Question

find the limit.
\\( \lim _{x \
ightarrow 225} \frac{\sqrt{x}-15}{x - 225} \\)
select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. \\( \lim _{x \
ightarrow 225} \frac{\sqrt{x}-15}{x - 225}= \\) (type an integer or a simplified fraction.)
b. the limit does not exist.

Explanation:

Step1: Rationalize the numerator

Multiply the numerator and denominator by $\sqrt{x}+15$.

$$\begin{align*} \lim_{x ightarrow225}\frac{\sqrt{x}-15}{x - 225}&=\lim_{x ightarrow225}\frac{(\sqrt{x}-15)(\sqrt{x}+15)}{(x - 225)(\sqrt{x}+15)}\\ \end{align*}$$

Using the difference - of - squares formula $(a - b)(a + b)=a^{2}-b^{2}$, the numerator becomes $x-225$.
So we have $\lim_{x
ightarrow225}\frac{x - 225}{(x - 225)(\sqrt{x}+15)}$

Step2: Simplify the expression

Cancel out the common factor $(x - 225)$ (for $x
eq225$, as $x
ightarrow225$ but $x
eq225$ in the limit process).
We get $\lim_{x
ightarrow225}\frac{1}{\sqrt{x}+15}$

Step3: Substitute $x = 225$

Substitute $x = 225$ into $\frac{1}{\sqrt{x}+15}$.
When $x = 225$, $\sqrt{x}=\sqrt{225}=15$. Then $\frac{1}{\sqrt{225}+15}=\frac{1}{15 + 15}=\frac{1}{30}$

Answer:

A. $\lim_{x
ightarrow225}\frac{\sqrt{x}-15}{x - 225}=\frac{1}{30}$