Question

Solve $\underset{x\to 0}{lim}\frac{2^x-1}{\sqrt{1+x}-1}$

Answer 1

Apply the limits on the given function,

$\underset{x\to 0}{lim}\frac{2^x-1}{\sqrt{1+x}-1}$

$=\frac{2^0-1}{\sqrt{1+0}-1}$

$=\frac{0}{0}$

This is the indeterminate form, therefore, applying L’Hospitals rule,

$\underset{x\to 0}{lim}\frac{2^x-1}{\sqrt{1+x}-1}=\underset{x\to 0}{lim}\frac{2^x\log 2}{\frac{1}{2\sqrt{1+x}}}$

$=\frac{2^0\log 2}{\frac{1}{2\sqrt{1+0}}}$

$=\frac{\log 2}{\frac{1}{2}}$

$=2\log 2$

$=\log 2^2$

$=\log 4$