# Help verify $\lim_{x\to 7} \frac {x^2+7x+49}{x^2+7x-98}$ .

So my question is "Evaluate the limit" $\displaystyle \lim_{x\to 7} \frac {x^2+7x+49}{x^2+7x-98}$

I know you can't factor the numerator but you can for denominator. But either way you can't divide by $0$. So I say my answer is D.N.E.

If anyone can verify that I got the right answer, I would be most grateful.

The two sided limit does not exist, when you approach from $7^{-} = - \infty$ and from $7^{+} = + \infty$.
• +Goodpoints. ${}{}{}$ – mrs Sep 5 '13 at 1:02
Observe $x^2+7x+49=x^2+7x-98+147$ hence:$$\lim_{x\to7}\frac{x^2+7x+49}{x^2+7x-98}=\lim_{x\to7}\frac{x^2+7x-98+147}{x^2+7x-98}=\lim_{x\to7}\left(1+\frac{147}{x^2+7x-98}\right)$$Now note $x^2+7x-98=x^2-7x+14x-98=(x+14)(x-7)$ hence as $x\to7$ our limit tends to $\pm\infty$.