative of e−xe^{-x} evaluated at x=1x=1, which equals −1e\frac{-1}{e}.
585 Easy
585 of 647
Application +4 | -1
If lim⁡x→0sin⁡(2+x)−sin⁡(2−x)x\lim \limits_{{x \rightarrow 0}} \frac{\sin (2+x)-\sin (2-x)}{x} is equal to pcos⁡qp \cos {q}, then sum of pp and qq is
View answer and explanation
Correct option: A
4
Option A is correct because the limit evaluates to 2cos(2), which means p+q = 2+2 = 4.

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586 Easy
586 of 647
Analyzing +4 | -1
lim⁡x→alog⁡(x−a)log⁡(ex−ea)\lim\limits_{x \to a} \frac{{\log \left( {x - a} \right)}}{{\log \left( {{e^x} - {e^a}} \right)}} is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because after applying L'Hopital's rule twice, the limit evaluates to 1.
587 Easy
587 of 647
Analyzing +4 | -1
If lim⁡x→0aex−bx=2\lim _{x \rightarrow 0} \frac{a e^x-b}{x}=2 then (a,b)(a, b) is equal to
View answer and explanation
Correct option: C
(2,2)(2,2)
Option C ((2,2)) is correct because by applying the limit definition and L'Hopital's rule, we can determine the values of a and b.
588 Easy
588 of 647
Application +4 | -1
The value of lim⁡x→1x+x2+⋯+xn−nx−1\lim _{x \rightarrow 1} \frac{x+x^2+\cdots+x^n-n}{x-1} is
View answer and explanation
Correct option: C
n(n+1)2\frac{n(n+1)}{2}
Option C (n(n+1)2\frac{n(n+1)}{2}) is correct because applying L'Hôpital's rule results in the sum of the first n natural numbers, which is n(n+1)2\frac{n(n+1)}{2}.
589 Easy
589 of 647
Analyzing +4 | -1
The value of lim⁡x→0(tan⁡({x}−1))sin⁡{x}{x}({x}−1)\lim \limits_{{x \rightarrow 0}} \frac{(\tan (\{x\}-1)) \sin \{x\}}{\{x\}(\{x\}-1)}, where {x}\{x\} denotes the fractional part function is
View answer and explanation
Correct option: D
Doesn't exist
Option D (Doesn't exist) is correct because the left-hand limit and right-hand limit are not equal, indicating that the limit does not exist.
590 Easy
590 of 647
Application +4 | -1
If [x][x] denotes the greatest integer less than or equal to xx, then
lim⁡n→∞[x]+[2x]+[3x]+…..+[nx]n2 equals \lim \limits_{{n \rightarrow \infty}} \frac{[x]+[2 x]+[3 x]+\ldots . .+[n x]}{n^2} \text { equals }
View answer and explanation
Correct option: A
x/2x / 2
Option A (x/2x/2) is correct because using the properties of the greatest integer function and the sandwich theorem, the limit evaluates to x/2x/2.
591 Easy
591 of 647
Application +4 | -1
If lim⁡x→0(1+ax)b/x=e4\lim \limits_{{x \rightarrow 0}}(1+a x)^{b / x}=e^4, where aa and bb are natural numbers then
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the condition ab=4 is derived from the limit, but none of the given options satisfy this condition.
592 Easy
592 of 647
Analyzing +4 | -1
lim⁡x→11+log⁡x−x1−2x+x2\lim \limits_{{x \rightarrow 1}} \frac{1+\log x-x}{1-2 x+x^2} equals
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because after applying L'Hopital's rule twice, the limit evaluates to -1/2.
593 Easy
593 of 647
Analyzing +4 | -1
lim⁡h→0log⁡e(1+2h)−2log⁡e(1+h)h2\lim \limits_{{h \rightarrow 0}} \frac{\log _e(1+2 h)-2 \log _e(1+h)}{h^2} is
View answer and explanation
Correct option: C
-1
Option C (-1) is correct because after applying L'Hôpital's rule, the limit evaluates to -1.
594 Easy
594 of 647
Application +4 | -1
lim⁡x→π/41−cot⁡3x2−cot⁡x−cot⁡3x\lim \limits_{{x \rightarrow \pi / 4}} \frac{1-\cot ^3 x}{2-\cot x-\cot ^3 x} is
View answer and explanation
Correct option: B
3/43 / 4
Option B (3/43 / 4) is correct because the expression is simplified by factoring and canceling terms, then the limit is evaluated by direct substitution.
595 Easy
595 of 647
Application +4 | -1
If f(x)={x2−1,0<x<22x+3,2≤x<3f(x)=\left\{\begin{array}{l}x^2-1,0 < x < 2 \\ 2 x+3,2 \leq x < 3\end{array}\right., then lim⁡x→2−f(x)\lim \limits_{x \rightarrow 2^{-}} f(x) and lim⁡x→2+f(x)\lim \limits_{x \rightarrow 2^{+}} f(x) are
View answer and explanation
Correct option: A
3,73, 7
Option A (3, 7) is correct because it represents the left-hand and right-hand limits of the function at x=2, respectively.
596 Easy
596 of 647
Analyzing +4 | -1
lim⁡x→0[8sin⁡x+xcos⁡x3tan⁡x+x2]\lim\limits_{x \to 0} \left[ {\frac{{8\sin x + x\cos x}}{{3\tan x + {x^2}}}} \right] is equal to
View answer and explanation
Correct option: A
3
Option A (3) is correct because the limit can be evaluated by dividing both the numerator and denominator by x and then applying the standard limits.
597 Easy
597 of 647
Application +4 | -1
lim⁡x→0(x1+x−1−x)\lim\limits_{x \to 0} \left( {\frac{x}{{\sqrt {1 + x} - \sqrt {1 - x} }}} \right) is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because by rationalizing the denominator, the limit evaluates to 1.
598 Easy
598 of 647
Analyzing +4 | -1
lim⁡x→0x(e(1+x2+x4−1)/x)−11+x2+x4−1\lim\limits_{x \to 0} \frac{{x({e^{\left( {\sqrt {1 + {x^2} + {x^4}} - 1)/x} \right)}} - 1}}{{\sqrt {1 + {x^2} + {x^4} - 1} }}
View answer and explanation
Correct option: B
is equal to 1
Option B is correct because by applying limits and L'Hopital's rule, the expression simplifies to 1.
599 Easy
599 of 647
Analyzing +4 | -1
lim⁡x→1nxn+1−(n+1)xn+1(ex−e)sin⁡πx\lim \limits_{{x \rightarrow 1}} \frac{n x^{n+1}-(n+1) x^n+1}{\left(e^x-e\right) \sin \pi x}, where n=100n=100 is equal to
View answer and explanation
Correct option: C
−5050πe-\frac{5050}{\pi e}
Option C (-5050πe\frac{5050}{\pi e}) is correct because the limit is evaluated using L'Hôpital's Rule and series expansion.
600 Easy
600 of 647
Application +4 | -1
lim⁡x→∞(2x−3)(3x−4)(4x−5)(5x−6)\lim\limits_{x \to \infty } \frac{{\left( {2x - 3} \right)\left( {3x - 4} \right)}}{{\left( {4x - 5} \right)\left( {5x - 6} \right)}} is equal to
View answer and explanation
Correct option: D
310\frac{3}{{10}}
Option D (310\frac{3}{{10}}) is correct because as x approaches infinity, the limit of the rational function is the ratio of the leading coefficients after applying L'Hopital's rule or dividing by the highest power of x.
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