2 calculus qestion
If fsubx(xsub0, ysub0) and fy(xsub0, ysub0) both exist, then f is continuous at (xsub0, ysub0). Prove or disprove
Hint consider the function defined by f(x,y) = (xy)/(x^2+y^2) , if (x,y) is not equal (0,0) or f(x,y) =0.
Let f(x,y) = (sin^2 (x-y))/ (abs(x) +abs(y) ).
Prove that lim as (x,y) approach (0,0) of f(x,y)=0.
Hint: for all real numbers m and n. abs(sin(m+n)) smaller than or equal to abo(m+n) smaller than or equal to abs(m) +abs(n).
note: abs(m) means absolution value of m.
Thank you very much
False. Use the example the book gives.
Originally Posted by littlemu
The function f(x,y) is defined on an open disk containing (0,0). Which means f is continous if and only if,
lim [(x,y)-->(0,0)] f(x,y) = f(0,0)=0
But if you choose the path x=y both approaching zero the limit of f(x,y) is then 1! It must always be zero if it continous.
This is Mine 55:):)th Post!!!
I shall use my own inequality which is simple but useful.
Let x,y>=0 then sqrt(x+y) <= sqrt(x)+sqrt(y)