# Find the limit using the polar coordinates `x=rcos(theta)` , `y=rsin(theta)` , `r=sqrt(x^2+y^2)` .` lim_(x,y->0,0) (x^2*y^2)/sqrt(x^2+y^2)` . We must know that r is approaching 0 from the right when (x,y)->(0,0).

## Expert Answers To answer this question, we'll need to evaluate the limit after substituting the polar coordinates for the Cartesian Coordinates:

`lim_(x,y->0,0) (x^2y^2)/sqrt(x^2+y^2) = lim_(r->0)(r^2cos^2thetar^2sin^2theta)/r`

You can reduce this both by recognizing the trig identity `costhetasintheta = 1/2sin(2theta)`

`=lim_(r->0)r^4/r*(1/2sin(2theta))^2`

Now, we simplify:

`= lim_(r->0) r^3 * 1/4sin^2 (2theta)`

Well, we have a couple of considerations here. The first is that `theta` is independent of `r`, so we can treat `1/4sin^2(2theta)` as a constant!

Now, I know what you might be thinking..."Why did we do that trig substitution if we're just going to treat the trig functions as constants?" My response would be that trig functions and their identities often hold the key to evaluating these problems, so getting used to seeing obvious identities will help you out a lot in terms of evaluating more complex versions of these problems later on.

Secondly, because we only need to look at the terms containing `r` , we can explicitly evaluate the limit as it is shown:

`lim_(r->0) r^3 * (1/4sin^2(2theta)) = 0`

We only care about the `r^3` term, and when r approaches 0, this term approaches 0 as well!

So, pretty clearly, our limit is 0. Hope that works out for you!

Approved by eNotes Editorial Team ## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

• 30,000+ book summaries
• 20% study tools discount
• Ad-free content
• PDF downloads
• 300,000+ answers
• 5-star customer support