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

Oct 2, 2016

Factor and reduce.

#### Explanation:

Because $x = - 1$ is a zero of ${x}^{3} + 1$, we can be certain that $x - \left(- 1\right)$ (that is: $x + 1$) is a factor of ${x}^{3} + 1$

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

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

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

$= - \frac{3}{2}$

If you haven't memorized how to factor the sum (and difference) of two cubes, use polynomial division to factor. If you don't know how to do polynomial division, you'll need to think it through.

${x}^{3} + 1 = \left(x + 1\right) \left(\text{ something }\right)$

Since we want the product to start with ${x}^{3}$, the $\text{ something }$ must start with ${x}^{2}$.

${x}^{3} + 1 = \left(x + 1\right) \left({x}^{2} + \text{ other stuff}\right)$

But now when we distribute the $+ 1$, we get an ${x}^{2}$ which does not appear in the product.

We cat fix this by putting in a $- x$ at the beginning of the $\text{ other stuff}$

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

We know we want to end with $+ 1$, so let's put a $+ 1$ in.

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

Multiply it out and we get ${x}^{3} + 1$.

Oct 2, 2016

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

#### Explanation:

We can factor the numerator and denominator then cancel the $\left(x + 1\right)$ factor in both...

$\frac{{x}^{3} + 1}{{x}^{2} - 1} = \frac{\left(\textcolor{red}{\cancel{\textcolor{b l a c k}{x + 1}}}\right) \left({x}^{2} - x + 1\right)}{\left(x - 1\right) \left(\textcolor{red}{\cancel{\textcolor{b l a c k}{x + 1}}}\right)} = \frac{{x}^{2} - x + 1}{x - 1}$

The derived rational function is identical to the original except that the original has a hole at $x = - 1$.

So we find:

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

Oct 2, 2016

$- \frac{3}{2}$

#### Explanation:

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

$\frac{{x}^{3} + 1}{{x}^{2} - 1} = x + \frac{x + 1}{{x}^{2} - 1} = x + \frac{1}{x - 1}$ then

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

Oct 2, 2016

$- \frac{3}{2}$.

#### Explanation:

We use a Standard Limit :${\lim}_{x \rightarrow a} \frac{{x}^{n} - {a}^{n}}{x - a} = n {a}^{n - 1} \ldots . . \left(\star\right)$.

Using $\left(\star\right) , {\lim}_{x \rightarrow - 1} \frac{{x}^{3} + 1}{x + 1}$

$= {\lim}_{x \rightarrow - 1} \frac{{x}^{3} - {\left(- 1\right)}^{3}}{x - \left(- 1\right)}$

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

$= 3. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(1\right)$.

Likewise, ${\lim}_{x \rightarrow - 1} \frac{{x}^{2} - 1}{x + 1} = {\lim}_{x \rightarrow - 1} \frac{{x}^{2} - {\left(- 1\right)}^{2}}{x - \left(- 1\right)}$

$= 2 {\left(- 1\right)}^{2 - 1} = - 2. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left(2\right)$.

Now, the Reqd. Limit $= {\lim}_{x \rightarrow - 1} \frac{{x}^{3} + 1}{{x}^{2} - 1}$

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

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

$= - \frac{3}{2} , \ldots \ldots \ldots \ldots \ldots \ldots . \text{[by (1) &, (2)]}$

Enjoy Maths.!