Question

Evaluate the limit : $\underset{x\to 0}{lim}\frac{3x+1}{x+3}$

Answer 1

$\underset{x\to 0}{lim}\frac{3x+1}{x+3}=\frac{3\left(0\right)+1}{0+3}=\frac{1}{3}$

Therefore,

$\underset{x\to 0}{lim}\frac{3x+1}{x+3}=\frac{1}{3}$