Question

evaluate the limit, if it exists. (if an answer does not exist, enter dne.)
\\(\lim_{h\to0}\frac{\sqrt{81 + h}-9}{h}\\)

Explanation:

Step1: Rationalize the numerator

Multiply numerator and denominator by $\sqrt{81 + h}+9$.
$\lim_{h
ightarrow0}\frac{\sqrt{81 + h}-9}{h}\times\frac{\sqrt{81 + h}+9}{\sqrt{81 + h}+9}$

Step2: Expand the numerator

Using $(a - b)(a + b)=a^{2}-b^{2}$, the numerator becomes $(81 + h)-81=h$.
$\lim_{h
ightarrow0}\frac{(81 + h)-81}{h(\sqrt{81 + h}+9)}=\lim_{h
ightarrow0}\frac{h}{h(\sqrt{81 + h}+9)}$

Step3: Simplify the fraction

Cancel out the $h$ terms.
$\lim_{h
ightarrow0}\frac{1}{\sqrt{81 + h}+9}$

Step4: Evaluate the limit

Substitute $h = 0$ into the expression.
$\frac{1}{\sqrt{81+0}+9}=\frac{1}{9 + 9}=\frac{1}{18}$

Answer:

$\frac{1}{18}$