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Easy delta/epsilon proof of a multivariable limitby pureza
Tags: limit definition 
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#1
Jan1812, 11:28 PM

P: 3

Hi,
I'm trying to wrap my head around epsilon/delta proofs for multivariable limits and it turns out I became stuck on an easy one! The limit is: [itex]\lim_{(x,y) \to (1,1)}\frac{xy}{x+y}[/itex] Obviously, the result is [itex]1/2[/itex], but I'm unable to prove it! Any hints? Thank you! 


#2
Jan1912, 09:02 AM

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P: 39,533

Well, what have you done? I assume you want to prove that
"Given [itex]\epsilon> 0[/itex] there exist [itex]\delta> 0[/itex] such that if [itex]\sqrt{(x1)^2+ (y1)^2}<\delta[/itex] then [tex]\sqrt{\frac{xy}{x+y} 1}<\epsilon[/tex]" 


#3
Jan1912, 09:48 AM

P: 3

Actually I think I might have solved it:
I want to prove that Given [itex]\epsilon> 0[/itex] there exist [itex]\delta> 0[/itex] such that if [itex]\sqrt{(x1)^2+ (y1)^2}<\delta[/itex] then [tex]\left\frac{xy}{x+y} \frac{1}{2}\right<\epsilon[/tex] Now, [tex]\left\frac{xy}{x+y} \frac{1}{2}\right=\left\frac{2xy(x+y)}{2(x+y)}\right=\left\frac{xy + xyxy)}{2(x+y)}\right=\left\frac{x(y1)+y(x1)}{2(x+y)}\right\leq\left\frac{x(y1)}{2(x+y)}\right+\left\frac{y(x1)}{2(x+y)}\right[/tex] by the triangle inequality. Now, I know that since [itex]{(x,y) \to (1,1)}[/itex] I can make [itex]x[/itex] and [itex]y[/itex] positive so that [itex]\left\frac{x}{x+y}\right\leq1[/itex] and [itex]\left\frac{y}{x+y}\right\leq1[/itex]. [I'm no sure about this argument] This way, [tex]\left\frac{x(y1)}{2(x+y)}\right+\left\frac{y(x1)}{2(x+y)}\right\leqy1+x1[/tex] But this is just a square with side 2 around the point [itex](x, y)[/itex]. This square is entirely contained by the circle with center [itex](x, y)[/itex] and radius [itex]\sqrt{2}[/itex]. So, when given some [itex]ε>0[/itex] if I set [itex]δ=min(1, ε\sqrt{2})[/itex] and I choose a point [itex](x, y)[/itex] such that [itex]\sqrt{(x1)^2+ (y1)^2}<ε\sqrt{2}[/itex], then certainly [itex]y1+x1<ε[/itex] and, by extension, [tex]\left\frac{xy}{x+y} \frac{1}{2}\right<ε[/tex] What's wrong with my rationale? :) 


#4
Jan1912, 11:43 AM

P: 1,622

Easy delta/epsilon proof of a multivariable limit



#5
Jan1912, 12:13 PM

P: 3

Obviously, you are right (thank you). Take [itex]ε=1[/itex]. I can choose a point in the circle with radius [itex]\sqrt{2}[/itex] that is not inside the square [itex]x1+y1<1[/itex].
How about this: if I choose a point inside the circle with radius ε, then that point is also inside the square with side 2ε, because the circle is contained within the square. Therefore, if I set [itex]δ\leq ε[/itex], then I know that any point inside the circle with radius δ verifies [itex]x1+y1<ε[/itex], which, if the first part of my derivation is correct, implies that [tex]\left\frac{xy}{x+y} \frac{1}{2}\right<ε[/tex] Is this correct? 


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