Q. 42 - 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.

$\underset{x\to 0}{\lim }\frac{\sin x}{x+\sin x}$.

Step-by-Step Solution

Answer

The exact value of the limit  $\underset{x\to 0}{\lim }\frac{\sin x}{x+\sin x}$ is, $\frac{1}{2}$.

Step 1 . Given information

$\underset{x\to 0}{\lim }\frac{\sin x}{x+\sin x}$.

Step 2 . The limit is given below:

$\underset{x\to 0}{\lim }\frac{\sin x}{x+\sin x}=\underset{x\to 0}{\lim }\frac{\cos x}{1+\cos x}$ [L'Hopital's rule]

                      $$\begin{gathered} =\frac{\cos 0}{1+\cos 0} \\ =\frac{1}{1+1} \\ =\frac{1}{2} \end{gathered}$$

Therefore, the exact value is,$\frac{1}{2}$.