Limiting f(h): Confirmation Needed - 1st Post!

[lilman7769]

i did this problem but i just wanted some confirmation if I am right...

lim (3+h)^-1 - (3)^-1 / h
h->0


i worked it out...but still got 0...maybe some confirmation and if u could show ur work that would be nice too...thx in advance



ps my first post!

 

[T@P]

if I am not mistaken that's a derivative in one of it (many) forms.
basically, it x' at the point 3... which would be 1 if i understood your notation right. (is the h under the whole fractoin or just the end?)

hope I am not confusing you too much.

 

[Townsend]

i did this problem but i just wanted some confirmation if I am right...

lim (3+h)^-1 - (3)^-1 / h
h->0


i worked it out...but still got 0...maybe some confirmation and if u could show ur work that would be nice too...thx in advance



ps my first post!

Based on what you are saying the function in question is a constant function. Meaning the function is of the form y=1/3, which is just a boring horizontal line. Now consider what a derivative is suppose to be. A derivative gives information about a functions slope at values of x in the domain of the function. Now consider your horizontal line, what can you tell me about its slope?

Best Regards

 

[HallsofIvy]

Based on what you are saying the function in question is a constant function. Meaning the function is of the form y=1/3, which is just a boring horizontal line. Now consider what a derivative is suppose to be. A derivative gives information about a functions slope at values of x in the domain of the function. Now consider your horizontal line, what can you tell me about its slope?

Best Regards

How did you get that interpretation?

Assuming the orginal post was lim ((3+h)^-1- 3^-1)/h
(notice the additional parentheses) then this is the derivative of x^-1 evaluated at x= 3. Assuming that the purpose of this is to actually calculate that derivative (so that you can't just use the derivative itself to get the limit!) then the best way to do it is to combine the fractions:

[tex]\frac{1}{3+h}- \frac{1}{3}= \frac{3- (3+h)}{3(3+h)}=\frac{-h}{3(3+h)}[/itex]
so the "difference quotient" becomes
[tex]\frac{-h}{3h(3+h)}[/itex].
As long as h is not 0 that is the same as
[tex]\frac{-1}{3(3+h)}[/itex]
and it should be easy to find the limit as h goes to 0.

 

[JasonRox]

I got zero on the work as well. Find the limit.

$$\frac{\frac{-1}{3(3+h)}}{h}$$

Simplify and use direct substitution.

I got zero.

I could be wrong though.

$$\frac{\frac{-1}{3(3+h)}}{h}$$

to

$$\frac{-h}{3(3+h)}$$

Now, substitute h~0.

Note: I excluded lim in my work for simplicities sake.

 

[JasonRox]

Also, I'm a Brock University student. ;)

 

[Tide]

Jason,

You dropped a factor of h in the numerator. Halls' analysis is correct.

 

[JasonRox]

We aren't looking for the rate of change at x=3.

We are looking for the limit.

The limit of 1/x is 0.

 

[HallsofIvy]

Brock University may get upset at you for using their name!

In the first place, "the limit of 1/x is 0" is meaningless- you have to say "limit of 1/x" as x goes to some specific value4. The only number that would give a limit of 0 for 1/x is infinity and infinity has nothing to do with the original problem.

"We aren't looking for the rate of change at x=3."

Perhaps you aren't but anyone who is trying to answer the original question is!

 

[JasonRox]

He/she wasn't looking for that. I am in the same program because we have the same assignment. The two questions she asked came from the same school.

About the limit mistake, there's an even bigger one. I thought about it last night while going to bed, and thought about what I said. First, it didn't make any sense, like you explained. Second, and I can't believe you didn't spot this, the limit doesn't really exist because the left hand limit doesn't equal the right hand limit.

Note: I have every right to say I'm a Brock student.

 

[JasonRox]

Just so you know the assignments have been handed in.

 

[Tide]

Just so you know the assignments have been handed in.

Well then be sure to report back to us when you find the "right" answer!

 

[T@P]

whoa
why doesn't the original poster just re-state the question so there's no more ambiguity?
I personally am not sure wether the last h refers to the whole limit or not...

 

[Townsend]

How did you get that interpretation?

I have no idea why I thought what I did but I am glad you were able to correct it. Sorry about that. Next time I try to help I will be a lot more careful before posting.

Regards