Question

evaluate the limit, if it exists. (if an answer does not exist, enter dne.)
lim_{x
ightarrow - 3}\frac{\frac{1}{3}+\frac{1}{x}}{3 + x}

Explanation:

Step1: Combine fractions in numerator

First, find a common - denominator for the fractions in the numerator. The common denominator of 3 and \(x\) is \(3x\). So, \(\frac{1}{3}+\frac{1}{x}=\frac{x + 3}{3x}\). Then the original limit \(\lim_{x
ightarrow - 3}\frac{\frac{1}{3}+\frac{1}{x}}{3 + x}\) becomes \(\lim_{x
ightarrow - 3}\frac{\frac{x + 3}{3x}}{3 + x}\).

Step2: Simplify the complex - fraction

Using the rule \(\frac{a/b}{c}=\frac{a}{bc}\), we can rewrite \(\frac{\frac{x + 3}{3x}}{3 + x}\) as \(\frac{x + 3}{3x(3 + x)}\).

Step3: Cancel out common factors

Since \(x
eq - 3\) (when taking the limit), we can cancel out the common factor \((x + 3)\) in the numerator and the denominator. The expression simplifies to \(\lim_{x
ightarrow - 3}\frac{1}{3x}\).

Step4: Evaluate the limit

Substitute \(x=-3\) into \(\frac{1}{3x}\). We get \(\frac{1}{3\times(-3)}=-\frac{1}{9}\).

Answer:

\(-\frac{1}{9}\)