Limit easy

Calculus Level 2

$\lim_{x,y \to 0,0}(\frac{xy}{\sqrt{x^2+y^2}} )$

1 0

\pi

Indeterminated -1

1 solution

Chris Juarez
Dec 23, 2017

$x=rcos\theta$ $y=rsin\theta$ $r=\sqrt{x^2+y^2}$ $=\lim_{x,y \to 0,0}\frac{rcos\theta*rsin\theta}{r}$ $=\lim_{x,y \to 0,0}\frac{r^2(cos\theta*sin\theta)}{r}$ $=\lim_{x,y \to 0,0}\frac{rcos\theta*sin\theta}{1}$ $=\lim_{x,y \to 0,0}\frac{\sqrt{x^2+y^2}*cos\theta*sin\theta}{1}$ $=\lim_{x,y \to 0,0}\sqrt{0+0}cos\theta*sin\theta$ $\lim_{x,y \to 0,0}=0$

I got to your line with r * cos(theta) * sin(theta). Then I thought to myself, for any infinitesimal radius epsilon, there is profound variation along theta. So the only way in which there is an unambiguous single answer is if we are actually at (0,0). But this seems to defy the concept of a limit, since a limit deals with what happens in proximity to a point, rather than what happens at the point. @Calvin Lin , am I totally wrong with this?

Steven Chase - 3 years, 5 months ago

This solution starts off "correct-ish", but just not written with the explanation of what is happening. He is parametrizing a path in the x-y plane with an endpoint at $(0,0)$ , by using polar coordinates.

Also, at the end, replacing $x$ and $y$ with 0 isn't kosher.

What he should have written is $\lim_{(x,y) \rightarrow 0} = \lim_{r \rightarrow 0 }$ . Then, it is clear that $\lim_{r \rightarrow 0 } r\cos\theta\sin\theta = 0$ .

Calvin Lin Staff - 3 years, 5 months ago

Limit easy

1 solution

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