Question

1. $limlimits_{x \to 0}\frac{x}{\frac{1}{x + 3}-\frac{1}{3}}$

Explanation:

Step1: Simplify the denominator

First, simplify \(\frac{1}{x + 3}-\frac{1}{3}\).

$$
LATEXBLOCK0
$$

So the original limit becomes \(\lim_{x
ightarrow0}\frac{x}{\frac{-x}{3(x + 3)}}\).

Step2: Simplify the limit expression

\(\lim_{x
ightarrow0}\frac{x}{\frac{-x}{3(x + 3)}}=\lim_{x
ightarrow0}\frac{3x(x + 3)}{-x}\).
Cancel out the non - zero \(x\) (since \(x
ightarrow0\) but \(x
eq0\) in the limit process), we get \(\lim_{x
ightarrow0}\frac{3(x + 3)}{-1}\).

Step3: Substitute \(x = 0\)

Substitute \(x = 0\) into \(\frac{3(x + 3)}{-1}\), we have \(\frac{3(0 + 3)}{-1}=-9\).

Answer:

\(-9\)