Question

evaluate the limit: $lim_{x
ightarrow0}\frac{sqrt{2x + 64}-8}{x}$

Explanation:

Step1: Rationalize the numerator

Multiply the fraction by $\frac{\sqrt{2x + 64}+8}{\sqrt{2x + 64}+8}$.
\[

$$\begin{align*} &\lim_{x ightarrow0}\frac{\sqrt{2x + 64}-8}{x}\times\frac{\sqrt{2x + 64}+8}{\sqrt{2x + 64}+8}\\ =&\lim_{x ightarrow0}\frac{(\sqrt{2x + 64})^2-8^2}{x(\sqrt{2x + 64}+8)}\\ =&\lim_{x ightarrow0}\frac{2x + 64 - 64}{x(\sqrt{2x + 64}+8)}\\ =&\lim_{x ightarrow0}\frac{2x}{x(\sqrt{2x + 64}+8)} \end{align*}$$

\]

Step2: Simplify the fraction

Cancel out the common - factor $x$ in the numerator and denominator.
\[

$$\begin{align*} &\lim_{x ightarrow0}\frac{2x}{x(\sqrt{2x + 64}+8)}\\ =&\lim_{x ightarrow0}\frac{2}{\sqrt{2x + 64}+8} \end{align*}$$

\]

Step3: Evaluate the limit

Substitute $x = 0$ into the simplified function.
\[

$$\begin{align*} &\frac{2}{\sqrt{2\times0+64}+8}\\ =&\frac{2}{\sqrt{64}+8}\\ =&\frac{2}{8 + 8}\\ =&\frac{2}{16}\\ =&\frac{1}{8} \end{align*}$$

\]

Answer:

$\frac{1}{8}$