Evaluate limit

  • Thread starter mvww
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  • #1
mvww
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Let U and t be independent real variable.
In the limit:
[itex]t/U \rightarrow 0[/itex]
Can I say that
[itex]t^2/U \rightarrow 0[/itex]
too? Or I can compare only same powers of both?

Regards.
 
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Answers and Replies

  • #2
DonAntonio
606
1
If
[itex]t/U \rightarrow 0[/itex]
can I say that
[itex]t^2/U \rightarrow 0[/itex]
too? Or I can compare only same powers of both?

Regards.


You don't say what is t, what is U, the limit of what tending to what...are we to guess?

DonAntonio
 
  • #3
mvww
9
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You don't say what is t, what is U, the limit of what tending to what...are we to guess?

DonAntonio

Fixed. thanks.
 
  • #4
Millennial
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0
First of all, you still did not explain anything. What does the limit tend to, what is t, what is U? Is t dependent or U, or for example they both are dependent on some arbitrary variable x?
 
  • #5
mvww
9
0
First of all, you still did not explain anything. What does the limit tend to, what is t, what is U? Is t dependent or U, or for example they both are dependent on some arbitrary variable x?

they are both independent variables. U is much bigger than t, in the sense that
[itex]t/U \rightarrow 0[/itex]
 
  • #6
DonAntonio
606
1
they are both independent variables. U is much bigger than t, in the sense that
[itex]t/U \rightarrow 0[/itex]


Ok, please DO PAY ATTENTION! You have to tell what in the name of Newton is tending where!
Is it [itex]\,t\to 0\,\,,\,t\to\infty\,\,,\,U\to 0...[/itex] ? Common...

DonAntonio
 
  • #7
SteveL27
799
7
Let U and t be independent real variable.
In the limit:
[itex]t/U \rightarrow 0[/itex]
Can I say that
[itex]t^2/U \rightarrow 0[/itex]
too? Or I can compare only same powers of both?

Regards.

[itex]\displaystyle\ \lim_{x\to\infty} \frac{x}{x^2} = 0[/itex]

but

[itex]\displaystyle\ \lim_{x\to\infty} \frac{x^2}{x^2} = 1[/itex]

Note that I specified what variable is going to what limit, which is an essential part of this question, as Don Antonio's pointing out.
 
  • #8
Bacle2
Science Advisor
1,089
10
I think this may have to see with the "Big O " class that the expression belongs to,

i.e., how fast does the expression go to 0 ?
 
  • #9
[itex]\displaystyle\ \lim_{x\to\infty} \frac{x}{x^2} = 0[/itex]

but

[itex]\displaystyle\ \lim_{x\to\infty} \frac{x^2}{x^2} = 1[/itex]

Note that I specified what variable is going to what limit, which is an essential part of this question, as Don Antonio's pointing out.

To use SteveL27's example: A limit is used to describe what value(s) a funcation approaches as x reaches a specific value. Steve's first example
[tex]\lim_{x\to\infty} \frac{x}{x^2}[/tex]
says that as x approaches infinity, for the function [itex]\frac{x}{x^2}[/itex] is equal to zero. although the function never quite reches this point, that's what value the function approaches, as x gets closer and closer to infinity. 1/12=1, 2/22=0.5, 3/32=0.33.3...155/1552=0.006451612903...12,347,222/12,347,2222=8.098987 X 10-8.
It gets closer and closer to zero. yet never acually gets to the point that f(x)=0.

Limits can be used for any part of any function, even simple ones (even thought there isn't any reason to waste time with evaluating limits for simple, continuous function), for example:

[tex]\lim_{x\to 1.5} 5x-{x^2}=5.2[/tex]

limits are usually used to describe breaks in non-continous functions, functions approaching (+ or -) infinity, or specific x values that make the y value spike up to (+ or -) infinity.

Here are some videos that may help:

http://www.youtube.com/watch?v=UkjgJQaGx98&feature=relmfu
 
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