Limit exist or not?

**nameck** · December 13th 2009, 12:21 AM

Find the following limit, if it exist:
$\lim_{x \to 1}\frac{x + 2}{x^2 + x - 2}$

apply L'Hospital's Rule,
i get, $\frac{1}{3}$

so the limit is exist..
correct?

**Prove It** · December 13th 2009, 12:46 AM

Originally Posted by nameck

Find the following limit, if it exist:
$\lim_{x \to 1}\frac{x + 2}{x^2 + x - 2}$

apply L'Hospital's Rule,
i get, $\frac{1}{3}$

so the limit is exist..
correct?

You can't apply L'Hospital's rule, because it does not tend to $\frac{0}{0}$ or $\frac{\infty}{\infty}$ .

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

$= \frac{1}{x - 1}$ .

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

Try graphing this function. I think you'll see that the left hand limit is $-\infty$ and the right hand limit is $\infty$ .

Since the left and right hand limits don't match, the limit does not exist.

**nameck** · December 13th 2009, 05:15 AM

Thanks Prove It!

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