Question

Evaluate ${lim}_{x\to 0}\frac{\log (a+x)-\log (a-x)}{x}$

Answer 1

${lim}_{x\to 0}\frac{\log (a+x)-\log (a-x)}{x}$

$={lim}_{x\to 0}\frac{\log \left(\frac{a+x}{a-x}\right)}{x}$ [Since, $\log a-\log b=log\left(\frac{a}{b}\right)$]

$={lim}_{x\to 0}\frac{log\left[\frac{(a-x)+2x}{a-x}\right]}{x}$ [Adding and subtracting $x$ in the numerator of $\log$ function ]
$={lim}_{x\to 0}\frac{\log (1+\frac{2x}{a-x})}{x}$
Multiplying $\frac{2(a-x)}{2(a-x)}$ in the denominator of above expression, we get
$={lim}_{x\to 0}\frac{\log (1+\frac{2x}{a-x})}{x\times \left[\frac{2(a-x)}{2(a-x)}\right]}$
$={lim}_{x\to 0}\frac{\log (1+\frac{2x}{a-x})}{\frac{2x}{a-x}}$ $\times {lim}_{x\to 0}\frac{2}{a-x}$
[Since, ${lim}_{x\to a}f\left(x\right)\times g\left(x\right)={lim}_{x\to a}f\left(x\right)\times {lim}_{x\to a}g\left(x\right)$]

Applying the formula of ${lim}_{t\to 0}\frac{\log (1+t)}{t}=1$
Where $t=\frac{2x}{a-x}$
Thus, above expression can be written as $=1\times \frac{2}{a-0}$
$=\frac{2}{a}$
Therefore, the value of given expression is $=\frac{2}{a}$