evaluate the limit: $lim_{x\rightarrow0}\frac{sqrt{2x + 64}-8}{x}$

Explanation:

Step1: Rationalize the numerator

Multiply the fraction by $\frac{\sqrt{2x + 64}+8}{\sqrt{2x + 64}+8}$. [ \begin{align*} &\lim_{x\rightarrow0}\frac{\sqrt{2x + 64}-8}{x}\times\frac{\sqrt{2x + 64}+8}{\sqrt{2x + 64}+8}\ =&\lim_{x\rightarrow0}\frac{(\sqrt{2x + 64})^2-8^2}{x(\sqrt{2x + 64}+8)}\ =&\lim_{x\rightarrow0}\frac{2x + 64 - 64}{x(\sqrt{2x + 64}+8)}\ =&\lim_{x\rightarrow0}\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\rightarrow0}\frac{2x}{x(\sqrt{2x + 64}+8)}\ =&\lim_{x\rightarrow0}\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:

$\frac{1}{8}$