Q. 42 - Calculus

Question

Calculate each of the limits in Exercises $29-70$.

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

Step-by-Step Solution

Answer

The value of $\underset{x\to 1^{+}}{\lim }\frac{1-\sqrt{x}}{1-x}$ is, $\frac{1}{2}$.

Step 1 . Given information

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

Step 2 . The limit is calculated as below:

$$\begin{gathered} \underset{x\to 1^{+}}{\lim }\frac{1-\sqrt{x}}{1-x}=\underset{x\to 1^{+}}{\lim }\frac{(1-\sqrt{x})(1+\sqrt{x})}{(1-x)(1+\sqrt{x})} \\ =\underset{x\to 1^{+}}{\lim }\frac{1^2-(\sqrt{x}{)}^2}{(1-x)(1+\sqrt{x})} \\ =\underset{x\to 1^{+}}{\lim }\frac{(1-x)}{(1-x)(1+\sqrt{x})} \\ =\underset{x\to 1^{+}}{\lim }\frac{1}{1+\sqrt{x}} \\ =\frac{1}{1+\sqrt{1}} \\ =\frac{1}{2} \end{gathered}$$