# How do you find the limit of (1/(x-1)+1/(x^2-3x+2)) as x->1?

Feb 9, 2017

$- 1$

#### Explanation:

${\lim}_{x \to 1} \left(\frac{1}{x - 1} + \frac{1}{{x}^{2} - 3 x + 2}\right)$

$= {\lim}_{x \to 1} \left(\frac{1}{x - 1} + \frac{1}{\left(x - 1\right) \left(x - 2\right)}\right)$

if we let $x - 1 = \delta$

$= {\lim}_{\delta \to 0} \left(\frac{1}{\delta} + \frac{1}{\delta \left(\delta - 1\right)}\right)$

$= {\lim}_{\delta \to 0} \left(\frac{\delta - 1}{\delta \left(\delta - 1\right)} + \frac{1}{\delta \left(\delta - 1\right)}\right)$

$= {\lim}_{\delta \to 0} \frac{\delta}{\delta \left(\delta - 1\right)}$

$= {\lim}_{\delta \to 0} \frac{1}{\delta - 1} = - 1$

Feb 9, 2017

${\lim}_{x \to 1} \frac{1}{x - 1} + \frac{1}{{x}^{2} - 3 x + 2} = - 1$

#### Explanation:

${\lim}_{x \to 1} \frac{1}{x - 1} + \frac{1}{{x}^{2} - 3 x + 2} = {\lim}_{x \to 1} \frac{1}{x - 1} + \frac{1}{\left(x - 1\right) \left(x - 2\right)}$
$\text{ } = {\lim}_{x \to 1} \frac{\left(x - 2\right) + 1}{\left(x - 1\right) \left(x - 2\right)}$
$\text{ } = {\lim}_{x \to 1} \frac{\left(x - 1\right)}{\left(x - 1\right) \left(x - 2\right)}$
$\text{ } = {\lim}_{x \to 1} \frac{1}{\left(x - 2\right)}$
$\text{ } = \frac{1}{\left(1 - 2\right)}$
$\text{ } = - 1$