# How do you evaluate the limit (x^2-7x+10)/(x-2) as x approaches 2?

Nov 30, 2016

$\therefore {\lim}_{x \rightarrow 2} \frac{{x}^{2} - 7 x + 10}{x - 2} = - 3$

#### Explanation:

${\lim}_{x \rightarrow 2} \frac{{x}^{2} - 7 x + 10}{x - 2} = {\lim}_{x \rightarrow 2} \frac{\left(x - 5\right) \left(x - 2\right)}{x - 2}$

When we evaluate the limit we look at the behaviour as $x$ approaches $2$ and we are not interested in what happens when $x = 2$ so we can cancel the $\left(x - 2\right)$ factor as $x \ne 2$

$\therefore {\lim}_{x \rightarrow 2} \frac{{x}^{2} - 7 x + 10}{x - 2} = {\lim}_{x \rightarrow 2} \left(x - 5\right)$
$\therefore {\lim}_{x \rightarrow 2} \frac{{x}^{2} - 7 x + 10}{x - 2} = \left(2 - 5\right)$
$\therefore {\lim}_{x \rightarrow 2} \frac{{x}^{2} - 7 x + 10}{x - 2} = - 3$

Nov 30, 2016

Factor the numerator to simplify the rational function

#### Explanation:

${x}^{2} - 7 x + 10 = \left(x - 2\right) \left(x - 5\right)$

${\lim}_{x \to 2} \frac{{x}^{2} - 7 x + 10}{x - 2} = {\lim}_{x \to 2} \frac{\left(x - 2\right) \left(x - 5\right)}{x - 2}$

$\therefore = {\lim}_{x \to 2} \left(x - 5\right) = - 3$