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