Question

Find \[\Large \lim_{(x, y) \rightarrow (0, 0)} \frac{x^2\sin^2(y)}{x^2 + 2y^2}\] if it exists. I have a feeling that the limit is 0, after checking paths along x-axis, y-axis, y=x and y = x^2, but I'm no quite sure how to prove it.

if \(\large 0 < \sqrt{x^2 + y^2} < \delta\) then \(\large \frac{x^2\sin^2(y)}{x^2 + 2y^2} < \epsilon\) (Dropped the absolute bars because they're all quadratics)

Since \(\large \sin^2(y) \leq 1\), \(\large \frac{x^2\sin^2(y)}{x^2 + 2y^2} \leq \frac{x^2}{x^2 + 2y^2}\)
But after this I'm stuck

Answer 1

...

Answer 2

I don't know how to do this

Answer 3

The limit does not exist, use 2-path test

Answer 4

replace sin^2(y) by y^2, since lim y->0 sin^y / y^2 =1 (single-variable calc stuff), now use the 2-path test

Answer 5

@Loser66: Well wait, I've just had a second thought. The limit actually exists and equals 0. Sorry I was just too careless.

Answer 6

|x^2 + 2y^2| >= sqrt(2)|xy|. So \[\lim_{(x,y) \rightarrow (0,0)}\left| \frac{ x ^{2}y^{2} }{ x^{2}+2y^{2} } \right|\le \lim_{(x,y) \rightarrow (0,0)}\frac{ 1 }{ \sqrt{2} }|xy|=0. \]

Answer 7

you should diffrentiate numerator and denominator independently wrt to any z till 0/0 form is removed

Answer 8

@rohankv: Are you going to say you are applying L'Hospital's Rule?

Answer 9

ofcourse....

Answer 10

Well, L'Hospital'S Rule is strictly a single-variable stuff, hence cannot be applied here.

Answer 11

L hospital rule concerns about rate of the c hange in in numerator expression & denominator expression as the variables/parameters tends to a finite or infinite value
so you can plot the graphs of the numeratorn & denominator expression as a function of one variable say" x" treating the oher variable constant and observing the graphs to draw result

Answer 12

Well said, but that's applicable only to single-variable functions and limits. What you are trying to do is iterating a limit, which, in general cases, is wrong as x approaches 0 and y approaches 0 at the same time.

Answer 13

Sorry for the late response, but I was in class so I could only check on my phone and not type a really large reply. I solved it in math class already though, using a similar approach as drawar :)

If anybody could check if this is correct it would be appreciated:

if \(0 < \sqrt{x^2 + y^2} < \delta\) then \(\large {x^2\sin^2y \over x^2 + 2y^2} < \epsilon\)
Since \(\large {x^2 \over x^2 + 2y^2} < 1\) and \(\sin^2 y < y^2\),
\[{x^2 \sin^2 y \over x^2 + 2y^2} < y^2 < x^2 + y^2 = \left(\sqrt{x^2 + y^2}\right)^2\]
So if we let \(\large \delta = \sqrt\epsilon\),
\[{x^2\sin^2y \over x^2 + 2y^2} < x^2 + y^2 < \delta^2 = \epsilon\]
So the limit exist and is 0

Answer 14

Proof by definition is always nice, but how do you justify this: \[\sin^{2}y<y^{2}\]?

Answer 15

sin(y) < y for every y != 0, so sin^2(y) < y^2

Answer 16

well, sin(y) < |y|, but squaring it makes it both positive

Answer 17

great, the last one makes sense to me :)

Answer 18

I liked your use of the squeeze theorem aswell :p