Question

evaluate the limit, if it exists. (if an answer does not exist, enter dne.)
lim_{x
ightarrow - 24}\frac{sqrt{x^{2}+49}-25}{x + 24}

Explanation:

Step1: Rationalize the numerator

Multiply by $\frac{\sqrt{x^{2}+49}+25}{\sqrt{x^{2}+49}+25}$.
\[

$$\begin{align*} &\lim_{x ightarrow - 24}\frac{\sqrt{x^{2}+49}-25}{x + 24}\times\frac{\sqrt{x^{2}+49}+25}{\sqrt{x^{2}+49}+25}\\ =&\lim_{x ightarrow - 24}\frac{(x^{2}+49)-625}{(x + 24)(\sqrt{x^{2}+49}+25)}\\ =&\lim_{x ightarrow - 24}\frac{x^{2}-576}{(x + 24)(\sqrt{x^{2}+49}+25)} \end{align*}$$

\]

Step2: Factor the numerator

Factor $x^{2}-576=(x + 24)(x - 24)$.
\[

$$\begin{align*} &\lim_{x ightarrow - 24}\frac{(x + 24)(x - 24)}{(x + 24)(\sqrt{x^{2}+49}+25)}\\ =&\lim_{x ightarrow - 24}\frac{x - 24}{\sqrt{x^{2}+49}+25} \end{align*}$$

\]

Step3: Substitute $x=-24$

\[

$$\begin{align*} &\frac{-24-24}{\sqrt{(-24)^{2}+49}+25}\\ =&\frac{-48}{\sqrt{576 + 49}+25}\\ =&\frac{-48}{\sqrt{625}+25}\\ =&\frac{-48}{25 + 25}\\ =&-\frac{24}{25} \end{align*}$$

\]

Answer:

$-\frac{24}{25}$