evaluate the limit, if it exists. (if an answer does not exist, enter dne.)\n lim_{x\rightarrow…

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

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

Answer

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\rightarrow - 3}\frac{\frac{1}{3}+\frac{1}{x}}{3 + x}) becomes (\lim_{x\rightarrow - 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\neq - 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\rightarrow - 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})