Question

The limit of $\{\frac{1}{x}\sqrt{1+x}-\sqrt{1+\frac{1}{x^2}}\}$ as $x\to 0$

• A. does not exist
• B. is equal to $\frac{1}{2}$
• C. is equal to $0$
• D. is equal to $1$

Answer 1

Answer: (B)

is equal to $\frac{1}{2}$

$\underset{x\to 0}{lim}\{\frac{1}{x}\sqrt{1+x}-\sqrt{1+\frac{1}{x^2}}\}$
$=\underset{x\to 0}{lim}\left\{\frac{\sqrt{1+x}-\sqrt{1+x^2}}{x}\right\}\left(\frac{0}{0}form\right)$
$=\underset{x\to 0}{lim}\frac{\frac{1}{2\sqrt{1+x}}-\frac{x}{\sqrt{1+x^2}}}{1}$
$=\frac{1}{2\sqrt{1+0}}-0=\frac{1}{2}$