Question

Evaluate the limit:
$\underset{x\to 0}{lim}\frac{5^x-1}{\sqrt{4+x}-2}$

Answer 1

We have:
$\underset{x\to 0}{lim}\frac{5^x-1}{\sqrt{4+x}-2}$

Rationalising the denominator, we get

$=\underset{x\to 0}{lim}\frac{(5^x-1)}{(\sqrt{4+x}-2)}\times \frac{\sqrt{4+x}+2}{\sqrt{4+x}+2}$

$=\underset{x\to 0}{lim}\left[\frac{(5^x-1)(\sqrt{4+x}+2)}{(\sqrt{4+x}{)}^2-2^2}\right]$

$[\because (a+b\left)\right(a-b)=a^2-b^2]$

$=\underset{x\to 0}{lim}\left[\frac{(5^x-1)(\sqrt{4+x}+2)}{4+x-4}\right]$

$=\underset{x\to 0}{lim}\left[\frac{(5^x-1)(\sqrt{4+x}+2)}{x}\right]$

$=\underset{x\to 0}{lim}[\left(\frac{5^x-1}{x}\right)\times (\sqrt{4+x}+2\left)\right]$

$=\ln 5\times (\sqrt{4+0}+2)[\because \underset{x\to 0}{lim}\frac{a^x-1}{x}=\ln a]$
$=\ln 5\times (2+2)=4\ln 5$

Therefore,
$\underset{x\to 0}{lim}\frac{5^x-1}{\sqrt{4+x}-2}=4\ln 5$