Question

If $\underset{x\to 0}{lim}\frac{a{e}^x-b\cos x+c{e}^{-x}}{x\sin x}=2$, then $a+b+c$ is equal to:

Answer 1

Step 1: Solve $\underset{x\to 0}{lim}\frac{a{e}^x-b\cos x+c{e}^{-x}}{x\sin x}$

$$\begin{gathered} \underset{x\to 0}{lim}\frac{a{e}^x-b\cos x+c{e}^{-x}}{x\sin x}=\frac{a{e}^0-b\cos 0+c{e}^{-0}}{0\sin 0} \\ =\frac{a-b+c}{0} \\ =\infty \end{gathered}$$

Step 2: Apply L' Hospital's Rule.

$\begin{array}{rcl}\underset{x\to 0}{lim}\frac{a{e}^x-b\cos x+c{e}^{-x}}{x\sin x}& =& \underset{x\to 0}{lim}\frac{a{e}^x+b\sin x-c{e}^{-x}}{\sin x+x\cos x}\\ & =& \frac{a{e}^0+b\sin 0-c{e}^0}{\sin 0+0\cos 0}\\ & =& \frac{a-c}{0}\\ & =& \infty \end{array}$

So, we will again apply L' Hospital's Rule.

$\begin{array}{rcl}\begin{array}{l}\underset{x\to 0}{lim}\end{array}\begin{array}{l}\frac{a{e}^x+b\sin x-c{e}^{-x}}{\sin x+x\cos x}\end{array}& =& \begin{array}{l}\underset{x\to 0}{lim}\end{array}\begin{array}{l}\frac{a{e}^x+b\cos x+c{e}^{-x}}{\cos x+\cos x-x\sin x}\end{array}\\ & =& \begin{array}{l}\frac{a{e}^0+b\cos 0+c{e}^0}{\cos 0+\cos 0-0\sin 0}\end{array}\\ & =& \frac{a+b+c}{1+1-0}\\ & =& \frac{a+b+c}{2}\end{array}$

Step 3: Find the value of $a+b+c$.

We have $\underset{x\to 0}{lim}\frac{a{e}^x-b\cos x+c{e}^{-x}}{x\sin x}=2$

$\begin{array}{rcl}& \Rightarrow & \frac{a+b+c}{2}=2\\ \Rightarrow a+b+c& =& 4\end{array}$

Hence, $a+b+c=4$