Question

Evaluate : $\underset{x\to 0}{lim}\frac{3^x-1}{\sqrt{1+x}-1}$.

Answer 1

$L={lim}_{x\to 0}\frac{3^x-1}{\sqrt{1+x}-1}$

Applying L' Hospital's rule

${lim}_{x\to 0}\frac{3^x\log 3}{\frac{1}{2\sqrt{x+1}}}={lim}_{x\to 0}2\sqrt{x+1}.3^x.\log 3=2\times \sqrt{0+1}\times 3^0.\log 3=2\log 3$