Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
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
add comment

2 Answers

up vote 4 down vote accepted

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


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 $\lim_{x \to \infty}, x≠-1$

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

share|improve this answer
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? –  kellax 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 ? –  kellax 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
add comment

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

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 $1$.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.