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. 

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 $\pm\infty$.
