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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.

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It is the correct result, but I am not sure how you reached your conclusion.

The two sided limit does not exist, when you approach from $7^{-} = - \infty$ and from $7^{+} = + \infty$.

So, be careful when doing these as I am not sure that is what you did.

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  • $\begingroup$ +Goodpoints. ${}{}{}$ $\endgroup$ – mrs Sep 5 '13 at 1:02
  • $\begingroup$ Needs another TU +1 $\endgroup$ – Namaste Sep 5 '13 at 1:13
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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 $\pm\infty$.

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