# What is the limit of (x^3 -1)/(x^2+2x-3) as x approaches 1?

Aug 1, 2015

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

#### Explanation:

In order to avoid the $\frac{0}{0}$ indeterminate form, which you would get if you tried to evaluate this limit for $x \to 1$, you can a little algebraic manipulation to rewrite your initial function.

The numerator of the fraction can be factored using the formula for the difference of two cubes

color(blue)(a^3 - b^3 = (a-b)^3 + 3ab(a-b)

This means that you can write

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

Now focus on the denominator, which can be factored to get

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

${\lim}_{x \to 1} \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 1\right)}}} \cdot \left[{\left(x - 1\right)}^{2} + 3 x\right]}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 1\right)}}} \left(x + 3\right)} = {\lim}_{x \to 1} \frac{{\left(x - 1\right)}^{2} + 3 x}{x + 3}$
As $x \to 1$, this limit will be equal to
${\lim}_{x \to 1} \frac{{\left(1 - 1\right)}^{2} + 3 \cdot 1}{1 + 3} = \textcolor{g r e e n}{\frac{3}{4}}$