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lim((ln(1+x*y^2))/(x^2+y^2) when (x,y) tends to (0,0) detailed please.

 Apr 18, 2016
 #1
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The limit is zero.  See below:

 

limit

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 Apr 18, 2016
 #2
avatar+33661 
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Here's a 3D graph to illustrate:

 

3D graph

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 Apr 18, 2016
 #3
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Alan, why can you say that theta tends to zero ?

Why can't (x, y) tend to (0, 0) along some direction other than the x-axis ?

(That's not to say that your answer is incorrect).

 Apr 18, 2016
 #4
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Good point Guest! You are quite right. A better derivation is needed. 

 

In fact, I drew the graph first, and could see that the limit was zero, so my derivation of the limit was done rather hastily!

Alan  Apr 18, 2016
 #6
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Better solution:

 

Let x  =  f*y  where f is a finite constant.

 

Then ln(1+x*y2)/(x2+y2) becomes  ln(1 + f*y3)/((f2+1)y2)

 

No matter what the size of f, eventually, as y tends to zero, f*y3 will become much smaller than 1, so ln(1 + f*y3) will tend to f*y3, hence ln(1 + f*y3)/((f2+1)y2)  →  f*y3/((f2 + 1)y2)  →  f*y/(f2+1)  →   0

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 Apr 19, 2016

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