Q. 45 - Calculus

Question

Calculate each of the limits in Exercises $21-48$. Some of these limits are made easier by L’Hopital’s rule, and some are not.

$\underset{x\to 0}{\lim }\frac{x\cos x}{1-e^x}$.

Step-by-Step Solution

Answer

The exact value of the limit $\underset{x\to 0}{\lim }\frac{x\cos x}{1-e^x}$ is, $-1$.

Step 1 . Given information

$\underset{x\to 0}{\lim }\frac{x\cos x}{1-e^x}$.

Step 2 . The limit using L'Hopital's rule is given below:

$\underset{x\to 0}{\lim }\frac{x\cos x}{1-e^x}=\underset{x\to 0}{\lim }\frac{x\cos x}{1-e^x}$ [ in the form of $\frac{0}{0}$; Using L'Hopital's rule]

                    $$\begin{gathered} =\underset{x\to 0}{\lim }\frac{-x\sin x+\cos x}{-e^x} \\ =\frac{0+\cos \left(0\right)}{-1} \\ =-1 \end{gathered}$$

Therefore, the exact value is, $-1$.