# Limit of equation as x tends to -1

I was given the following expression and had to find the limit as: $$x \rightarrow 1, x \rightarrow - 1, x \rightarrow \infty$$ $$\lim_{x \to -1} \frac{x^2 +3x +2}{x^2 -1} = \lim_{x \to -1} \frac{\frac{x^2}{x^2} + \frac{3x}{x^2} + \frac{2}{x^2}}{\frac{x^2}{x^2} - \frac{1}{x^2}} = \lim_{x \to -1} \frac{\frac{1}{1} + 0 + 0}{\frac{1}{1} - 0} = \lim_{x \to -1} \frac{1}{1} = 1$$

So for $-1$, I got 1. However the text book says it's $1/2$. I tried pluging in -1 but I don't get $1/2$, no matter how I shift this.

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Note that the fractions above with $x$ or $x^2$ in the denominator don't tend to $0$ when $x \rightarrow -1$ (that's the case when $x \rightarrow \infty$). –  Quintofron Sep 4 '12 at 16:16

$$\lim_{x \to -1} \frac{x^2 +3x +2}{x^2 -1}$$ $$=\lim_{x \to -1} \frac{(x+1)(x+2)}{(x+1)(x-1)}$$ $$=\lim_{x \to -1} \frac{(x+2)}{(x-1)}$$ $$\text{as}:x \to -1, x≠-1$$

$$=\frac{-1+2}{-1-1}=-\frac{1}{2}$$

Now $\lim_{x \to 1} \frac{x^2 +3x +2}{x^2 -1}$ $=\lim_{x \to 1} \frac{(x+2)}{(x-1)}$ as $\lim_{x \to 1}, x≠-1$

$\lim_{x \to 1^{+}} \frac{x^2 +3x +2}{x^2 -1}=\infty$

$\lim_{x \to 1^{-}} \frac{x^2 +3x +2}{x^2 -1}=-\infty$

SO, the limit does not exist at $x=1$(as identified by Quintofron )

Now $\lim_{x \to \infty} \frac{x^2 +3x +2}{x^2 -1}$ $=\lim_{x \to \infty} \frac{(x+2)}{(x-1)}$ as $\lim_{x \to \infty}, x≠-1$

$=\lim_{x \to \infty}\frac{1+\frac{2}{x}}{1-\frac{1}{x}}=1$

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WoW so simple need to pay more attention next time so split, cancel out and plug-in. Thanks a lot mate. Just 1 more question for x -> 1, i would do the same approach but as i would get 3/0 the limit is undefined? or 0? –  Sterling Duchess Sep 4 '12 at 16:13
For x -> 1 how come its infinity if pluging 1 in to (x+2)/(x-1) gives 3/0 ? –  Sterling Duchess Sep 4 '12 at 16:27
In the case of $x$ approaching $1$, the two-sided limit does not exist. You can examine one-sided limits though (as $x \rightarrow 1^{-}$ and $x \rightarrow 1^{+}$). –  Quintofron Sep 4 '12 at 16:56
@kellax: Isn't $\frac 30$ equal to infinity? –  Gigili Sep 4 '12 at 17:21

$$\require{cancel} \begin{equation*} \lim \frac{x^2 +3x +2}{x^2 -1}= \lim \frac{\cancel{(x+1)}(x+2)}{\cancel{(x+1)}(x-1)}= \lim \frac{x+2}{x-1}= \begin{cases} -\frac 12 & \text{if x \to -1,} \\ +\infty &\text{if x \to 1.} \end{cases} \end{equation*}$$

When $x \to \infty$ you should consider the terms with the biggest power of the main variable in both denominator and numerator, which is $x$ and $x$ here, so the answer would be $1$.

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