12.2 Finding limits: properties of limits  (Page 3/5)

Factor where possible, and simplify.

Find the LCD for the denominators of the two terms in the numerator, and convert both fractions to have the LCD as their denominator.

$\begin{array}{ll}\underset{x\to 0}{\lim }\left(\frac{\sqrt{25-x}-5}{x}\right)=\underset{x\to 0}{\lim }\left(\frac{\left(\sqrt{25-x}-5\right)}{x}\cdot \frac{\left(\sqrt{25-x}+5\right)}{\left(\sqrt{25-x}+5\right)}\right)\hfill & \text{Multiply numerator and denominator by the conjugate}.\hfill \\ =\underset{x\to 0}{\lim }\left(\frac{\left(25-x\right)-25}{x\left(\sqrt{25-x}+5\right)}\right)\hfill & \text{Multiply: }\left(\sqrt{25-x}-5\right)\cdot \left(\sqrt{25-x}+5\right)=\left(25-x\right)-25.\hfill \\ =\underset{x\to 0}{\lim }\left(\frac{-x}{x\left(\sqrt{25-x}+5\right)}\right)\hfill & \text{Combine like terms}.\hfill \\ =\underset{x\to 0}{\lim }\left(\frac{-\cancel{x}}{\cancel{x}\left(\sqrt{25-x}+5\right)}\right)\hfill & \text{Simplify }\frac{-x}{x}=-1.\hfill \\ =\frac{-1}{\sqrt{25-0}+5}\hfill & \text{Evaluate}.\hfill \\ =\frac{-1}{5+5}=-\frac{1}{10}\hfill & \hfill \end{array}$

$$\begin{array}{ll}\underset{x\to 4}{\lim }(\frac{4-x}{\sqrt{x}-2})=\underset{x\to 4}{\lim }(\frac{(2+\sqrt{x})(2-\sqrt{x})}{\sqrt{x}-2})\hfill & \text{Factor.}\hfill \\ =\underset{x\to 4}{\lim }(\frac{(2+\sqrt{x})\cancel{(2-\sqrt{x})}}{-\cancel{(2-\sqrt{x})}})\hfill & \text{Factor }\mathrm{-1}\text{ out of the denominator}\text{. Simplify}.\hfill \\ =\underset{x\to 4}{\lim }-(2+\sqrt{x})\hfill & \text{Evaluate}.\hfill \\ =-(2+\sqrt{4})\hfill & \hfill \\ =-4\hfill & \hfill \end{array}$$

The function is undefined at $x=7,$ so we will try values close to 7 from the left and the right.

Left-hand limit: $\frac{\left|6.9-7\right|}{6.9-7}=\frac{\left|6.99-7\right|}{6.99-7}=\frac{\left|6.999-7\right|}{6.999-7}=-1$

Right-hand limit: $\frac{\left|7.1-7\right|}{7.1-7}=\frac{\left|7.01-7\right|}{7.01-7}=\frac{\left|7.001-7\right|}{7.001-7}=1$

Since the left- and right-hand limits are not equal, there is no limit.