Let $\alpha $ and $\beta $ be the roots of $a\;{x^2} + b\;x + c = 0$, then $\mathop {\lim }\limits_{x \to a} \frac{{1 - \cos \left( {a\;{x^2} + bx + c} \right)}}{{{{\left( {x - \alpha } \right)}^2}}}$ is equal to

$\frac{1}{2}{\left( {\alpha - \beta } \right)^2}$

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JEE • MATHEMATICS • LIMITS

Limits practice · Medium

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Showing 1-30 of 267 questions

1 Medium

1 of 267

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}}\left(\frac{1}{x^2}-\operatorname{cosec}^2 x\right)

is equal to

View answer and explanation

Correct option: C

\frac{-1}{3}

Option C (

\frac{-1}{3}

) is correct because after rewriting the expression and applying L'Hôpital's rule multiple times, the limit evaluates to

\frac{-1}{3}

2 Medium

2 of 267

Analyzing +4 | -1

\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x}{(1-\sin x)^{2 / 3}}=

View answer and explanation

Correct option: C

\infty

Option C (∞) is correct because as x approaches pi/2, the expression tends towards infinity.

3 Medium

3 of 267

Application +4 | -1

\lim \limits_{{x \rightarrow \infty}} x\left(e^{1 / x}-1\right)

is equal to

View answer and explanation

Correct option: C

Option C (1) is correct because by rewriting the expression and applying L'Hopital's rule, the limit evaluates to 1.

4 Medium

4 of 267

Application +4 | -1

\lim\limits_{n \to \infty } {\left( {1 + \sin \frac{a}{n}} \right)^n}

equals

View answer and explanation

Correct option: A

{e^a}

Option A (

{e^a}

) is correct because by rewriting the expression in the form of the exponential limit, the limit can be evaluated to

{e^a}

5 Medium

5 of 267

Analyzing +4 | -1

The value of

\lim \limits_{{x \rightarrow a}}\left[\left\{\left(\frac{\sqrt{a}+\sqrt{x}}{a^{1 / 4}-x^{1 / 4}}\right)^{-1}-\frac{2(a x)^{1 / 4}}{x^{3 / 4}-a^{1 / 4} x^{1 / 2}+a^{1 / 2} x^{1 / 4}-a^{3 / 4}}\right\}^{-1}\right.\left.-(\sqrt{2})^{\log _4 a}\right]^8

, is

View answer and explanation

Correct option: C

a^2

Option C (

a^2

) is correct because after simplification and taking the limit, the expression evaluates to

a^2

6 Medium

6 of 267

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}} \frac{\log _e \cot \left(\frac{\pi}{4}-K_1 x\right)}{\tan K_2 x}=1

, then

View answer and explanation

Correct option: B

2 K_1=K_2

The correct answer is B (

2 K_1=K_2

) because by applying logarithmic properties and limits, the relationship between

K_1

and

K_2

is found to be

2K_1 = K_2

7 Medium

7 of 267

Application +4 | -1

\lim \limits_{{x \rightarrow 0}}(\cos x+\sin x)^{1 / x}

equals

View answer and explanation

Correct option: A

e

Option A (

e

) is correct because by taking the logarithm and applying L'Hopital's rule, the limit is found to be

e

8 Medium

8 of 267

Analyzing +4 | -1

Let

m

and

n

be two positive integers greater than 1. If

\lim \limits_{{\alpha \rightarrow 0}}\left(\frac{e^{\cos \left(\alpha^n\right)}-e}{\alpha^m}\right)=-\frac{e}{2}

, than the value of

\frac{m}{n}

, is

View answer and explanation

Correct option: A

Option A (2) is correct because the limit can be evaluated using L'Hopital's rule or Taylor series expansion, leading to the condition m = 2n, and thus m/n = 2.

9 Medium

9 of 267

Application +4 | -1

\lim \limits_{{x \rightarrow \infty}} f(x)

, where

\frac{2 x-3}{x} < f(x) < \frac{2 x^2+5 x}{x^2}

, is-

View answer and explanation

Correct option: B

Option B (2) is correct because the limit is evaluated using the Squeeze Theorem (Sandwich Theorem).

10 Medium

10 of 267

Analyzing +4 | -1

L=\lim \limits_{{x \rightarrow 4}}\left(\frac{\log _5\left(\frac{5 x}{4}\right)}{\log _5 \frac{20}{x}}\right)^{\left(\log _5 \frac{x}{4}\right)^{-1}}

,then [L] is (where [.] denotes greatest integer function)

View answer and explanation

Correct option: B

Option B (7) is correct because the limit L evaluates to

e^2

, and the greatest integer function of

e^2

is 7.

11 Medium

11 of 267

Analyzing +4 | -1

The value of

\lim\limits_{x \to 0} \frac{{1 + \sin x - \cos x + \log \left( {1 - x} \right)}}{{{x^3}}}

View answer and explanation

Correct option: C

- \frac{1}{2}

Option C (-

\frac{1}{2}

) is correct because using Taylor series expansions for the trigonometric and logarithmic functions, the limit evaluates to -

\frac{1}{2}

12 Medium

12 of 267

Application +4 | -1

{l_1} = \lim\limits_{x \to - 2} \left( {x + \left| x \right|} \right),\;{l_2} = \lim\limits_{x \to - 2} \left( {2x + \left| x \right|} \right)

and

{l_3} = \lim\limits_{x \to \pi /2} \frac{{\cos x}}{{x - \pi /2}}

, then

View answer and explanation

Correct option: B

{l_2} < {l_3} < {l_1}

Option B (

{l_2} < {l_3} < {l_1}

) is correct because it accurately compares the limits calculated by considering the absolute value and the limit of the trigonometric function.

13 Medium

13 of 267

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}}\left[\frac{\sin (\operatorname{sgn}(x))}{\operatorname{sgn}(x)}\right]

, where [·] denotes the greatest integer function is equal to

View answer and explanation

Correct option: A

Option A (0) is correct because the limit from both the right and left sides equals 0, considering the greatest integer function.

14 Medium

14 of 267

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 0}}\left(\frac{1}{x}-\cot x\right)

is equal to

View answer and explanation

Correct option: A

Option A (0) is correct because after applying L'Hôpital's rule twice, the limit evaluates to 0.

15 Medium

15 of 267

Application +4 | -1

\lim _{x \rightarrow \infty}\left(1+\frac{a}{x}+\frac{b}{x^2}\right)^{2 x}=e^2

, then the values of

a

and

b

, are

View answer and explanation

Correct option: B

a=1, b \in \mathrm{R}

Option B (a=1, b ∈ R) is correct because the limit depends only on 'a', and 'b' can take any real value.

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16 Medium

16 of 267

Analyzing +4 | -1

The value of

\lim\limits_{x \to \infty } \sqrt {{a^2}{x^2} + ax + 1} - \sqrt {{a^2}{x^2} + 1}

View answer and explanation

Correct option: A

\frac{1}{2}

Option A (

\frac{1}{2}

) is correct because after rationalizing and taking the limit as x approaches infinity, the expression simplifies to 1/2.

17 Medium

17 of 267

Application +4 | -1

The value of the limit

\lim \limits_{{x \rightarrow 2}} \frac{e^{3 x-6}-1}{\sin (2-x)}

View answer and explanation

Correct option: C

-3

Option C (-3) is correct because L'Hopital's rule is applied after substituting to get the 0/0 form.

18 Medium

18 of 267

Analyzing +4 | -1

\lim \limits_{{x \rightarrow 1}} \frac{\int\limits_{1}^x|t-1| d t}{\sin (x-1)}

is equal to

View answer and explanation

Correct option: A

Option A (0) is correct. By applying L'Hopital's rule (or Leibnitz's rule) and evaluating the limit, we find that the limit is 0.

19 Medium

19 of 267

Analyzing +4 | -1

Let

{a}_m

be the

{m}^{\text {th }}

term of the sequence

\left(\frac{2^2}{1^2}-\frac{2}{1}\right)^{-1},\left(\frac{3^3}{2^3}-\frac{3}{2}\right)^{-2},\left(\frac{4^4}{3^4}-\frac{4}{3}\right)^{-3}, \ldots .

. , then the value of

\lim \limits_{{m \rightarrow \infty}}\left(a_m\right)^{\frac{1}{m}}

is equal to

View answer and explanation

Correct option: C

\frac{1}{{e}-1}

Option C,

\frac{1}{{e}-1}

, is correct because the limit of the given expression as m approaches infinity simplifies to this value using properties of limits and the definition of e.

20 Medium

20 of 267

Application +4 | -1

\lim\limits_{x \to 0} {\left\{ {\tan \left( {\frac{\pi }{4} + x} \right)} \right\}^{\frac{1}{x}}} =

View answer and explanation

Correct option: B

e^2

Option B (

e^2

) is correct because the limit can be evaluated by using the tangent addition formula and then applying the standard limit formula.

21 Medium

21 of 267

Application +4 | -1

f(1)=1

and

f^{\prime}(1)=4

, then the value of

\lim \limits_{{x \rightarrow 1}} \frac{\sqrt{f(x)}-1}{\sqrt{x}-1}

View answer and explanation

Correct option: A

Option A (4) is correct because it is the limit calculated using L'Hopital's rule after recognizing the indeterminate form.

22 Medium

22 of 267

Analyzing +4 | -1

Let

\alpha

and

\beta

be the roots of

a\;{x^2} + b\;x + c = 0

, then

\lim\limits_{x \to a} \frac{{1 - \cos \left( {a\;{x^2} + bx + c} \right)}}{{{{\left( {x - \alpha } \right)}^2}}}

is equal to

View answer and explanation

Correct option: C

\frac{{{\alpha ^2}}}{2}{\left( {\alpha - \beta } \right)^2}

Option C (

\frac{{{{\alpha ^2}}}}{2}{\left( {\alpha - \beta } \right)^2}

) is correct because after applying trigonometric identities and L'Hopital's rule, the limit evaluates to the given expression.

23 Medium

23 of 267

Application +4 | -1

f(9)=9, f^{\prime}(9)=4

, then

\lim \limits_{{x \rightarrow 9}} \frac{\sqrt{f(x)}-3}{\sqrt{x}-3}=

View answer and explanation

Correct option: B

Option B (4) is correct because by applying L'Hôpital's rule, the limit simplifies to 4.

24 Medium

24 of 267

Analyzing +4 | -1

f(x)=\lim \limits_{{n \rightarrow \infty}} \sum\limits_{{r=0}}^n \frac{\tan \frac{x}{2^{r+1}}+\tan ^3 \frac{x}{2^{r+1}}}{1-\tan ^2 \frac{x}{2^{r+1}}}

then

\lim \limits_{{x \rightarrow 0}} \frac{f(x)}{x}

is equal to

View answer and explanation

Correct option: A

Option A (1) is correct because after simplifying the summation and using the limit definition of tan(x)/x as x approaches 0, the limit evaluates to 1.

25 Medium

25 of 267

Application +4 | -1

Let

\alpha=\lim \limits_{{n \rightarrow \infty}} \frac{\left(1^3-1^2\right)+\left(2^3-2^2\right)+\ldots+\left(n^3-n^2\right)}{n^4}

, then

\alpha

is equal to:

View answer and explanation

Correct option: B

\frac{1}{4}

Option B (1/4) is correct because the limit of the given expression as n approaches infinity simplifies to 1/4 after evaluating the sums.

26 Medium

26 of 267

Application +4 | -1

The value of

\lim \limits_{{x \rightarrow 2 a}} \frac{\sqrt{x-2 a}+\sqrt{x}-\sqrt{2 a}}{\sqrt{x^2-4 a^2}}

View answer and explanation

Correct option: B

\frac{1}{2 \sqrt{a}}

Option B (

\frac{1}{2 \sqrt{a}}

) is correct because the limit is evaluated by rationalizing and simplifying the expression.

27 Medium

27 of 267

Application +4 | -1

The value of the limit

\lim\limits_{x \to 1} \frac{{\sin \left( {{e^{x - 1}} - 1} \right)}}{{\log x}}

View answer and explanation

Correct option: D

Option D (1) is correct because by substituting and using standard limits, the expression simplifies to 1.

28 Medium

28 of 267

Application +4 | -1

The value of

\lim\limits_{x \to 0} {\left( {\frac{{1 + 5{x^2}}}{{1 + 3{x^2}}}} \right)^{1/{x^2}}}

View answer and explanation

Correct option: A

{e^2}

Option A (

{e^2}

) is correct because the limit can be transformed into the form

{e^{\lim f(x)}}

and the exponent evaluates to 2.

29 Medium

29 of 267

Understanding +4 | -1

The left hand limit of the function

f(x) = \left\{ {\begin{array}{llllllllllllllllllll} {\frac{{|x - 4|}}{{x - 4}},}&{x \ne 4}\\ {0,}&{x = 4} \end{array}} \right.

x=4

, is

View answer and explanation

Correct option: B

-1

Option A (Increasing on

(0, \pi / 2)

) is correct because the derivative of

f(x)

is positive on this interval, indicating that

f(x)

is monotonically increasing.

30 Medium

30 of 267

Analyzing +4 | -1

f(x)

is differentiable and

f^{\prime}(0) \neq 0

, then the value of

\lim _{x \rightarrow 0} \frac{f\left(x^2\right)-f(x)}{f(x)-f(0)}

View answer and explanation

Correct option: C

-1

Option C (-1) is correct because after applying L'Hôpital's rule, the limit simplifies to -1, given that f'(0) is not zero.

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Key Takeaways

✓ This set contains 267 single-correct-answer MCQs on LIMITS.
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