Proving that lim(x,y)->(0,0) (x^2 + x|y|)/sqrt(x^2 + y^2) = 0

[Ale_ygt]

Hi, I'm stuck with calculating the following limit (or rather proving it, since I already know the result):
\(\displaystyle
\lim_{(x,y) \to (0,0)}\frac{x^{2}+x|y|}{\sqrt{x^{2}+y^{2}}}=0
\)

I have no idea about how to proceed, I was thinking about using the squeeze theorem but the non-absolute x in the numerator apparently makes it a void attempt:

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

1st limit:
\(\displaystyle
\frac{x^{2}}{\sqrt{x^{2}+y^{2}}} \leq \frac{x^{2} + y^{2}}{\sqrt{x^{2}+y^{2}}} =\sqrt{x^{2}+y^{2}}
\)

So by the squeeze theorem, the 1st limit tends to 0; as for the 2nd, the same method doesn't apply:

\(\displaystyle
\frac{x|y|}{\sqrt{x^{2}+y^{2}}}
= \frac{x\sqrt{y^{2}}}{\sqrt{x^{2}+y^{2}}}
\)

Thanks in advance to whoever can help

 

[topsquark]

Hi, I'm stuck with calculating the following limit (or rather proving it, since I already know the result):
\(\displaystyle
\lim_{(x,y) \to (0,0)}\frac{x^{2}+x|y|}{\sqrt{x^{2}+y^{2}}}=0
\)

I have no idea about how to proceed, I was thinking about using the squeeze theorem but the non-absolute x in the numerator apparently makes it a void attempt.
Thanks in advance to whoever can help

This one just screams "polar coordinates" to me. Try subbing \(\displaystyle x =r~cos( \theta )\) and \(\displaystyle y = r~sin( \theta )\). Then you simply have a limit as \(\displaystyle r \to 0\). Give it a try and get back to us.

And remember, this is a 2-D limit. It has to hold for any direction you approach (0, 0) from. How can you make an argument for that?

-Dan

 

[Ale_ygt]

I edited the 1st post adding some details on how I approached the issue without realizing that I got a reply meanwhile, my bad.

And remember, this is a 2-D limit. It has to hold for any direction you approach (0, 0) from. How can you make an argument for that?

It might be because it's a bit late over here, but I'm not sure what do you mean... I did the slope-substitution f(x,mx) and it seemed to hold, even though I know it's a necessary (and not sufficient) condition to prove that the limit exists.
I might as well get some rest and give it another try tomorrow, possibly with the polar coordinates method you suggested (even though we haven't been taught that yet)

 

[topsquark]

I edited the 1st post adding some details on how I approached the issue without realizing that I got a reply meanwhile, my bad.


It might be because it's a bit late over here, but I'm not sure what do you mean... I did the slope-substitution f(x,mx) and it seemed to hold, even though I know it's a necessary (and not sufficient) condition to prove that the limit exists.
I might as well get some rest and give it another try tomorrow, possibly with the polar coordinates method you suggested (even though we haven't been taught that yet)

Yes, you can approach it that way as well. So long as your result comes out to be "0" no matter what m is (what about approaching with a vertical line?) then it should work just fine.

-Dan

 

[Ale_ygt]

I *think* I solved it using the limit definition; supposing that

\(\displaystyle
\lim_{(x,y) \to (0,0)}\frac{x^{2}+x|y|}{\sqrt{x^{2}+y^{2}}}=0
\)

We have:

\(\displaystyle
\mid x^{2}+x|y| - 0\mid \leq \mid x^{2} \mid + \mid x|y| \mid
= x^{2} + \mid xy \mid
\)

Then we have:
1)
\(\displaystyle
\frac{x^{2}}{\sqrt{x^{2}+y^{2}}} \leq \frac{x^{2} + y^{2}}{\sqrt{x^{2}+y^{2}}} =\sqrt{x^{2}+y^{2}}
\)

2)
\(\displaystyle
\frac{ \mid xy \mid}{\sqrt{x^{2}+y^{2}}} \leq \frac{x^{2} + y^{2}}{2 \cdot \sqrt{x^{2}+y^{2}}} = \frac{ \sqrt{x^{2}+y^{2}}}{2} \leq \sqrt{x^{2}+y^{2}}
\)

Summing the left and right sides of inequalities 1) and 2) yields
\(\displaystyle
\frac{ x^{2}+ \mid xy \mid }{\sqrt{x^{2}+y^{2}}} \leq 2 \cdot \sqrt{x^{2}+y^{2}}
\)

Therefore, if we choose \(\displaystyle \delta =\frac{\varepsilon }{2}\), the limit is verified:
\(\displaystyle
\sqrt{x^{2}+y^{2}} < \delta \Rightarrow \frac{\mid x^{2}+x|y| - 0\mid}{\sqrt{x^{2}+y^{2}}} < \varepsilon
\)

Is it correct?