Limit of x as it approaches a variable

  • Thread starter Wyatt
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    limits
  • #1
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1. Problem statment

Q. Use limit to find the instantaneous velocity at time t if the postition is p(t) at time t.

p(t) = t + (1/t) at x = t

2. Homework Equations

(Sorry dashes are in there to keep everything where it should be. I don't know how to make fractions in this and it was deleting spaces.)


lim------f ( x+h ) - f (x)
x->0 ---_____________ (instantaneous velocity)
------------------h



The Attempt at a Solution




(Sorry dashes are in there to keep everything where it should be. I don't know how to make fractions in this and it was deleting spaces.)


-First I plugged in t
lim------f ( t+h ) - f (t)
x->0---_____________
----------------h

-Then I plugged in what the equation gives you when you plug those in


lim------[( t + h ) + (1 / ( t + h )] - [ t + ( 1 / t) ]
x->0---______________________________
--------------------------------h
-simplified it


lim-----t + h + ( 1 / t ) + ( 1 / h ) - t + ( 1 / t )--------lim-----h + ( 1 / h )
x->0 -_____________________________ = x->0---___________
-------------------------h-----------------------------------------------------h
- so then to get the fraction out of the numerator i multiplied numerator and denominator by h/h giving me

lim----h^2 + 1
x->0 -_______
-----------h^2
-this is where im stuck. I'm pretty sure I cant stop here because if I plug in zero then I get DNE. But I'm also not sure if I messed up somewhere or just dont know something
[/B]
 

Answers and Replies

  • #2
RUber
Homework Helper
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I think you might have the wrong definition...
You are finding the instantaneous velocity, which is defined as:
##\lim_{h\to 0} \frac{f(t+h) - f(t)}{h}##
This should make sense, if you recall simple position functions like p(t) = t would have a constant velocity of 1:
##\lim_{h\to 0} \frac{t+h - t}{h} = \frac{h}{h} = 1.##
 
  • #3
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This is a few of the problems we worked in class. Via the examples I thought the goal was to make it so that you could solve for h and have it be anything but DNE
https://learn.unt.edu/bbcswebdav/pid-3914721-dt-content-rid-51752337_1/xid-51752337_1 [Broken]
https://learn.unt.edu/bbcswebdav/pid-3914722-dt-content-rid-51752338_1/xid-51752338_1 [Broken]
 
Last edited by a moderator:
  • #4
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oh it deleted the pictures
 
  • #5
7
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https://learn.unt.edu/bbcswebdav/pid-3914721-dt-content-rid-51752337_1/courses/MATH.1710.114-NT752.1168.1/9-1%20Problem%208.jpg [Broken]
https://learn.unt.edu/bbcswebdav/pid-3914722-dt-content-rid-51752338_1/courses/MATH.1710.114-NT752.1168.1/9-1%20Problem%2010.jpg [Broken]
 
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  • #7
Stephen Tashi
Science Advisor
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(Sorry dashes are in there to keep everything where it should be. I don't know how to make fractions in this and it was deleting spaces.)

Studying how to use LaTex instructions in forum posts is an activity in itself, but if you have a knack for computer programming, you can probably pick-up LaTex quickly. If you reply to a forum post that has mathematics symbols in it, you can see how the LaTex instructions have been formed. (You don't have to really reply to a post. You can press the "reply" icon, look at the quoted post, and then abort the process.)

-simplified it

lim-----t + h + ( 1 / t ) + ( 1 / h ) - t + ( 1 / t )--------lim-----h + ( 1 / h )
x->0 -_____________________________ = x->0---___________
-------------------------h-----------------------------------------------------h




It looks to me like you are using the incorrect algebra ##\frac{1}{t + h} = \frac{1}{t} + \frac{1}{h} ##. That's not valid. Common denominator - remember ?
 
  • #8
RUber
Homework Helper
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344
In your original work, you assumed that x went to zero, or equivalently that t went to zero. What you should be doing is assuming that the difference between time 1 and time 2 go to zero for any time 1 = t and time 2 = t+h.
When you set this up, you should get a form that looks like
##\lim_{h\to 0} \frac{ t +h + \frac{1}{t+h} - t - \frac1t}{h} ##
Then use appropriate algebra to simplify.
 
  • #9
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Ok so I can take the t's out because they subtract from each other to 0. Then can i just take the h's out because they simplify and my answer is just
1/ (t + h) - (1/t)?
 
  • #10
RUber
Homework Helper
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344
No, that's not proper algebra.
t-t = 0, that step is good. Taking out the h's doesn't take them all into account.
##\frac{h + \frac{1}{t+h} - \frac1t}{h}##
You can break apart the fraction to:
##\frac{h}{h} +\frac{ \frac{1}{t+h} - \frac1t}{h}##
The first fraction should be simple to reduce. The second one requires you to combine the fractions in the numerator. How do you add or subtract fractions with different denominators?
##\frac{1}{t+h} - \frac1t = ???##
Take your time and work through it.
 
  • #11
RUber
Homework Helper
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344
@Wyatt, Have you made any progress on this problem?
 
  • #12
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Yes sorry. h/h =1
Then 1 / ( t+h ) - 1 / t = [t / t(t+h)] -[ ( t+h) / t ( t +h) ]
simplified it becomes: h / t^2+h
simplifying the whole problem it become t^2+h
then when h is approaching 0 the answer is t^2
 
  • #13
33,999
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Yes sorry. h/h =1
Then 1 / ( t+h ) - 1 / t = [t / t(t+h)] -[ ( t+h) / t ( t +h) ]
simplified it becomes: h / t^2+h
simplifying the whole problem it become t^2+h
then when h is approaching 0 the answer is t^2
This might be right, but it's not the whole problem, so you're not done.

This is the limit you're trying to evaluate: ##\lim_{h\to 0} \frac{ t +h + \frac{1}{t+h} - t - \frac1t}{h} ## (from RUber's work in post #8)
 

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