limx→∞ 2m+x1m-2n+x1nx is equal to

limx→02m+x1m-2n+x1nx = limx→0 2m+x1m-2x -limx→0 2n+x1n-2x = lima→2 a-2am-2m -limb→2 b-2bn-2n Putting 2m+x=am and 2n+x=bπ =1m2m-1-1m2n-1

limx→∞ 2m+x1m-2n+x1nx is equal to

limx→02m+x1m-2n+x1nx = limx→0 2m+x1m-2x -limx→0 2n+x1n-2x = lima→2 a-2am-2m -limb→2 b-2bn-2n Putting 2m+x=am and 2n+x=bπ =1m2m-1-1m2n-1

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$\lim_{x \to \infty} \frac{{(2^{m} + x)}^{\frac{1}{m}} - {(2^{n} + x)}^{\frac{1}{n}}}{x}$ is equal to

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$\frac{1}{m 2^{m}} - \frac{1}{n 2^{n}}$

$\frac{1}{m 2^{m}} + \frac{1}{n 2^{n}}$

$\frac{1}{m 2^{m - 1}} - \frac{1}{n 2^{n = 1}}$

$\frac{1}{m 2^{m - 1}} + \frac{1}{n 2^{n - 1}}$

answer is C.

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Detailed Solution

$\begin{array}{l} \lim_{x \to 0} \frac{{(2^{m} + x)}^{\frac{1}{m}} - {(2^{n} + x)}^{\frac{1}{n}}}{x} \\ = \lim_{x \to 0} \frac{{(2^{m} + x)}^{\frac{1}{m}} - 2}{x} \underset{x \to 0}{- \lim} \frac{{(2^{n} + x)}^{\frac{1}{n}} - 2}{x} \\ = \lim_{a \to 2} \frac{a - 2}{a^{m} - 2^{m}} - \lim_{b \to 2} \frac{b - 2}{b^{n} - 2^{n}} \\ [Putting 2^{m} + x = a^{m} and 2^{n} + x = b^{π}] \\ = \frac{1}{m 2^{m - 1}} - \frac{1}{m 2^{n - 1}} \end{array}$