Question

9.

$$ lim _ { x ightarrow 0 } \frac { \frac { 1 } { x - 6 } + \frac { 1 } { 6 } } { x } $$

Explanation:

Step1: Simplify the numerator

First, find a common denominator for \(\frac{1}{x - 6}+\frac{1}{6}\). The common denominator is \(6(x - 6)\).

$$
LATEXBLOCK0
$$

So the original limit becomes \(\lim_{x
ightarrow0}\frac{\frac{x}{6(x - 6)}}{x}\).

Step2: Simplify the complex - fraction

When we have \(\frac{\frac{x}{6(x - 6)}}{x}\), using the rule \(\frac{a/b}{c}=\frac{a}{b\times c}\) (\(b
eq0,c
eq0\)), we get \(\frac{x}{6(x - 6)\times x}\). Cancel out the non - zero \(x\) (since \(x
ightarrow0\) but \(x
eq0\) in the limit process), and we have \(\lim_{x
ightarrow0}\frac{1}{6(x - 6)}\).

Step3: Evaluate the limit

Substitute \(x = 0\) into \(\frac{1}{6(x - 6)}\).

$$ \frac{1}{6(0 - 6)}=\frac{1}{- 36}=-\frac{1}{36} $$

Answer:

\(-\frac{1}{36}\)