Prove the following

Question:

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

Solution:

$\lim _{x \rightarrow 0} \frac{a x-\left(e^{4 x}-1\right)}{a x\left(e^{4 x}-1\right)} \quad\left(\frac{0}{0}\right)$

$=\lim _{x \rightarrow 0} \frac{\operatorname{ax}-\left(e^{4 x}-1\right)}{a x \cdot 4 x} \quad$ Use $\lim _{x \rightarrow 0} \frac{e^{4 x}-1}{4 x}=1$

Apply L'Hospital Rule

$=\lim _{x \rightarrow 0} \frac{a-4 e^{4 x}}{8 a x} \quad\left(\frac{a-4}{0}\right.$ form $)$

limit exists only when $a-4=0 \Rightarrow a=4$

$=\lim _{x \rightarrow 0} \frac{4-4 e^{4 x}}{32 x}$

$=\lim _{x \rightarrow 0} \frac{1-e^{4 x}}{8 x} \quad\left(\frac{0}{0}\right)$

$=\lim _{x \rightarrow 0} \frac{-e^{4 x} \cdot 4}{8}=-\frac{1}{2} \Rightarrow b=-\frac{1}{2}$

$a-2 b=4-2\left(-\frac{1}{2}\right)$

$=5$

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