Limits

magdaddy101

New member
Joined
Oct 4, 2009
Messages
6
This is a true/false question, and I have to say why it's true or false:

lim (x,y)-->(0,0) |x^2 - y^2| / (x^2 + y^2) = 1


just in case it's hard to interpret the question, it says that the limit as (x,y) approaches (0,0) of |x^2 - y^2| / (x^2 + y^2) is equal to 1.

I'm not allowed to use L'Hospital's rule to explain this.
Thanks in advance for your help.
 
Given: lim(x,y)(0,0)x2y2x2+y2 = 1, To prove true or false.\displaystyle Given: \ \lim_{(x,y)\to(0,0)}\frac{|x^{2}-y^{2}|}{x^{2}+y^{2}} \ = \ 1, \ To \ prove \ true \ or \ false.

If (x,y)>(0,0) along the xaxis, then y = 0, so\displaystyle If \ (x,y)->(0,0) \ along \ the \ x-axis, \ then \ y \ = \ 0, \ so

lim(x,y)(0,0)f(x,y) = lim(x,y)(0,0)f(x,0) = lim(x,y)(0,0)x2x2 = lim(x,y)(0,0)1 = 1.\displaystyle \lim_{(x,y)\to(0,0)}f(x,y) \ = \ \lim_{(x,y)\to(0,0)}f(x,0) \ = \ \lim_{(x,y)\to(0,0)}\frac{x^{2}}{x^{2}} \ = \ \lim_{(x,y)\to(0,0)}1 \ = \ 1.

If (x,y)>(0,0) along the yaxis, then x = 0, so\displaystyle If \ (x,y)->(0,0) \ along \ the \ y-axis, \ then \ x \ = \ 0, \ so

lim(x,y)(0,0)f(x,y) = lim(x,y)(0,0)f(0,y) = lim(x,y)(0,0)y2y2 = lim(x,y)(0,0)1 = 1.\displaystyle \lim_{(x,y)\to(0,0)}f(x,y) \ = \ \lim_{(x,y)\to(0,0)}f(0,y) \ = \ \lim_{(x,y)\to(0,0)}\frac{|-y^{2}|}{y^{2}} \ = \ \lim_{(x,y)\to(0,0)}1 \ = \ 1.

But if (x,y)>(0,0) along y = x, then\displaystyle But \ if \ (x,y)->(0,0) \ along \ y \ = \ x, \ then

lim(x,y)(0,0)f(x,y) = lim(x,y)(0,0)f(x,x) = lim(x,y)(0,0)x2x2x2+x2 = lim(x,y)(0,0)02x2 = lim(x,y)(0,0)0 =0.\displaystyle \lim_{(x,y)\to(0,0)}f(x,y) \ = \ \lim_{(x,y)\to(0,0)}f(x,x) \ = \ \lim_{(x,y)\to(0,0)}\frac{|x^{2}-x^{2}|}{x^{2}+x^{2}} \ = \ \lim_{(x,y)\to(0,0)}\frac{0}{2x^{2}} \ = \ \lim_{(x,y)\to(0,0)}0 \ =0.

Hence, the limit doesnt exist and the answer is false.\displaystyle Hence, \ the \ limit \ doesn't \ exist \ and \ the \ answer \ is \ false.

Note: Calculating limits along specific curves is a useful technique for showing that a limit\displaystyle Note: \ Calculating \ limits \ along \ specific \ curves \ is \ a \ useful \ technique \ for \ showing \ that \ a \ limit

 does not exist because it is only necessary to find two curves in which the limits differ. However,\displaystyle \ does \ not \ exist \ because \ it \ is \ only \ necessary \ to \ find \ two \ curves \ in \ which \ the \ limits \ differ. \ However,

 this method is of no use for showing that the limit exists because it is impossible to examine all\displaystyle \ this \ method \ is \ of \ no \ use \ for \ showing \ that \ the \ limit \ exists \ because \ it \ is \ impossible \ to \ examine \ all

 possible curves.\displaystyle \ possible \ curves.
 
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