Question

$\displaystyle\lim _{x \rightarrow 0} \frac{\left(2^{m}+x\right)^{1 / m}-\left(2^{n}+x\right)^{1 / n}}{x}$ is equal to

• A. $\frac{1}{m 2^{m}}-\frac{1}{n 2^{n}}$
• B. $\frac{1}{m 2^{m}}+\frac{1}{n 2^{n}}$
• C. $\frac{1}{m 2^{m-1}}-\frac{1}{n 2^{n-1}}$
• D. None of these

Answer 1

Answer: (C)

$\displaystyle\lim _{x \rightarrow 0} \frac{\left(2^{m}+x\right)^{1 / m}-\left(2^{n}+x\right)^{1 / n}}{x}$
$=\displaystyle\lim _{x \rightarrow 0} \frac{\left(2^{m}+x\right)^{1 / m}-2}{x}-\displaystyle\lim _{x \rightarrow 0} \frac{\left(2^{n}+x\right)^{1 / n}-2}{x}$
$=\displaystyle\lim _{a \rightarrow 2} \frac{a-2}{a^{m}-2^{m}}-\displaystyle\lim _{b \rightarrow 2} \frac{b-2}{b^{n}-2^{n}}$
[Putting $2^{m}+x=a^{m}$ and $2^{n}+x=b^{n}$ ]
$=\frac{1}{m 2^{m-1}}-\frac{1}{n 2^{n-1}}$