# Having trouble finding this limit

• Sep 13th 2011, 08:23 PM
terrorsquid
Having trouble finding this limit
Find the limit, if it exists, of the following:

$\displaystyle \lim_{x \to 2}\left(\frac{1}{x-2}-\frac{1}{x^2-4}\right)$

I have tried factoring this in a number of different ways, combining the factions with various denominators and can't seem to get it into a form where the denominator is not 0. I am also unsure as to how I would prove that it has no limit if that is the case.
• Sep 13th 2011, 08:49 PM
TKHunny
Re: Having trouble finding this limit
I'd try splitting that right-hand piece into partial fractions. Something magic may happen.
• Sep 13th 2011, 09:37 PM
terrorsquid
Re: Having trouble finding this limit
so for the partial fraction of $\displaystyle \frac{1}{(x+2)(x-2)}$ I get:

$\displaystyle -\frac{\frac{1}{4}}{(x+2)}+\frac{\frac{1}{4}}{(x-2)}$

so subbing that into the original equation:

$\displaystyle \frac{1}{x-2}-\left(-\frac{\frac{1}{4}}{(x+2)}+\frac{\frac{1}{4}}{(x-2)}\right)$

$\displaystyle \frac{1}{x-2}+\frac{\frac{1}{4}}{x+2}-\frac{\frac{1}{4}}{x-2}$

$\displaystyle \frac{\frac{3}{4}}{x-2}+\frac{\frac{1}{4}}{x+2}$

I still can't get the x-2 out of the denominator :S
• Sep 13th 2011, 10:25 PM
TutorMeNate
Re: Having trouble finding this limit
Hi Terrorsquid,

This problem has no limit. The simple reasoning is that if you graph it in your calculator, you'll find that it goes towards infinite as you approach 2 from the right, and it goes towards negative infinite as you approach 2 from the left.

The in depth reasoning is this:

If you plugged in 2.0000001, you would get 1 divided by a very small number for the left term, which would be infinite. For the second term you would get the same thing, but the closer you got towards 2 (example 2.000000000000000001), the smaller the second term would be compared to the first term. Therefore, it would approach infinite.

If you plugged in 1.99999999, you would get the same thing, except it would be negative.

Since the left and right never approach the same value, the limit does not exist.

Hopefully this made sense. Let me know if you need me to explain any of it more clearly.

-Nathan
• Sep 14th 2011, 10:32 AM
skeeter
Re: Having trouble finding this limit
Quote:

Originally Posted by terrorsquid
Find the limit, if it exists, of the following:

$\displaystyle \lim_{x \to 2}\left(\frac{1}{x-2}-\frac{1}{x^2-4}\right)$

I have tried factoring this in a number of different ways, combining the factions with various denominators and can't seem to get it into a form where the denominator is not 0. I am also unsure as to how I would prove that it has no limit if that is the case.

$\displaystyle \frac{1}{x-2} - \frac{1}{x^2-4} =$

$\displaystyle \frac{x+2}{(x+2)(x-2)} - \frac{1}{(x+2)(x-2)} =$

$\displaystyle \frac{x+1}{(x+2)(x-2)}$

the rational function represented by the expression above has a vertical asymptotes at $\displaystyle x = \pm 2$, so ...

$\displaystyle \lim_{x \to 2} \frac{x+1}{(x+2)(x-2)}$ does not exist