We are asked to evaluate two limits. a) $\lim_{x \to 3} \frac{1}{x-3}$ b) $\lim_{x \to 0} \frac{2x+2}{3x+5}$

1. Problem Description

We are asked to evaluate two limits.

a) $\lim_{x \to 3} \frac{1}{x-3}$

b) $\lim_{x \to 0} \frac{2x+2}{3x+5}$

2. Solution Steps

a) We need to evaluate $\lim_{x \to 3} \frac{1}{x-3}$.

As $x$ approaches 3, the denominator $x-3$ approaches

0. If $x$ approaches 3 from the right (i.e., $x > 3$), then $x-3$ is a small positive number, so $\frac{1}{x-3}$ approaches $+\infty$.

If $x$ approaches 3 from the left (i.e., $x < 3$), then $x-3$ is a small negative number, so $\frac{1}{x-3}$ approaches $-\infty$.

Since the limit from the right and the limit from the left are not equal, the limit does not exist.

b) We need to evaluate $\lim_{x \to 0} \frac{2x+2}{3x+5}$.

We can find the limit by direct substitution because the function is continuous at $x=0$.

Substituting $x=0$ into the expression, we get:

$\frac{2(0) + 2}{3(0) + 5} = \frac{0+2}{0+5} = \frac{2}{5}$

3. Final Answer

a) The limit does not exist.

b) $\frac{2}{5}$