Find the following limit, if it exist:
$\displaystyle \lim_{x \to 1}\frac{x + 2}{x^2 + x - 2}$
apply L'Hospital's Rule,
i get, $\displaystyle \frac{1}{3}$
so the limit is exist..
correct?
You can't apply L'Hospital's rule, because it does not tend to $\displaystyle \frac{0}{0}$ or $\displaystyle \frac{\infty}{\infty}$.
$\displaystyle \frac{x + 2}{x^2 + x - 2} = \frac{x + 2}{(x + 2)(x - 1)}$
$\displaystyle = \frac{1}{x - 1}$.
So $\displaystyle \lim_{x \to 1}\frac{x + 2}{x^2 + x - 2} = \lim_{x \to 1}\frac{1}{x - 1}$
Try graphing this function. I think you'll see that the left hand limit is $\displaystyle -\infty$ and the right hand limit is $\displaystyle \infty$.
Since the left and right hand limits don't match, the limit does not exist.