Q. 43 - 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.

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

Step-by-Step Solution

Answer

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

Step 1 . Given information

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

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

$\underset{x\to 0}{\lim }\frac{1-\cos x}{\tan x}=\underset{x\to 0}{\lim }\frac{0-(-\sin x)}{{\sec }^{2}x}$ [ L'Hopital's rule]

                       $$\begin{gathered} =\underset{x\to 0}{\lim }\frac{\sin x}{{\sec }^{2}x} \\ =\frac{0}{1} \\ =0 \end{gathered}$$

The exact value of the limit is, $0$.