1
$\begingroup$

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.

  • 0
    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 $x \rightarrow \infty$).2012-09-04

2 Answers 2

4

$ \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 $\lim_{x \to \infty}, x≠-1$

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

  • 0
    @kellax: Isn't $\frac 30$ equal to infinity?2012-09-04
3

$ \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 $1$.