# How do you find the limit of (x^2-3x+2)/(x^3-4x) as x approaches 2 from the right, as x approaches -2 from the right, as x approaches 0 from the left, and as x approaches 1 from the right?

May 5, 2015

Start by factoring the expression,

$\frac{{x}^{2} - 3 x + 2}{{x}^{3} - 4 x} = \frac{\left(x - 2\right) \left(x - 1\right)}{x \left(x + 2\right) \left(x - 2\right)} = \frac{\left(x - 1\right)}{x \left(x + 2\right)}$

Now start taking the required limits.

${\lim}_{x \setminus \rightarrow {0}^{-}} \left[\frac{\left(x - 1\right)}{x \left(x + 2\right)}\right]$

$= {\lim}_{x \setminus \rightarrow {0}^{-}} \left[\frac{\left(x - 1\right)}{\left(x + 2\right)}\right] \cdot {\lim}_{x \setminus \rightarrow {0}^{-}} \left[\frac{1}{x}\right]$

$= \frac{\left(0 - 1\right)}{\left(0 + 2\right)} \cdot \left(- \setminus \infty\right) = - \infty$

${\lim}_{x \setminus \rightarrow {0}^{-}} \left[\frac{1}{x}\right] = - \infty$ because we are approaching zero from the negative side of the number line.

For the next limit,

${\lim}_{x \setminus \rightarrow - {2}^{+}} \left[\frac{\left(x - 1\right)}{x \left(x + 2\right)}\right]$

$= {\lim}_{x \setminus \rightarrow - {2}^{+}} \left[\frac{\left(x - 1\right)}{x}\right] \cdot {\lim}_{x \setminus \rightarrow - {2}^{+}} \left[\frac{1}{\left(x + 2\right)}\right]$

$= \frac{- 3}{- 2} \cdot \left(+ \infty\right) = + \infty$

${\lim}_{x \setminus \rightarrow - {2}^{+}} \left[\frac{1}{\left(x + 2\right)}\right] = + \infty$ because as $\left(x \setminus \rightarrow - {2}^{+}\right)$, $\left(x + 2\right)$ gets very small, but stays positive.

For the next limit,

${\lim}_{x \setminus \rightarrow {1}^{+}} \left[\frac{\left(x - 1\right)}{x \left(x + 2\right)}\right] = {\lim}_{x \setminus \rightarrow {1}^{+}} \left[\frac{1}{x \left(x + 2\right)}\right] {\lim}_{x \setminus \rightarrow {1}^{+}} \left[\left(x - 1\right)\right]$
$= \frac{1}{1 \cdot 3} \cdot 0 = 0$

For the last limit,

${\lim}_{x \setminus \rightarrow {2}^{+}} \left[\frac{\left(x - 1\right)}{x \left(x + 2\right)}\right] = \frac{3}{2 \cdot 4} = \frac{3}{8}$

Even though the denominator of the original expression went to zero at $x = 2$, the limit is still finite because the numerator went to zero just as quickly. Recall, at the start we were able to cross out the $\left(x - 2\right)$ in the denominator (which was causing the singularity) with another $\left(x - 2\right)$ factor in the numerator.