Question

Solve : $\underset{x\to 2}{lim}\frac{x^2-5x+6}{x^2-7x+10}$

Answer 1

Now,

$\underset{x\to 2}{lim}\frac{x^2-5x+6}{x^2-7x+10}$ $\frac{0}{0}$ form

Now applying L'Hospital's rule we get,

$=\underset{x\to 2}{lim}\frac{2x-5}{2x-7}$

$=\frac{2.2-5}{2.2-7}$

$=\frac{1}{3}$.