Question

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

Explanation:

Step1: Combine the fractions in the numerator

First, find a common - denominator for $\frac{1}{3}+\frac{1}{x}$. The common denominator 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

$\frac{\frac{x + 3}{3x}}{3 + x}=\frac{x + 3}{3x(3 + x)}$. Since $x
eq - 3$ (when taking the limit, we consider values of $x$ approaching $-3$ but not equal to $-3$), we can cancel out the common factor $(x + 3)$ in the numerator and the denominator. So, $\frac{x + 3}{3x(3 + x)}=\frac{1}{3x}$.

Step3: Evaluate the limit

Now, we find $\lim_{x
ightarrow - 3}\frac{1}{3x}$. Substitute $x=-3$ into $\frac{1}{3x}$. We get $\frac{1}{3\times(-3)}=-\frac{1}{9}$.

Answer:

$-\frac{1}{9}$