3. $limlimits_{x \to 0}\frac{x}{\frac{1}{x + 3}-\frac{1}{3}}$
3. $limlimits_{x \to 0}\frac{x}{\frac{1}{x + 3}-\frac{1}{3}}$
Answer
Explanation:
Step1: Simplify the denominator
First, simplify (\frac{1}{x + 3}-\frac{1}{3}). [ \begin{align*} \frac{1}{x + 3}-\frac{1}{3}&=\frac{3-(x + 3)}{3(x + 3)}\ &=\frac{3-x - 3}{3(x + 3)}\ &=\frac{-x}{3(x + 3)} \end{align*} ] So the original limit becomes (\lim_{x\rightarrow0}\frac{x}{\frac{-x}{3(x + 3)}}).
Step2: Simplify the limit expression
(\lim_{x\rightarrow0}\frac{x}{\frac{-x}{3(x + 3)}}=\lim_{x\rightarrow0}\frac{3x(x + 3)}{-x}). Cancel out the non - zero (x) (since (x\rightarrow0) but (x\neq0) in the limit process), we get (\lim_{x\rightarrow0}\frac{3(x + 3)}{-1}).
Step3: Substitute (x = 0)
Substitute (x = 0) into (\frac{3(x + 3)}{-1}), we have (\frac{3(0 + 3)}{-1}=-9).
Answer:
(-9)