Question

Finding limits: Find the given limits if they exist. If a limit does not exist, explain why.

Short answer

Step-by-step solution

Step 1. Given information

Step 2. Finding limit for limt→0 1/(3+t)−1/3t

$\underset{t\to 0}{lim} \frac{1/(3+t)-1/3}{t}=-\frac{1}{3(t+3)}$

$\underset{t\to 0}{lim} \frac{1/(3+t)-1/3}{t}=-\frac{1}{3(0+3)}$

$\underset{t\to 0}{lim} \frac{1/(3+t)-1/3}{t}=-\frac{1}{9}$

Step 3. Finding limit for limt→0 (3+t)2−9t

$\underset{t\to 0}{lim} \frac{(3+t{)}^2-9}{t}=\underset{t\to 0}{lim} t+6$

$\underset{t\to 0}{lim} \frac{(3+t{)}^2-9}{t}=0+6=6$