Question

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

Answer 1

Answer: 5

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