Limits of Multi-Variable Functions
Date: 11/12/2004 at 03:24:36 From: Tom Subject: Limits and continuity Find the limit, if it exists, or show that the limit doesn't exist. lim (x*y*cos(y))/(3*x^2 + y^2) as (x,y) ==> (0,0) I don't really know how to approach this. The main problem is that I don't know what makes it so that the limit doesn't exist. There was something in our lecture notes where for these types of questions we were meant to find the limit of the function using different paths. For example, you would substitute y = 0 into the function and find the limit as x approaches 0. Doing that with this function I get lim 0/(3*x^2)? I don't really know how to interpret this.
Date: 11/13/2004 at 18:09:12 From: Doctor Jordan Subject: Re: Limits and continuity Hi Tom, For a limit to exist, the function must be approaching the same value no matter which way it is approaching the point we are taking the limit at. For example, if we want to find the limit of a function z(x,y) as (x,y) approach the origin (0,0), then we could come at the origin from the y-axis, come at the origin from the x-axis, come at the origin from the line y = x, etc. If we get different values for any of these approaches (i.e. the limit as you approach from, say the x-axis is different from the limit as you approach from the line y = x), then we say the limit does not exist at the point we're considering. If we substitute y = 0 in, that means we are approaching along the x- axis. Does that make sense? It's like we're saying "y is already there, x you come over now". So you could try setting y = x (e.g. replace every y with an x and just look at the function as x approaches 0), or x = 0 and replace every x in the function with a 0 and just look at it as y approaches 0. You could try other approaches, but these are the easiest ones to check and the most common in exercises. If you try those methods and still can't figure it out, write me back. (Note: You should only start trying to prove that the limit doesn't exist when you've tried to figure out a limit, but failed each time.) - Doctor Jordan, The Math Forum http://mathforum.org/dr.math/
Date: 11/16/2004 at 11:04:32 From: Tom Subject: Thank you (Limits and continuity) Thank you for your help. :)
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