9. $$ lim _ { x \rightarrow 0 } \frac { \frac { 1 } { x - 6 } + \frac { 1 } { 6 } } { x } $$
9. $$ lim _ { x \rightarrow 0 } \frac { \frac { 1 } { x - 6 } + \frac { 1 } { 6 } } { x } $$
Answer
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)). [ \begin{align*} \frac{1}{x - 6}+\frac{1}{6}&=\frac{6+(x - 6)}{6(x - 6)}\ &=\frac{6+x - 6}{6(x - 6)}\ &=\frac{x}{6(x - 6)} \end{align*} ] So the original limit becomes (\lim_{x\rightarrow0}\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\neq0,c\neq0)), we get (\frac{x}{6(x - 6)\times x}). Cancel out the non - zero (x) (since (x\rightarrow0) but (x\neq0) in the limit process), and we have (\lim_{x\rightarrow0}\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})