Evaluate the limit or explain why it does not exist.

1.) (1/x - 1/2 )(1/(x-2)) as x approaches 2

2.) 2x/(x+4) + 8/(x+4) as x approaches -4

I concluded that both 1 and 2 do not exist. *crosses fingers*

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- Sep 2nd 2008, 01:51 PMNotEinsteinSum and product of limits
Evaluate the limit or explain why it does not exist.

1.) (1/x - 1/2 )(1/(x-2)) as x approaches 2

2.) 2x/(x+4) + 8/(x+4) as x approaches -4

I concluded that both 1 and 2 do not exist. *crosses fingers* - Sep 2nd 2008, 02:00 PMo_O
How did you conclude that?

#1: $\displaystyle \lim_{x \to 2} \left(\frac{1}{x} - \frac{1}{2}\right)\left(\frac{1}{x-4}\right) = \lim_{x \to 2} \left(\frac{1}{x(x-4)} - \frac{1}{2(x-4)}\right) = \hdots $

#2: $\displaystyle \lim_{x \to -4} \left(\frac{2x}{x+4} + \frac{8}{x+4}\right) = \lim_{x \to -4} \left(\frac{2x+8}{x+4}\right) = \hdots$

Factor out the numerator and what cancels (Surprised)? - Sep 2nd 2008, 02:01 PMmr fantastic
No on both counts.

Obviously $\displaystyle \frac{1}{x} - \frac{1}{2} \rightarrow 0$ as x --> 2 ....

The second one can be re-witten as $\displaystyle \frac{2x + 8}{x+4} = \frac{2(x+4)}{x+4} ....$

Knowledge of basic algebra and its application is essential when studying limits. - Sep 2nd 2008, 02:03 PMNotEinstein
- Sep 2nd 2008, 02:09 PMmr fantastic
- Sep 2nd 2008, 03:32 PMNotEinstein
- Sep 2nd 2008, 03:44 PMmr fantastic