51) $lim_{x \to 0}\frac{sqrt{64 + x}-8}{x}$

Explanation:

Step1: Rationalize the numerator

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

Step2: Simplify the fraction

Cancel out the common - factor $x$ in the numerator and denominator. [ \begin{align*} &\lim_{x\rightarrow0}\frac{x}{x(\sqrt{64 + x}+8)}\ =&\lim_{x\rightarrow0}\frac{1}{\sqrt{64 + x}+8} \end{align*} ]

Step3: Evaluate the limit

Substitute $x = 0$ into the simplified function. [ \begin{align*} &\frac{1}{\sqrt{64+0}+8}\ =&\frac{1}{8 + 8}\ =&\frac{1}{16} \end{align*} ]

Answer:

$\frac{1}{16}$