Question

Evaluate the given limit :
$\underset{x\to 4}{lim}\frac{4x+3}{x-2}$

Answer 1

Given : $\underset{x\to 4}{lim}\frac{4x+3}{x-2}$
Substituting $x=4$
$=\frac{4\left(4\right)+3}{4-2}$
$=\frac{16+3}{2}$
$=\frac{19}{2}$
$\therefore \underset{x\to 4}{lim}\frac{4x+3}{x-2}=\frac{19}{2}$