Simple Limit Problem

1. Jun 4, 2006

swears

lim x=>2

(1/x) - (1/2)
------------
x-2

I was looking through my notes and found this problem. It shows the answer to be -.25, but I don't see how they got that. I know I want to cancel out the (x-2) on both sides of the division bar, but I'm not sure how to do this. Can anyone help me out? Thanks!

2. Jun 4, 2006

arildno

Have you learnt about L'hopital's rule yet?
If you haven't, let your first step be to write your numerator as a single fraction.

3. Jun 4, 2006

HallsofIvy

L'Hopital's rule is "overkill" here!

Arildno's second suggestion, writing
$$\frac{1}{x}- \frac{1}{2}$$
as a single fraction is best. What is the common denominator?

4. Jun 4, 2006

swears

No, never heard of him, sorry.

I only see a common numerator of 1 not denominator.

Can I do this:

$$\frac{-1}{x-2}$$ ?

Last edited: Jun 4, 2006
5. Jun 4, 2006

arildno

HORROR OF HORRORS!!!!!!!!

NEVER EVER MISTREAT FRACTIONS IN THAT MANNER!

6. Jun 4, 2006

swears

umm, OK.

Can someone else help me out?

7. Jun 4, 2006

arildno

Do you even know what a fraction is, or what the symbol x stands for?

8. Jun 4, 2006

swears

yes, x is a variable. 1/2 is a fraction.

9. Jun 4, 2006

arildno

No. You have completely misunderstood it.

10. Jun 4, 2006

swears

Riight. Well, are you going to correct me, or just keep criticizing?

11. Jun 4, 2006

arildno

Why don't you sit back and re-think how we add or subtract fractions together?

12. Jun 4, 2006

swears

I've been looking at this problem for an hour, I obviously don't know how to do it.

13. Jun 4, 2006

arildno

Look at $\frac{1}{x}-\frac{1}{2}$
How many fractions do you have in this expression?

14. Jun 4, 2006

swears

2 fractions

15. Jun 4, 2006

arildno

Correct!
Now, what is a common denominator for those fractions?

16. Jun 4, 2006

swears

-"this message is too short"- 1?

17. Jun 4, 2006

i'll give you a hint or two
$$\frac{\frac{1}{x} - \frac{1}{2}}{x-2}$$
$$=\frac{\frac{2}{2x} - \frac{x}{2x}}{x-2}$$
$$=\frac{\frac{2-x}{2x}}{x-2}$$
$$=\frac{\frac{-(x-2)}{2x}}{x-2}$$
$$=-\frac{1}{2x}$$
where i have used the following
$$\frac{\frac{a}{b}}{c}=\frac{a}{bc}$$
$$2-x=-(x-2)$$
$$\frac{a}{b}-\frac{c}{d}=\frac{ad-bc}{bd}$$
$$\frac{a}{c} - \frac{b}{c} = \frac{a-b}{c}$$

Last edited: Jun 4, 2006
18. Jun 4, 2006

swears

Thanks for your help. I'm going to try and soak this in and figure out how u did that.

Last edited: Jun 4, 2006
19. Jun 4, 2006

Math Is Hard

Staff Emeritus
20. Jun 4, 2006

yeah you can do that because for example
$$\frac{1\times 2}{2\times 2} = \frac{2}{4}=\frac{1}{2}$$