Question

Limit X tends to zero e raised to a x - e raise to minus A X upon log 1 + bx

Answer 1

Question :- Evaluate

$\rm \: \displaystyle\lim_{x \to 0}\rm \: \frac{ {e}^{ax} - {e}^{ - ax} }{log(1 + bx)}$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \displaystyle\lim_{x \to 0}\rm \: \frac{ {e}^{ax} - {e}^{ - ax} }{log(1 + bx)}$

If we substitute directly x = 0, we get

$\rm \: = \: \dfrac{ {e}^{0} - {e}^{0}}{log1}$

$\rm \: =  \:\dfrac{1 - 1}{0}$

$\rm \: =  \:\dfrac{0}{0}$

which is indeterminant form.

So, Consider again

$\rm \: \displaystyle\lim_{x \to 0}\rm \: \frac{ {e}^{ax} - {e}^{ - ax} }{log(1 + bx)}$

can be rewritten as

$\rm \: = \: \displaystyle\lim_{x \to 0}\rm \bigg({e}^{ax} - \dfrac{1}{{e}^{ax}} \bigg) \times \displaystyle\lim_{x \to 0}\rm \frac{1}{log(1 + bx)}$

$\rm \: = \: \displaystyle\lim_{x \to 0}\rm \bigg(\dfrac{{e}^{2ax} - 1}{{e}^{ax}} \bigg) \times \displaystyle\lim_{bx \to 0}\rm \frac{bx}{log(1 + bx)} \times \frac{1}{bx}$

We know,

$\boxed{ \rm{ \:\displaystyle\lim_{x \to 0}\rm \frac{log(1 + x)}{x} = 1 \: }}$

So, using this, we get

$\rm \: = \:\displaystyle\lim_{x \to 0}\rm \frac{1}{{e}^{ax}} \times \displaystyle\lim_{x \to 0}\rm \bigg(\dfrac{{e}^{2ax} - 1}{bx} \bigg) \times 1$

$\rm \: = \dfrac{1}{b} \times \:1 \times \displaystyle\lim_{ax \to 0}\rm \bigg(\dfrac{{e}^{2ax} - 1}{2ax} \bigg) \times 2a$

We know,

$\boxed{ \rm{ \:\displaystyle\lim_{x \to 0}\rm \frac{{e}^{x} - 1}{x} = 1 \: }}$

So, using this, we get

$\rm \: = \: \dfrac{1}{b} \times 1 \times 2a$

$\rm \: = \: \dfrac{2a}{b}$

Hence,

$\rm\implies \:\rm \: \displaystyle\lim_{x \to 0}\rm \: \frac{ {e}^{ax} - {e}^{ - ax} }{log(1 + bx)} = \frac{2a}{b}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{sinx}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{tanx}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{log(1 + x)}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{ {e}^{x} - 1}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{ {a}^{x} - 1}{x} = loga}\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}$