Question

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

Answer 1

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

Rationalising the denominator, we get:

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