Question

evaluate the limit
lim_{x
ightarrow3}\frac{\frac{1}{x + 6}-\frac{1}{9}}{x - 3}
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Explanation:

Step1: Simplify the numerator

First, find a common - denominator for the fraction in the numerator. $\frac{1}{x + 6}-\frac{1}{9}=\frac{9-(x + 6)}{9(x + 6)}=\frac{9 - x - 6}{9(x + 6)}=\frac{3 - x}{9(x + 6)}$.
So the original limit becomes $\lim_{x
ightarrow3}\frac{\frac{3 - x}{9(x + 6)}}{x - 3}$.

Step2: Rewrite the complex - fraction

$\lim_{x
ightarrow3}\frac{\frac{3 - x}{9(x + 6)}}{x - 3}=\lim_{x
ightarrow3}\frac{3 - x}{9(x + 6)(x - 3)}$.
Factor out a $-1$ from the numerator: $\lim_{x
ightarrow3}\frac{-(x - 3)}{9(x + 6)(x - 3)}$.

Step3: Cancel out the common factor

Cancel out the $(x - 3)$ terms (since $x
eq3$ when taking the limit), we get $\lim_{x
ightarrow3}\frac{-1}{9(x + 6)}$.

Step4: Substitute the value of $x$

Substitute $x = 3$ into $\frac{-1}{9(x + 6)}$, we have $\frac{-1}{9\times(3 + 6)}=\frac{-1}{9\times9}=-\frac{1}{81}$.

Answer:

$-\frac{1}{81}$