MHB Fixing Limits: Ensuring Correct Solutions with Proper Resolution Techniques

[Fabio010]

I know that the solution is correct. But i do not know if i resolved it in the correct way.

The images are attached.

 

[Sudharaka]

I know that the solution is correct. But i do not know if i resolved it in the correct way.

The images are attached.

Hi fabio010,

Your image is not clear enough to give any help. Learn a bit of LaTeX to post in this forum.

http://www.mathhelpboards.com/forumdisplay.php?26-LaTeX-Help

 

[CaptainBlack]

Let's assume the first one is:

\[ \lim_{x,y\to 0} (x^2+y^2) \sin\left( \frac{1}{xy}\right) \]

Then yes the limit is zero by the squeeze theorem:

\[ -(x^2+y^2) \le (x^2+y^2) \sin \left( \frac{1}{xy}\right) \le (x^2+y^2) \]

and \( \displaystyle \lim_{x,y \to 0}(x^2+y^2)=0\).You should consider either inproving your hand writing (probably essential if you use handwritting for your exams) or the LaTeX type setting system.

CB

 

[CaptainBlack]

The second is also correct, but there is a flaw in your method, you assume that:

\[ \lim_{x \to 0}_{ y \to 2} \frac{\sin(xy)}{x}=\lim_{x \to 0} \left[ \lim_{y \to 2} \frac{\sin(xy)}{x}\right] \]

which the next example shows you cannot (in general without further justification) do.

Here you can put \(z=xy\), and the limit then becomes:

\[ \lim_{z \to 0}_{ y \to 2}\; y\; \frac{\sin(z)}{z} \]

and as both of the limits: \(\displaystyle \lim_{z \to 0}_{ y \to 2} y=2\) and \( \displaystyle \lim_{z \to 0}_{ y \to 2} \frac{\sin(z)}{z}=1\) we have:\[ \lim_{z \to 0}_{ y \to 2} y \frac{\sin(z)}{z}=\left[ \lim_{z \to 0}_{ y \to 2}\; y \right]\;\left[ \lim_{z \to 0}_{ y \to 2}\; \frac{\sin(z)}{z}\right] =2 \times 1=2\]

CB

 

[CaptainBlack]

Both the third and fourth are correct, but you need more explanation. What you are showing is that taking the limits first wrt x and then wrt y and the other way around give different results, which implies that the limits do not exist (since if they did exist any path in the (x,y) plane to the limit will give the same result).

CB

 

[CaptainBlack]

The last two are correct, but I see no justification for the argument for the first, and I don't understand what you are doing in the second going from the first line to the second.

CB

 

[Fabio010]

Thanks a lot for correct the limits.

First of all sorry for the writing, it is so ugly because i wrote it in paint.
Next time i am going to try to use LATEX.

The last two, in first one i just used the notable limit of e^k
the second one is wrong... x^2/(x^2+y^2) is not equal to 1/(x+y^2) -_-

When all paths get as result the same limit, we can use the polar coordinates too prove that the limit is limit of all paths, right?

 

[CaptainBlack]

Thanks a lot for correct the limits.

First of all sorry for the writing, it is so ugly because i wrote it in paint.
Next time i am going to try to use LATEX.

The last two, in first one i just used the notable limit of e^k
the second one is wrong... x^2/(x^2+y^2) is not equal to 1/(x+y^2) -_-

You are assuming that the limit exists and so the limits may be taken in any order.

When all paths get as result the same limit, we can use the polar coordinates too prove that the limit is limit of all paths, right?

Still does not look right

CB