Q. 40 - 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.

$\underset{x\to 0^{+}}{\lim }\frac{\ln x}{\ln x-x+1}$.

Step-by-Step Solution

Answer

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

Step 1 . Given information

$\underset{x\to 0^{+}}{\lim }\frac{\ln x}{\ln x-x+1}$.

Step 2 . The limit is given below:

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

                            $$\begin{gathered} =\underset{x\to 0^{+}}{\lim }\frac{\frac{1}{1-x}}{x} \\ =\underset{x\to 0^{+}}{\lim }\frac{x}{1-x} \\ =0 \end{gathered}$$

The exact value is, $0$.