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

Step 1: Apply L'Hospital Rule.

As $\underset{x\to 0}{lim}\frac{ax-\left(e^{4x}-1\right)}{ax\left(e^{4x}-1\right)}$is in $\frac{0}{0}$form so, we will apply L'Hospital rule, according to which $\underset{x\to c}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to c}{lim}\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}$.

So,

$\underset{x\to 0}{lim}\frac{ax-\left(e^{4x}-1\right)}{ax\left(e^{4x}-1\right)}=\underset{x\to 0}{lim}\frac{a-4e^{4x}}{a\left(e^{4x}-1\right)+ax\left(4e^{4x}\right)}$

By applying limit we get

$$\begin{gathered} \frac{a-4e^0}{a\left(e^0-1\right)+a\left(0\right)\left(4e^0\right)} \\ =\frac{a-4}{0} \end{gathered}$$

Step 2: Find the value of $a$.

For $\underset{x\to 0}{lim}\frac{ax-\left(e^{4x}-1\right)}{ax\left(e^{4x}-1\right)}$to exist.

$\Rightarrow a=4$

Step 3: Find the value of $b$.

By applying L'Hospital Rule to $\underset{x\to 0}{lim}\frac{a-4e^{4x}}{a\left(e^{4x}-1\right)+ax\left(4e^{4x}\right)}$, we get

$\underset{x\to 0}{lim}\frac{a-4e^{4x}}{a\left(e^{4x}-1\right)+ax\left(4e^{4x}\right)}=\underset{x\to 0}{lim}\frac{-16e^{4x}}{a\left(4e^{4x}\right)+a\left(4e^{4x}\right)+ax\left(16e^{4x}\right)}$

By applying limit, we get

$\begin{array}{rcl}\frac{-16\left(e^0\right)}{a\left(4e^0\right)+a\left(4e^0\right)+a\left(0\right)\left(16e^0\right)}& =& b\\ \Rightarrow \frac{-16}{8a}& =& b\\ \Rightarrow \frac{-16}{8\left(4\right)}& =& b\\ \Rightarrow b& =& -\frac{1}{2}\end{array}$

Step 4: Find the value of $a-2b$.

$\begin{array}{rcl}a-2b& =& 4-2\left(-\frac{1}{2}\right)\\ & =& 5\end{array}$

Therefore, $a-2b=5$