Question

Evaluate
$l=\underset{x\to 2}{lim}(\frac{1}{(x^2-5x+5)}+\frac{1}{(2x^2-7x+6)})$

Answer 1

We have,

$l={lim}_{x\to 2}(\frac{1}{(x^2-5x+5)}+\frac{1}{(2x^2-7x+6)})$

Taking limit and we get,

$l=(\frac{1}{(2^2-5\left(2\right)+5)}+\frac{1}{(2{\left(2\right)}^2-7\left(2\right)+6)})$

$=(\frac{1}{4-10+5}+\frac{1}{8-14+6})$

$=(\frac{1}{-1}+\frac{1}{0})$

$=\infty$Hence, this is the answer.