anonymous
  • anonymous
Evaluate the limit or show it does not exist.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\lim_{(x,y) \rightarrow (0,0)}\frac{ x^2-y^2 }{ x^2+y^2 }\]
anonymous
  • anonymous
Just show it approaching from different paths?
anonymous
  • anonymous
Try polar coordinates.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Makes it harder.
anonymous
  • anonymous
I get: \[\lim_{r \rightarrow \theta} \cos^2(\theta)-\sin^2(\theta)\]
anonymous
  • anonymous
r approaches 0 sorry not theta.
anonymous
  • anonymous
Yeah, which is equal to \[ \\cos^2(\theta)-\sin^2(\theta) \]Now convert it back to Cartesian coordinates.
anonymous
  • anonymous
Ohh... :P .
anonymous
  • anonymous
Sec.
anonymous
  • anonymous
Actually that doesn't really help, nevermind.
anonymous
  • anonymous
I can't though.
anonymous
  • anonymous
Yeah :P
anonymous
  • anonymous
I think evaluating it to \[ \cos^2(\theta)-\sin^2(\theta) \]Does have some meaning though.
anonymous
  • anonymous
No I got a better idea actually.
anonymous
  • anonymous
Got it!
anonymous
  • anonymous
Try the path y=x and y=0. THe limits dont match so the limit does not exist.
anonymous
  • anonymous
Thanks though :) .

Looking for something else?

Not the answer you are looking for? Search for more explanations.