Question

using the factorization $x - 36 = (sqrt{x}-6)(sqrt{x}+6)$, we have $m_{\tan}=lim_{x
ightarrow36}\frac{sqrt{x}-6}{(sqrt{x}-6)(sqrt{x}+6)}=lim_{x
ightarrow36}\frac{1}{square}=square$.

Explanation:

Step1: Simplify the fraction

Cancel out the common factor $\sqrt{x}-6$ in the numerator and denominator of $\frac{\sqrt{x} - 6}{(\sqrt{x}-6)(\sqrt{x}+6)}$. We get $\lim_{x
ightarrow36}\frac{1}{\sqrt{x}+6}$.

Step2: Evaluate the limit

Substitute $x = 36$ into $\frac{1}{\sqrt{x}+6}$. Since $\sqrt{36}=6$, then $\frac{1}{\sqrt{36}+6}=\frac{1}{6 + 6}=\frac{1}{12}$.

Answer:

$\frac{1}{12}$