# Math Help - Sum and product of limits

1. ## Sum 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*

2. How did you conclude that?

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

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

Factor out the numerator and what cancels ?

3. Originally Posted by NotEinstein
Evaluate the limit or explain why it does not exist.

1.) (1/x - 1/2 )(1/(x-4)) 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*
No on both counts.

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

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

Knowledge of basic algebra and its application is essential when studying limits.

4. Originally Posted by o_O
How did you conclude that?

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

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

Factor out the numerator and what cancels ?
for #1 it was actually x-2 in the denominator of the 2nd function. i edited it, but i will look at your work and go from there

5. Originally Posted by NotEinstein
for #1 it was actually x-2 in the denominator of the 2nd function. i edited it, but i will look at your work and go from there
Then some simple algebra (get under a common denominator etc.) will re-arrange it into $\frac{2-x}{2x(x - 2)} = ....$

6. Originally Posted by mr fantastic
Then some simple algebra (get under a common denominator etc.) will re-arrange it into $\frac{2-x}{2x(x - 2)} = ....$
Is there any way to justify this using the product limit law? first function being f(x) 2nd being g(x)? Or is that implied already?

7. Originally Posted by mr fantastic
Then some simple algebra (get under a common denominator etc.) will re-arrange it into $\frac{2-x}{2x(x - 2)} = ....$
Originally Posted by NotEinstein
Is there any way to justify this using the product limit law? first function being f(x) 2nd being g(x)? Or is that implied already?
Actually I was expecting that you'd cancel the common factor and then take the limit .....