Proving the limit of a multivariable function

**mrmackey** · October 10th 2012, 01:42 AM

Hi there! I'm newly getting introduced to multivariable functions and their limits. I had an assignment of 6 or 7 of them, and I got the majority except for the last. I've tried a few paths of approach and have concluded that the limit is indeed equal to 0. I just cannot prove it.

$\lim_{\{x,y\}\to \{1,0\}} \, \frac{(x-1)^2 \ln (x)}{(x-1)^2+y^2}$

If $0<\left\|(x-1,y)\right\| = \sqrt{(x-1)^2+y^2}<\delta$ , FIXED AFTER EDIT.

then $\left|g(x,y) - 0\right|<\epsilon$
$\left|g(x,y)\right| = \left|\frac{(x-1)^2 \ln (x)}{(x-1)^2+y^2}\right|= \newline \left|\frac{(x-1)^2}{(x-1)^2+y^2}\right|\left|\ln[x]\right| \leq \left|\frac{(x-1)^2}{(x-1)^2}\right| \left|\ln[x]\right| = \left| \ln[x] \right|$

What would be a good choice for $\delta$ so that $\ln[x] < \epsilon$ ?

But now that I look at it, it seems like I've done the process a tedious way as I'm not employing the IF statement (simply divided it out).

Any advice? Thanks in advance.

**hollywood** · October 10th 2012, 08:04 AM

You want $|\ln{x}|<\epsilon$ , so:

$-\epsilon < \ln{x}<\epsilon$

$e^{-\epsilon}<x<e^{\epsilon}$

$e^{-\epsilon}-1<x-1<e^{\epsilon}-1$ .

So choose $\delta=\min(1-e^{-\epsilon},e^{\epsilon}-1)$ .

- Hollywood

**mrmackey** · October 10th 2012, 09:56 AM

Hi Hollywood,

Thanks for replying, By the way, I fixed the IF statement in my OP because of a small mistake.

Secondly, I'm not sure how this choice of $\delta$ helps. In fact, I'm not sure if any choice of it will help. As I mentioned, I think my process to end up at just $\left|\ln[x]\right|$ is unwise since that means the $0 < \sqrt{(x-1)^2+y^2}<\delta$ becomes unused in the proof.

Plus, the $\sqrt{(x-1)^2+y^2}$ is actually $\frac{1}{\sqrt{(x-1)^2+y^2}}$ in the proof. Taking the reciprocal of the $\sqrt{(x-1)^2+y^2}$ would only give a $\geq \frac{1}{\delta}$ and not a $\leq \frac{1}{\delta}$ part. And we don't want to bound from below with $\epsilon$ , we want to do so from above.

Maybe I should restart the process all together? Any advice on how to approach this?

**hollywood** · October 11th 2012, 12:00 AM

To show that the limit is zero:

Given $\epsilon>0$ , set $\delta=\min(1-e^{-\epsilon},e^{\epsilon}-1)$ .

Then whenever $\sqrt{(x-1)^2+y^2}<\delta$ , we have $|x-1|<\delta$ , so
$e^{-\epsilon}-1<x-1<e^{\epsilon}-1$
$e^{-\epsilon}<x<e^{\epsilon}$
$-\epsilon < \ln{x}<\epsilon$
$|\ln{x}|<\epsilon$

And as a result,
$\left|g(x,y) - 0\right|=$
$\left|g(x,y)\right| = \left|\frac{(x-1)^2 \ln (x)}{(x-1)^2+y^2}\right|=$
$\left|\frac{(x-1)^2}{(x-1)^2+y^2}\right|\left|\ln[x]\right| \leq$
$\left|\frac{(x-1)^2}{(x-1)^2}\right| \left|\ln[x]\right| =$
$\left| \ln[x] \right| < \epsilon$ .

I think that's a sound argument.

The $y^2$ in the denominator always pushes the function closer to zero, so that's why it's not a significant part of the proof.

- Hollywood

**mrmackey** · October 11th 2012, 02:39 AM

That works perfectly Hollywood! Thanks a bunch for taking the time to help me and for explaining the process (so that I learn for similar problems in the future).

If someone can mark this as solved, it would be great. Again, thank you!

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