Limits, what they are and what they do

  • #26
Thoth
Let’s say f (x) =x 2 and we want to see if there is a limit L=4 as x approaches c=2. At this point the problem might state that we want to choose a point so close to c that from x to c the absolute distance is three places of zero (&delta;=. 001) So we use the Limit definition and we found that |x-2|<. 001 or 1.999<x<2.001. Here we are approaching 2 from its left side (1.999) and from its right side (2.001).

We see that f (1.999)&asymp; 3.996 and f (2.001)&asymp; 4.004. So |f (x) –L| is 4.004-4= .004 and 3.996-4=-.004 hence &epsilon;= |. 004|=. 004 or &epsilon;=4&sigma;

We see that as we approach point B from its left and from its right point B approaches a Limit number L from both directions then the limit exists and point B can be reached.

This is like saying that number 2 between number 1 and number 3 exists because if one approaches number 2 from number 1 (left side) or from number 3 (right side) one still reaches number 2.:wink:
 
  • #27
163
0
A puzzling limit paradox

Hello String,
While in graduate school and engulfed in Kaplan's "Advanced Calculus Theory" I ran into a function that may have been unique because it occurred for only one integer exponent in the set of so-called "power series" of the form: f(x) = k / x^n where n is an integer. As I recall, at all values of "n" the limit as x -> 0 is asymptotic to the y axis; however, for at least one integer the area under the curve between zero and any x is finite.
This is really not show boating but rather a challenge for you calculus students to ponder over. I believe that the integral dx/x = ln x has somthing to do with understanding this mystery. Cheers, Jim
 
  • #28
Hurkyl
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17
&int;0..x t^n dt

only exists for n > -1, so what you're remembereing must be something different.. I can't think of what that might be.

PS: For those interested, the series:

(&Sigma;n=1..&infin; a-n 1/xn) + a0 + (&Sigma;n=1..&infin; an xn)

is called a Laurent series.
 
  • #29
163
0
You could be right Hurkyl

I really don't remember the exact function but the detailed discussion that occurred in class was real. I'm pretty sure that we were in the textbook area having to do with powers of the independent variable in the denominator. My Kaplan disappeared at least 35 years ago and I'm too busy to browse the two or so cubic feet of notes that I think are somewhere in a box. I thank you for calling out my mis-thought. Maybe I'll take time to browse my distant past later. Cheers, Jim
 
  • #30
Hurkyl
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Well, the "n > -1" criterion holds for all real n. So, for example with n = -1/2:

&intt=0..1 1 / t^(1/2) dt = 2
 
  • #31
163
0
Thanks again

Hi Hurkyl,
By Jove,I think you've got it! Cheers, Jim
 

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