We are asked to solve two limit problems. 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 solve two limit problems.

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. Therefore, the limit will either be $+\infty$, $-\infty$, or it may not exist. Let us consider the limit from the right side, where $x>3$. Then $x-3>0$ and $x-3$ is approaching 0, so $\frac{1}{x-3}$ approaches $+\infty$. Let us now consider the limit from the left side, where $x<3$. Then $x-3<0$ and $x-3$ is approaching 0, so $\frac{1}{x-3}$ approaches $-\infty$. Since the left and right limits are not equal, the limit does not exist.

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

Since the function $\frac{2x+2}{3x+5}$ is continuous at $x=0$, we can just substitute $x=0$ into the expression to evaluate the limit.

$\lim_{x\to 0} \frac{2x+2}{3x+5} = \frac{2(0)+2}{3(0)+5} = \frac{2}{5}$.

3. Final Answer

a) The limit does not exist.

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