Q. 46 - 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.

$\underset{x\to 1}{\lim }\frac{\sin \left(\ln x\right)}{x-1}$.

Step-by-Step Solution

Answer

The exact value of the limit $\underset{x\to 1}{\lim }\frac{\sin \left(\ln x\right)}{x-1}$ is, $1.$

Step 1 . Given information

$\underset{x\to 1}{\lim }\frac{\sin \left(\ln x\right)}{x-1}$.

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

$\underset{x\to 1}{\lim }\frac{\sin \left(\ln x\right)}{x-1}=\underset{x\to 1}{\lim }\frac{\sin \left(\ln x\right)}{x-1}$ [ in the form of $\frac{0}{0}$; Using L'Hopital's rule]

                      $$\begin{gathered} =\underset{x\to 1}{\lim }\frac{\cos \left(\ln x\right)·\frac{1}{x}}{1-0} \\ =\underset{x\to 1}{\lim }\frac{\cos \left(\ln x\right)·\frac{1}{x}}{1} \\ =\cos \left(\ln 1\right)·\frac{1}{1} \\ =1.1 \\ =1 \end{gathered}$$

Therefore, the exact value is, $1.$