Possible to use analytic methods on lim x-->-2 x^2/(x+2) ?

**lamp23** · February 23rd 2011, 05:40 PM

Evaluate:

I understand the answer is negative infinity but only by using the table method. Is there an analytic method possible?

**Ackbeet** · February 23rd 2011, 05:51 PM

Well, the numerator is always positive, right? And if x < -2, which it is, then the denominator is always negative. Since the numerator is bounded below by 4, and the denominator is going to zero, the limit is negative infinity.

How's that?

**lamp23** · February 23rd 2011, 06:29 PM

Since I couldn't factor and even the derivative is undefined at -2, I just wanted to make sure there wasn't anything else I could do.

**DrSteve** · February 23rd 2011, 08:21 PM

The limit has the form $\frac{4}{0}$ . This is not an indeterminate form. So no special "tricks" are needed. A one sided limit in this case will always be positive or negative infinity - you just need to check which.

**FernandoRevilla** · February 23rd 2011, 08:37 PM

Perhaps you are thinking about the L'Hopital's rule. In such a case, take into account that

$\displaystyle\lim_{x\to -2^-}\;x^2=4\;,\quad\displaystyle\lim_{x\to -2^-}(x+2)=0$

so, the corresponding hypothesis are not satisfied.

Edited: Sorry, I didn't see DrSteve's post.

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