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JEE • MATHEMATICS • LIMITS

Limits practice · Easy

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241 Easy
241 of 647
Application +4 | -1
lim⁡x→π6  2sin⁡2x+sin⁡x−12sin⁡2x−3sin⁡x+1\lim\limits_{x \to \frac{\pi }{6}} \,\,\frac{{2{{\sin }^2}x + \sin x - 1}}{{2{{\sin }^2}x - 3\sin x + 1}} is equal to
View answer and explanation
Correct option: B
−3- 3
Option B (-3) is correct because the limit is evaluated using L'Hopital's rule or by factoring and simplifying the expression.
242 Easy
242 of 647
Analyzing +4 | -1
lim⁡x→−∞(2x2−3−5x)\lim _{x \rightarrow-\infty}\left(\sqrt{2 x^2-3}-5 x\right) is
View answer and explanation
Correct option: B
+∞+\infty
Option B (+∞) is correct because as x approaches negative infinity, the expression tends towards positive infinity.
243 Easy
243 of 647
Application +4 | -1
If f′(2)=2,f′′(2)=1f'\left( 2 \right) = 2,f''\left( 2 \right) = 1, then lim⁡x→22x2−4f′(x)x−2\lim\limits_{x \to 2} \frac{{2{x^2} - 4f'\left( x \right)}}{{x - 2}}, is
View answer and explanation
Correct option: A
4
Option A (4) is correct because applying L'Hopital's rule and using the given values of f'(2) and f''(2) results in this value.
244 Easy
244 of 647
Analyzing +4 | -1
If 0<a<b0 < a < b, then lim⁡n→∞an+bnan−bn\lim\limits_{n \to \infty } \frac{{{a^n} + {b^n}}}{{{a^n} - {b^n}}}
View answer and explanation
Correct option: B
Equals −1- 1
Option B (-1) is correct because as n approaches infinity, (a/b)^n approaches 0, simplifying the expression.
245 Easy
245 of 647
Analyzing +4 | -1
lim⁡x→0sin⁡(πcos⁡2x)x2\lim \limits_{{x \rightarrow 0}} \frac{\sin \left(\pi \cos ^2 x\right)}{x^2} equals
View answer and explanation
Correct option: B
π\pi
Option B (π\pi) is correct because the limit is evaluated using trigonometric identities and the standard limit lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1.
246 Easy
246 of 647
Application +4 | -1
Let a=min⁡{x2+2x+3:x∈R}a=\min \left\{x^2+2 x+3: x \in R\right\} and b=lim⁡θ→01−cos⁡θθ2b=\lim \limits_{{\theta \rightarrow 0}} \frac{1-\cos \theta}{\theta^2}. Then, the value of ∑r=0arbn−r\sum\limits_{{r=0}} a^r b^{n-r}, is
View answer and explanation
Correct option: C
4n+1−13×2n\frac{4^{n+1}-1}{3 \times 2^n}
Option C, 4n+1−13×2n\frac{4^{n+1}-1}{3 \times 2^n}, is correct because it is the result of simplifying the expression and using the formula for the sum of a geometric series.
247 Easy
247 of 647
Application +4 | -1
Let L=lim⁡n→∞1n3∑k=1nk2eknL=\lim \limits_{{n \rightarrow \infty}} \frac{1}{n^3} \sum\limits_{{k=1}}^n k^2 e^{\frac{k}{n}}, then find the value of e−Le-L.
View answer and explanation
Correct option: A
2
Option A (2) is correct because the limit L is evaluated as a definite integral, and then e-L is calculated.
248 Easy
248 of 647
Application +4 | -1
The value of lim⁡x→∞(3x−43x+4)(x+13)\lim\limits_{x \to \infty } {\left( {\frac{{3x - 4}}{{3x + 4}}} \right)^{\left( {\frac{{x + 1}}{3}} \right)}}, is
View answer and explanation
Correct option: A
e−2/3{e^{ - 2/3}}
Option A (e−2/3{e^{ - 2/3}}) is correct because after simplifying the expression and applying the limit, the result is e−2/3{e^{ - 2/3}}.
249 Easy
249 of 647
Analyzing +4 | -1
lim⁡x→0(1−cos⁡2x)(3+cos⁡x)xtan⁡4x\lim \limits_{{x \rightarrow 0}} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x} is equal to
View answer and explanation
Correct option: A
2
Option A (2) is correct because the limit is evaluated using trigonometric identities and standard limit results, resulting in 2.
250 Easy
250 of 647
Analyzing +4 | -1
If for some real number kk lim⁡x→0  kx  cosec(x)=lim⁡x→0  xcosec(kx)\lim\limits_{x \to 0} \,\,kx\,\,\mathrm{cosec}\left( x \right) = \lim\limits_{x \to 0} \,\,x\mathrm{cosec}\left( {kx} \right), then the possible values of kk are
View answer and explanation
Correct option: A
1,−11, - 1
Option A (1,−11, -1) is correct because these are the values of k that satisfy the given limit equation after simplification.
251 Easy
251 of 647
Application +4 | -1
The value of lim⁡x→1(2−x)tan⁡πx2\lim\limits_{x \to 1} {\left( {2 - x} \right)^{\tan \frac{{\pi x}}{2}}} is equal to
View answer and explanation
Correct option: C
e2/π{e^{2/\pi }}
Option C (e2/π{e^{2/\pi }}) is correct because the limit can be evaluated by using L'Hopital's rule after rewriting the expression.
252 Easy
252 of 647
Analyzing +4 | -1
If α>β>0\alpha > \beta > 0 are the roots of the equation ax2+bx+1=0a x^{2}+b x+1=0, and lim⁡x→1α(1−cos⁡(x2+bx+a)2(1−αx)2)12=1k(1β−1α)\lim \limits_{x \rightarrow \frac{1}{\alpha}}\left(\frac{1-\cos \left(x^{2}+b x+a\right)}{2(1-\alpha x)^{2}}\right)^{\frac{1}{2}}=\frac{1}{k}\left(\frac{1}{\beta}-\frac{1}{\alpha}\right), then kk is equal to
View answer and explanation
Correct option: C
2α2 \alpha
Option C (2α) is correct because by using L'Hopital's rule and the relationship between the roots and coefficients of the quadratic equation, we can find the value of k.
253 Easy
253 of 647
Application +4 | -1
If lim⁡x→0(1+xlog⁡(1+b2))1/x=2bsin⁡2θ,b>0\lim \limits_{{x \rightarrow 0}}\left(1+x \log \left(1+b^2\right)\right)^{1 / x}=2 b \sin ^2 \theta, b> 0 and θ∈[−π,π]\theta \in[-\pi, \pi], then the value of θ\theta is
View answer and explanation
Correct option: D
±π2\pm \frac{\pi}{2}
Option D (±π2\pm \frac{\pi}{2}) is correct because by evaluating the limit and equating it to 2bsin²θ, we find that sin²θ = 1, which implies θ = ±π2\pm \frac{\pi}{2}.
254 Easy
254 of 647
Application +4 | -1
Evaluate lim⁡x→1(21−x2+1x−1).\lim\limits_{x \to 1} \left( {\frac{2}{{1 - {x^2}}} + \frac{1}{{x - 1}}} \right).
View answer and explanation
Correct option: C
1/21/2
Option C (1/2) is correct because after simplifying the expression and applying the limit, the result is 1/2.
255 Easy
255 of 647
Application +4 | -1
lim⁡x→0x(ex−1)1−cos⁡x=\lim \limits_{{x \rightarrow 0}} \frac{x\left(e^{x}-1\right)}{1-\cos x}=
View answer and explanation
Correct option: D
2
Option D (2) is correct because applying L'Hopital's rule or using series expansions, the limit evaluates to 2.

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256 Easy
256 of 647
Application +4 | -1
If f(9)=9,  f′(9)=4f\left( 9 \right) = 9,\;f'\left( 9 \right) = 4, then lim⁡x→9f(x)−3x−3\lim\limits_{x \to 9} \frac{{\sqrt {f\left( x \right)} - 3}}{{\sqrt x - 3}} equals
View answer and explanation
Correct option: A
4
The correct answer is A (4) because applying L'Hopital's rule and using the given values of f(9) and f'(9) leads to the solution.
257 Easy
257 of 647
Application +4 | -1
Evaluate lim⁡x→1f(x)\lim \limits_{{x \rightarrow 1}} f(x), where, 2x2≤f(x)≤x(x2+1),∀x2 x^2 \leq f(x) \leq x\left(x^2+1\right), \forall x that are near to 1 but not equal to 1.
View answer and explanation
Correct option: A
2
Option A (2) is correct because by the Sandwich theorem, the limit of f(x) as x approaches 1 is squeezed between two functions that both approach 2.
258 Easy
258 of 647
Application +4 | -1
lim⁡x→0(1−ex)sin⁡xx2+x3\lim\limits_{x \to 0} \frac{{\left( {1 - {e^x}} \right)\sin x}}{{{x^2} + {x^3}}} is equal to
View answer and explanation
Correct option: A
−1- 1
Option A (−1-1) is correct because the limit can be evaluated by using Taylor series expansions for exe^x and sin⁡x\sin x or by applying L'Hopital's rule.
259 Easy
259 of 647
Application +4 | -1
The value of lim⁡x→0{ax+bx+cx3}1/x\lim \limits_{{x \rightarrow 0}}\left\{\frac{a^x+b^x+c^x}{3}\right\}^{1 / x}, is
View answer and explanation
Correct option: B
(abc)1/3(a b c)^{1 / 3}
Option B ((abc)1/3(a b c)^{1 / 3}) is correct because the limit can be evaluated using the exponential form of limits and the standard limit lim⁡x→0ax−1x=ln⁡a\lim_{x \rightarrow 0} \frac{{a^x - 1}}{{x}} = \ln a.
260 Easy
260 of 647
Application +4 | -1
lim⁡x→−1x3−2x−1x5−2x−1=\lim \limits_{{x \rightarrow-1}} \frac{x^3-2 x-1}{x^5-2 x-1}=
View answer and explanation
Correct option: B
13\frac{1}{3}
Option B (13\frac{1}{3}) is correct because by applying L'Hopital's rule or factoring, the limit evaluates to 13\frac{1}{3}.
261 Easy
261 of 647
Analyzing +4 | -1
lim⁡x→∞(3x2+2x+1x2+x+2)6x+13x+2\lim\limits_{x \to \infty } {\left( {\frac{{3{x^2} + 2x + 1}}{{{x^2} + x + 2}}} \right)^{\frac{{6x + 1}}{{3x + 2}}}} is equal to
View answer and explanation
Correct option: C
9
Option C (9) is correct because the limit can be evaluated by finding the limits of the base and exponent separately and then combining them.
262 Easy
262 of 647
Analyzing +4 | -1
Among
(S 1): lim⁡n→∞1n2(2+4+6+…+2n)=1\lim \limits_{{n \rightarrow \infty}} \frac{1}{n^{2}}(2+4+6+\ldots+2 n)=1
(S2): lim⁡n→∞1n16(115+215+315+…+n15)=116\lim \limits_{{n \rightarrow \infty}} \frac{1}{n^{16}}\left(1^{15}+2^{15}+3^{15}+\ldots+n^{15}\right)=\frac{1}{16}
View answer and explanation
Correct option: A
Both (S1) and (S2) are true
Option A (Both (S1) and (S2) are true) is correct because both limits are evaluated correctly using summation formulas and limit properties.
263 Easy
263 of 647
Application +4 | -1
The value of lim⁡x→0eax−ebxx\lim \limits_{{x \rightarrow 0}} \frac{e^{a x}-e^{b x}}{x}, is
View answer and explanation
Correct option: B
a−ba-b
Option B (a−ba-b) is correct because the limit can be evaluated using the standard limit lim⁡x→0ex−1x=1\lim_{x \to 0} \frac{e^x - 1}{x} = 1.
264 Easy
264 of 647
Analyzing +4 | -1
If lim⁡x→0{(a−n)nx−tan⁡x}sin⁡nxx2=0\lim \limits_{{x \rightarrow 0}} \frac{\{(a-n) n x-\tan x\} \sin n x}{x^2}=0, where nn is aa non-zero real number, then aa 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 by applying L'Hopital's rule or simplifying the limit expression, we find that a must equal n+1nn+\frac{1}{n} for the limit to be 0.
265 Easy
265 of 647
Analyzing +4 | -1
If lim⁡x→0(1+a3)+8e1/x1+(1−b3)e1/x=2\lim \limits_{{x \rightarrow 0}} \frac{\left(1+a^3\right)+8 e^{1 / x}}{1+\left(1-b^3\right) e^{1 / x}}=2, then
View answer and explanation
Correct option: A
a=1,b=(−3)1/3a=1, b=(-3)^{1 / 3}
Option A (a=1,b=(−3)1/3a=1, b=(-3)^{1 / 3}) is correct because by comparing coefficients after taking the limit as x approaches 0, we find that a=1a=1 and b=(−3)1/3b=(-3)^{1/3}.
266 Easy
266 of 647
Analyzing +4 | -1
If f(1)=g(1)=2f\left( 1 \right) = g\left( 1 \right) = 2, then lim⁡x→1f(1)g(x)−f(x)g(1)−f(1)+g(1)f(x)−g(x)\lim\limits_{x \to 1} \frac{{f\left( 1 \right)g\left( x \right) - f\left( x \right)g\left( 1 \right) - f\left( 1 \right) + g\left( 1 \right)}}{{f\left( x \right) - g\left( x \right)}} is equal to
View answer and explanation
Correct option: D
−2- 2
Option D (-2) is correct because simplifying the expression and applying the given conditions leads to the answer -2.
267 Easy
267 of 647
Application +4 | -1
If f(1)=2f\left( 1 \right) = 2 and f′(1)=1f'\left( 1 \right) = 1, then value of lim⁡x→1  2x−f(x)x−1\lim\limits_{x \to 1} \,\,\frac{{2x - f\left( x \right)}}{{x - 1}} is
View answer and explanation
Correct option: C
1
Option C (1) is correct because by applying L'Hopital's rule and using the given values of f(1) and f'(1), the limit evaluates to 1.
268 Easy
268 of 647
Application +4 | -1
The value of lim⁡h→0(a+h)2sin⁡(a+h)−a2sin⁡ah\lim\limits_{h \to 0} \frac{{{{\left( {a + h} \right)}^2}\sin \left( {a + h} \right) - {a^2}\sin a}}{h}, is
View answer and explanation
Correct option: A
2asin⁡a+a2cos⁡a2a\sin a + {a^2}\cos a
Option A (2asin⁡a+a2cos⁡a2a\sin a + {a^2}\cos a) is correct because it is the derivative of the function x2sin⁡xx^2\sin x evaluated at x=ax=a.
269 Easy
269 of 647
Application +4 | -1
lim⁡x→0∫0xtsin⁡(10t)dtx\lim\limits_{x \to 0} \frac{{\int_0^x {t\sin (10t)dt} }}{x} is equal to
View answer and explanation
Correct option: A
0
Option B (A−6IA - 6I) is correct because it is the result of using the characteristic equation and Cayley-Hamilton theorem to find 10A−110A^{-1}.
270 Easy
270 of 647
Application +4 | -1
If lim⁡x→1x2−ax+bx−1=5\lim _{x \rightarrow 1} \frac{x^2-a x+b}{x-1}=5; then a+ba+b is equal to:
View answer and explanation
Correct option: C
-7
Option C (-7) is correct because it is the sum of a and b that satisfies the given limit condition after applying L'Hopital's Rule.
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