Question

Column - I (A) \(\lim _{x \rightarrow \infty}(\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}})\) equals (B) The value of the limit, $\lim _{x \rightarrow 0} \frac{\sin 2 x-2 \tan x}{\ln \left(1+x^{3}\right)}$ is (C) $\lim _{x \rightarrow 0^{-}}\left(\ln \sin ^{3} x-\ln \left(x^{4}+e x^{3}\right)\right)$ equals (D) Let tan \((2 \pi|\sin \theta|)=\cot (2 \pi|\cos \theta|)\), where $\theta \in \mathbb{R}$ and \(\mathrm{f}(\mathrm{x})=(|\sin \theta|+\cos \theta \mid)^{\mathrm{x}} .\) The value of $\lim _{\mathrm{x} \rightarrow \infty}\left[\frac{2}{\mathrm{f}(\mathrm{x})}\right]$ equals (Here [] represents greatest integer function) Column - II (P) \(-2\) (Q) \(-1\) (R) 0 (S) 1

Short answer

Question: Match the limits with their corresponding solutions. A. \(\lim _{x \rightarrow \infty}(\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}})\) B. \(\lim _{x \rightarrow 0} \frac{\sin 2 x-2 \tan x}{\ln \left(1+x^{3}\right)}\) C. \(\lim _{x \rightarrow 0^{-}}\left(\ln \sin ^{3} x-\ln \left(x^{4}+e x^{3}\right)\right)\) D. \(\lim _{x \rightarrow \infty}\left[\frac{2}{(|\cos \theta - \sin \theta|)^x}\right]\) Solutions: (Q) -1 (R) 0 (S) 1 Answer: Limit A corresponds to (S) Limit B corresponds to (Q) Limit C corresponds to (R) Limit D corresponds to (S)

Step-by-step solution

Examining Limit A

First, let us examine the limit A: \(\lim _{x \rightarrow \infty}(\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}})\) To simplify the expression, we'll try rationalizing the numerator. Multiplying the term inside the limit by \(\frac{\sqrt{x+\sqrt{x}}+\sqrt{x-\sqrt{x}}}{\sqrt{x+\sqrt{x}}+\sqrt{x-\sqrt{x}}}\), we get: \(\lim _{x \rightarrow \infty}\frac{x}{\sqrt{x+\sqrt{x}}+\sqrt{x-\sqrt{x}}}\)

Solving Limit A

As x approaches infinity, \(\sqrt{x}\) approaches infinity as well. Therefore, we can divide each term inside the limit by \(\sqrt{x}\): \(\lim _{x \rightarrow \infty}\frac{1}{\sqrt{1+\frac{1}{\sqrt{x}}}+\sqrt{1-\frac{1}{\sqrt{x}}}} = \frac{1}{\sqrt{1+0}+\sqrt{1-0}} = 1\) So, limit A corresponds to (S).

Examining Limit B

Now, let us examine the limit B: \(\lim _{x \rightarrow 0} \frac{\sin 2 x-2 \tan x}{\ln \left(1+x^{3}\right)}\)

Solving Limit B

We can rewrite the limit using L'Hopital's Rule: \(\lim _{x \rightarrow 0} \frac{\sin 2x - 2\tan x}{\ln(1+x^3)} = \lim _{x \rightarrow 0} \frac{2\cos 2x - 2\sec^2 x}{3x^2/(1+x^3)}\) Using the fact that \(\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1\), we get: \(\lim _{x \rightarrow 0} \frac{2\cos 2x - 2(1+\tan^2 x)}{3x^2} = \frac{2\cos 0- 2}{0} = -1\) So, limit B corresponds to (Q).

Examining Limit C

Moving on to limit C: \(\lim _{x \rightarrow 0^{-}}\left(\ln \sin ^{3} x-\ln \left(x^{4}+e x^{3}\right)\right)\)

Solving Limit C

We can simplify the expression inside the limit as follows: \(\lim _{x \rightarrow 0^{-}}\left(\ln \frac{\sin ^{3} x}{x^{4}+e x^{3}}\right)\) Using L'Hopital's Rule, we differentiate the numerator and the denominator inside the limit with respect to x: \(\lim _{x \rightarrow 0^{-}}\frac{3\sin^2x\cos x}{4x^3+3ex^2} = \frac{0}{0+0} = 0\) So, limit C corresponds to (R).

Examining Limit D

Finally, let us work on limit D: The given expression is: \(tan (2 \pi|\sin \theta|)=\cot (2 \pi|\cos \theta|),f(x)=(|\sin \theta|+\cos \theta)^{x}\) The limit to be found is: \(\lim _{x \rightarrow \infty}\left[\frac{2}{f(x)}\right]\)

Solving Limit D

Using the given equation and the fact that \(\cot x = \frac{1}{\tan x}\), we have: \(\tan (2 \pi|\sin \theta|) = \frac{1}{\tan (2 \pi|\cos \theta|)}\) Squaring both sides, we get: \(\tan ^2 (2 \pi|\sin \theta|) = \frac{1}{\tan ^2 (2 \pi|\cos \theta|)}\) Since \(\tan^2 x = \frac{\sin^2 x}{\cos^2 x}\), we obtain: \(\frac{\sin ^2 (2 \pi|\sin \theta|)}{\cos ^2 (2 \pi|\sin \theta|)} = \frac{\cos ^2 (2 \pi|\cos \theta|)}{\sin ^2 (2 \pi|\cos \theta|)}\) Now, we compute the limit: \(\lim _{x \rightarrow \infty}\left[\frac{2}{(|\cos \theta - \sin \theta|)^x}\right] = \left[\frac{2}{1}\right] = 1\) So, limit D corresponds to (S). In conclusion, we have: - Limit A corresponds to (S) - Limit B corresponds to (Q) - Limit C corresponds to (R) - Limit D corresponds to (S)