Question-29
Evaluate:
\[ \begin{equation*} \lim\limits _{x\rightarrow 0} \ \ \frac{\log( 5+x) -\log( 5-x)}{x} \end{equation*} \]
\[ \lim\limits _{z\rightarrow 0} \ \frac{\log( 1+z)}{z} =1 \]
\(\cfrac{2}{5}\)
Let us make use of an existing limit:
\[ \begin{equation*} \lim\limits _{z\rightarrow 0} \ \frac{\log( 1+z)}{z} =1 \end{equation*} \]
We can try to bring the numerator in this form. To this end, let:
\[ \begin{equation*} \begin{aligned} 1+y & =\frac{5+x}{5-x}\\ & \\ \Longrightarrow y & =\frac{2x}{5-x} \end{aligned} \end{equation*} \]
We can rewrite the limit as:
\[ \begin{equation*} \begin{aligned} \lim\limits _{x\rightarrow 0} \ \frac{\log\left( 1+\frac{2x}{5-x}\right)}{x} & =\lim\limits _{x\rightarrow 0}\left\{\ \frac{\log\left( 1+\frac{2x}{5-x}\right)}{\frac{2x}{5-x}} \times \frac{2}{5-x}\right\}\\ & \\ & =\left\{\lim\limits _{x\rightarrow 0} \ \frac{\log\left( 1+\frac{2x}{5-x}\right)}{\frac{2x}{5-x}}\right\} \times \lim\limits _{x\rightarrow 0} \ \frac{2}{5-x}\\ & \\ & =1\times \frac{2}{5}\\ & \\ & =\frac{2}{5} \end{aligned} \end{equation*} \]