Question

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

Answer 1

Answer: (C)

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

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

$=\underset{x\to 0}{lim}\frac{(2^m+x{)}^{\frac{1}{m}}-2}{x}-\underset{x\to 0}{lim}\frac{(2^n+x{)}^{\frac{1}{n}}-2}{x}$
$=\underset{a\to 2}{lim}\frac{a-2}{a^m-2^m}-\underset{b\to 2}{lim}\frac{b-2}{b^m-2^m}\text{[Putting}{2}^m+x=a^m\text{and}{2}^n+x=b^n\text{]}$
$=\frac{1}{m{2}^{m-1}}-\frac{1}{n{2}^{n-1}}$
Hence, option 'C' is correct.