Question

$\underset{x\to 0}{\lim }\frac{{\left(2^m+x\right)}^{1/m}-{\left(2^n+x\right)}^{1/n}}{x}$ 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)

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