Question

Evaluate the given limit :
$\underset{x\to 1}{lim}\frac{a{x}^2+bx+c}{c{x}^2+bx+a},a+b+c\ne 0$

Answer 1

Given : $\underset{x\to 1}{lim}\frac{a{x}^2+bx+c}{c{x}^2+bx+a}$
Substituting $x=1$
$=\frac{a(1{)}^2+b(1)+(c)}{c(1{)}^2+b(1)+a}$
$=\frac{a+b+c}{a+b+c}$
$=1(\because a+b+c\ne 0)$
$\therefore \underset{x\to 1}{lim}\frac{a{x}^2+bx+c}{c{x}^2+bx+a}=1$