Question

Evaluate the $\underset{x\to 0}{\lim }\left(\frac{\log (1+5x)}{x}\right)$.

Answer 1

Evaluate for the required value

Given: $\underset{x\to 0}{\lim }\left(\frac{\log (1+5x)}{x}\right)$

applying L' Hospital Rule, where $\underset{x\to a}{\lim }\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{f\text{'}\left(a\right)}{g\text{'}\left(a\right)}$

Put $f\left(x\right)=\log (1+5x),g\left(x\right)=x$

$\underset{x\to 0}{\lim }\left(\frac{\log (1+5x)}{x}\right)=\frac{\left({\displaystyle \frac{1}{1+5x}}\right)\left(5\right)}{1}$ at $x=0$

$$\begin{gathered} \Rightarrow \underset{x\to 0}{\lim }\left(\frac{\log (1+5x)}{x}\right)=\frac{5}{1+5\times 0} \\ \Rightarrow \underset{x\to 0}{\lim }\left(\frac{\log (1+5x)}{x}\right)=5 \end{gathered}$$

Hence, the required answer is $5$.