Question

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

Answer 1

Here $\underset{x\to 1}{\text{lim}}\frac{a{x}^2+bx+c}{c{x}^2+bx+a}$

= $\frac{a\times (1{)}^2+b\times 1+c}{c\times (1{)}^2+b\times 1+a}=\frac{a+b+c}{c+b+a}=1$.