Question

If $\displaystyle\lim _{x \rightarrow 0} \frac{ax-\left( e ^{4 x}-1\right)}{ax\left( e ^{4 x }-1\right)}$ exists and is equal to $b ,$ then the value of $a -2 b$ is _______.

Answer 1

$\displaystyle\lim _{x \rightarrow 0} \frac{a x-\left(e^{4 x}-1\right)}{ax\left(e^{4 x}-1\right)} \left(\frac{0}{0}\right)$
$=\displaystyle\lim _{x \rightarrow 0} \frac{ax-\left(e^{4 x}-1\right)}{ax \cdot 4 x} $ Use $\displaystyle\lim _{x \rightarrow 0} \frac{e^{4 x}-1}{4 x}=1$
Apply L'Hospital Rule
$=\displaystyle\lim _{x \rightarrow 0} \frac{a-4 e^{4 x}}{8 a x} \left(\frac{a-4}{0}\right.$ form $)$
limit exists only when $a-4=0$
$ \Rightarrow a=4$
$=\displaystyle\lim _{x \rightarrow 0} \frac{4-4 e ^{4 x }}{32 x }$
$-\displaystyle\lim _{x \rightarrow 0} \frac{1- e ^{4 x }}{8 x } \left(\frac{0}{0}\right)$
$=\displaystyle\lim _{x \rightarrow 0} \frac{- e ^{4 x } \cdot 4}{8}=-\frac{1}{2} $
$\Rightarrow b =-\frac{1}{2}$
$a -2 b =4-2\left(-\frac{1}{2}\right)$
$=5$