Question

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

Answer 1

$\underset{x\to 0}{lim}\frac{ax-(e^{4x}-1)}{ax(e^{4x}-1)}$
Applying L'Hospital Rule
$\underset{x\to 0}{lim}\frac{a-4e^{4x}}{a(e^{4x}-1)+ax\left(4e^{4x}\right)}$
So for limit to exist, $a=4$
Applying L′Hospital Rule
$\underset{x\to 0}{lim}\frac{-16e^{4x}}{a\left(4e^{4x}\right)+a\left(4e^{4x}\right)+ax\left(16e^{4x}\right)}$
$=\frac{-16}{4a+4a}=\frac{-16}{32}=-\frac{1}{2}=b$
$\therefore a-2b=4-2\left(\frac{-1}{2}\right)=4+1=5$