# How do you find the limit of (x^2-3x+2)/(x^3-4x) as x approaches 2?

Nov 30, 2016

${\lim}_{x \to 2} \frac{{x}^{2} - 3 x + 2}{{x}^{3} - 4 x} = \frac{1}{8}$

#### Explanation:

Factorize the numerator and denominator:

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

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

You can see you can simplify the rational function, making it continuous in $x = 2$:

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

As in this form the function is continuous, the limit equals the value:

${\lim}_{x \to 2} \frac{x - 1}{x \left(x + 2\right)} = \frac{2 - 1}{2 \left(2 + 2\right)} = \frac{1}{8}$

Alternatively, for instance in the case of polynomials that are harder to factorize, you can use L'Hôpital's rule, stating that as the limit is in the indeterminate form $\frac{0}{0}$ and as numerator and denominator are differentiable around $x = 2$,

${\lim}_{x \to 2} \frac{{x}^{2} - 3 x + 2}{{x}^{3} - 4 x} = {\lim}_{x \to 2} \frac{\frac{d}{\mathrm{dx}} \left({x}^{2} - 3 x + 2\right)}{\frac{d}{\mathrm{dx}} \left({x}^{3} - 4 x\right)}$

${\lim}_{x \to 2} \frac{{x}^{2} - 3 x + 2}{{x}^{3} - 4 x} = {\lim}_{x \to 2} \frac{2 x - 3}{3 {x}^{2} - 4} = \frac{2 \cdot 2 - 3}{3 \cdot {2}^{2} - 4} = \frac{1}{8}$