Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
Determine whether the limit lim?
Determine whether the limit lim as (x,y) approaches (0,0) xysin(1/(x^2+y^2)) exists or not. If it exists, compute it.
Please give the steps and details.
1 Answer
- kbLv 79 years agoFavorite Answer
Use polar coordinates.
So, the limit transforms to
lim(r→0+) (r cos t)(r sin t) sin(1/r^2)
= lim(r→0+) r^2 sin(1/r^2) * cos t sin t.
This equals 0 by the Squeeze Theorem. (Details below.)
Since sine and cosine have range [-1, 1], we have
-1 ≤ sin(1/r^2) * cos t sin t ≤ 1 for all t and nonzero r
==> -r^2 ≤ r^2 sin(1/r^2) * cos t sin t ≤ r^2 for all t and nonzero r.
Since lim(r→0+) ±r^2 = 0, the conclusion now follows from the Squeeze Theorem.
I hope this helps!
