# Sum and product of limits

• Sep 2nd 2008, 01:51 PM
NotEinstein
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*
• Sep 2nd 2008, 02:00 PM
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 (Surprised)?
• Sep 2nd 2008, 02:01 PM
mr fantastic
Quote:

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.
• Sep 2nd 2008, 02:03 PM
NotEinstein
Quote:

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 (Surprised)?

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 :)
• Sep 2nd 2008, 02:09 PM
mr fantastic
Quote:

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 $\frac{2-x}{2x(x - 2)} = ....$
• Sep 2nd 2008, 03:32 PM
NotEinstein
Quote:

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?
• Sep 2nd 2008, 03:44 PM
mr fantastic
Quote:

Originally Posted by mr fantastic
Then some simple algebra (get under a common denominator etc.) will re-arrange it into $\frac{2-x}{2x(x - 2)} = ....$

Quote:

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 .....