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.
 
\(\displaystyle Given: \ \lim_{(x,y)\to(0,0)}\frac{|x^{2}-y^{2}|}{x^{2}+y^{2}} \ = \ 1, \ To \ prove \ true \ or \ false.\)

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

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

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

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

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

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

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

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

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

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

\(\displaystyle \ possible \ curves.\)
 
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