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JEE • MATHEMATICS • LIMITS

Limits practice · Easy

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Showing 331-360 of 647 questions

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331 Easy
331 of 647
Analyzing +4 | -1
Let f,gf, g and hh be the real valued functions defined of R\mathbb{R} as

f(x){x∣x∣,x≠01,x=0,   g(x){sin⁡(x+1)(x+1),x≠−11,x=−1f(x)\left\{\begin{array}{ll} \frac{x}{|x|}, & x \neq 0 \\ 1, & x=0 \end{array}, \,\,\, g(x) \begin{cases}\frac{\sin (x+1)}{(x+1)}, & x \neq-1 \\ 1, & x=-1\end{cases}\right.

and h(x)=2[x]−f(x)h(x)=2[x]-f(x), where [x][x] is the greatest integer ≤x\leq x.

Then the value of lim⁡x→1g(h(x−1))\lim \limits_{{x \rightarrow 1}} g(h(x-1)) is
View answer and explanation
Correct option: A
1
Option A is correct because the limit of g(h(x−1))g(h(x-1)) as xx approaches 1 is 1.
332 Easy
332 of 647
Analyzing +4 | -1
lim⁡x→1sin⁡2(x3+x2+x−3)1−cos⁡(x2−4x+3)\lim \limits_{{x \rightarrow 1}} \frac{\sin ^2\left(x^3+x^2+x-3\right)}{1-\cos \left(x^2-4 x+3\right)} has the value equal to
View answer and explanation
Correct option: A
18
Option A (18) is correct because after applying L'Hopital's rule and simplifying, the limit evaluates to 18.
333 Easy
333 of 647
Understanding +4 | -1
If f(x)=∣x∣−5f(x)=|x|-5, then the value of lim⁡x→5f(x)\lim \limits_{{x \rightarrow 5}} f(x) is
View answer and explanation
Correct option: A
0
Option A (0) is correct because as x approaches 5, |x| approaches 5, and |x| - 5 approaches 0.
334 Easy
334 of 647
Application +4 | -1
If G(x)=−25−x2G(x)=-\sqrt{25-x^2}, then
lim⁡x→1(G(x)−G(1)x−1)=\lim _{x \rightarrow 1}\left(\frac{G(x)-G(1)}{x-1}\right)=
View answer and explanation
Correct option: A
126\frac{1}{2 \sqrt{6}}
Option A (126\frac{1}{2 \sqrt{6}}) is correct because, after applying L'Hopital's rule or rationalizing, the limit of the expression as x approaches 1 is 126\frac{1}{2 \sqrt{6}}.
335 Easy
335 of 647
Application +4 | -1
Let f:R→Rf:R \to R be a differentiable function and f(1)=4f\left( 1 \right) = 4. Then, the value of lim⁡x→1  ∫4f(x)  2 tx−1  dt\lim\limits_{x \to 1} \,\,\int \nolimits_4^{f\left( x \right)} \,\,\frac{{2\,t}}{{x - 1}}\,\,dt〗is
View answer and explanation
Correct option: A
8  f′  (1)8\,\,f'\,\,\left( 1 \right)
Option A (8f′(1)8 f'(1)) is correct because it is the result of applying L'Hopital's rule and the Fundamental Theorem of Calculus to the given limit.
336 Easy
336 of 647
Analyzing +4 | -1
If 0<p<q0 < p < q, then lim⁡n→∞(qn+pn)1/n\lim\limits_{n \to \infty } {\left( {{q^n} + {p^n}} \right)^{1/n}} is equal to
View answer and explanation
Correct option: C
qq
Option C (qq) is correct because as nn approaches infinity, the term with the larger base (qq) dominates, and the limit approaches qq.
337 Easy
337 of 647
Application +4 | -1
lim⁡x→1(1+cos⁡πxtan⁡2πx)=\lim\limits_{x \to 1} \left( {\frac{{1 + \cos \pi x}}{{{{\tan }^2}\pi x}}} \right) =
View answer and explanation
Correct option: D
12\frac{1}{2}
Option D (12\frac{1}{2}) is correct because after simplification and applying the limit, the expression evaluates to 1/2.
338 Easy
338 of 647
Application +4 | -1
lim⁡x→0(1−cos⁡2x)sin⁡5xx2sin⁡3x\lim\limits_{x \to 0} \frac{{\left( {1 - \cos 2x} \right)\sin 5x}}{{{x^2}\sin 3x}} equals
View answer and explanation
Correct option: A
103\frac{{10}}{3}
Option A (103\frac{{10}}{3}) is correct because by applying trigonometric identities and standard limits, the expression simplifies to this value.
339 Easy
339 of 647
Analyzing +4 | -1
If lim⁡x→0[(a−n)nx−tan⁡x]sin⁡nxx2=0\lim\limits_{x \to 0} \frac{{\left[ {\left( {a - n} \right)nx - \tan x} \right]\sin nx}}{{{x^2}}} = 0, where nn is non-zero real number, then a is equal to
View answer and explanation
Correct option: D
n+1nn + \frac{1}{n}
Option D (n+1nn + \frac{1}{n}) is correct because substituting the limit into the given equation and solving for 'a' yields this result.
340 Easy
340 of 647
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.
341 Easy
341 of 647
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}.
342 Easy
342 of 647
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.
343 Easy
343 of 647
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.
344 Easy
344 of 647
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.
345 Easy
345 of 647
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}.

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs
346 Easy
346 of 647
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.
347 Easy
347 of 647
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}.
348 Easy
348 of 647
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.
349 Easy
349 of 647
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.
350 Easy
350 of 647
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.
351 Easy
351 of 647
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}.
352 Easy
352 of 647
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.
353 Easy
353 of 647
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).
354 Easy
354 of 647
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.
355 Easy
355 of 647
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.
356 Easy
356 of 647
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.
357 Easy
357 of 647
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.
358 Easy
358 of 647
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.
359 Easy
359 of 647
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.
360 Easy
360 of 647
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.
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