Solve Simple Limit Problem: (1/x)-(1/2)÷(x-2)

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Homework Help Overview

The problem involves evaluating the limit of the expression (1/x) - (1/2) divided by (x-2) as x approaches 2. Participants discuss the method to simplify the expression and the application of L'Hopital's rule.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the simplification of the numerator into a single fraction and question the use of L'Hopital's rule. There is discussion about identifying common denominators and the proper treatment of fractions.

Discussion Status

Some participants have provided guidance on rewriting the expression and identifying common denominators. There is a mix of understanding among participants, with some expressing confusion about fraction operations and others attempting to clarify these concepts.

Contextual Notes

There appears to be some misunderstanding regarding the manipulation of fractions, and participants are encouraged to reconsider their approach to adding and subtracting fractions. The original poster expresses difficulty in solving the problem after an extended period of consideration.

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lim x=>2

(1/x) - (1/2)
------------
x-2I 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!
 
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Have you learned about L'hopital's rule yet?
If you haven't, let your first step be to write your numerator as a single fraction.
 
L'Hopital's rule is "overkill" here!

Arildno's second suggestion, writing
[tex]\frac{1}{x}- \frac{1}{2}[/tex]
as a single fraction is best. What is the common denominator?
 
arildno said:
Have you learned about L'hopital's rule yet?
If you haven't, let your first step be to write your numerator as a single fraction.

No, never heard of him, sorry.

HallsofIvy said:
L'Hopital's rule is "overkill" here!

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

I only see a common numerator of 1 not denominator.

Can I do this:

[tex]\frac{-1}{x-2}[/tex] ?
 
Last edited:
swears said:
No, never heard of him, sorry.



I only see a common numerator of 1 not denominator.

Can I do this:

[tex]\frac{-1}{x-2}[/tex] ?
HORROR OF HORRORS!

NEVER EVER MISTREAT FRACTIONS IN THAT MANNER! :mad: :mad: :mad: :mad:
 
umm, OK.

Can someone else help me out?
 
Do you even know what a fraction is, or what the symbol x stands for?
 
yes, x is a variable. 1/2 is a fraction.
 
No. You have completely misunderstood it.
 
  • #10
Riight. Well, are you going to correct me, or just keep criticizing?
 
  • #11
Why don't you sit back and re-think how we add or subtract fractions together?
 
  • #12
I've been looking at this problem for an hour, I obviously don't know how to do it.
 
  • #13
Look at [itex]\frac{1}{x}-\frac{1}{2}[/itex]
How many fractions do you have in this expression?
 
  • #14
arildno said:
Look at [itex]\frac{1}{x}-\frac{1}{2}[/itex]
How many fractions do you have in this expression?

2 fractions
 
  • #15
Correct!
Now, what is a common denominator for those fractions?
 
  • #16
-"this message is too short"- 1?
 
  • #17
i'll give you a hint or two
[tex]\frac{\frac{1}{x} - \frac{1}{2}}{x-2}[/tex]
[tex]=\frac{\frac{2}{2x} - \frac{x}{2x}}{x-2}[/tex]
[tex]=\frac{\frac{2-x}{2x}}{x-2}[/tex]
[tex]=\frac{\frac{-(x-2)}{2x}}{x-2}[/tex]
[tex]=-\frac{1}{2x}[/tex]
then substitue x=2 in there to get your answer of -.25
where i have used the following
[tex]\frac{\frac{a}{b}}{c}=\frac{a}{bc}[/tex]
[tex]2-x=-(x-2)[/tex]
[tex]\frac{a}{b}-\frac{c}{d}=\frac{ad-bc}{bd}[/tex]
[tex]\frac{a}{c} - \frac{b}{c} = \frac{a-b}{c}[/tex]
 
Last edited:
  • #18
Thanks for your help. I'm going to try and soak this in and figure out how u did that.
 
Last edited:
  • #20
yeah you can do that because for example
[tex]\frac{1\times 2}{2\times 2} = \frac{2}{4}=\frac{1}{2}[/tex]
 
  • #21
Cool, Thanks so much. I totally forgot about that Lowest Common Denominator thing.
 

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