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

As $x$ approaches 3, $x-3$ approaches

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

If $x$ approaches 3 from the left (i.e., $x < 3$), $x-3$ approaches 0 from the negative side, so $\frac{1}{x-3}$ approaches $-\infty$.

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

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

Since the function $f(x) = \frac{2x+2}{3x+5}$ is continuous at $x=0$, we can directly substitute $x=0$ into the expression to find the limit.

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

3. Final Answer

a) Limit does not exist.

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