# Proving the limit of a multivariable function

• Oct 10th 2012, 01:42 AM
mrmackey
Proving the limit of a multivariable function
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.

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

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

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

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

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).

• Oct 10th 2012, 08:04 AM
hollywood
Re: Proving the limit of a multivariable function
You want $\displaystyle |\ln{x}|<\epsilon$, so:

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

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

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

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

- Hollywood
• Oct 10th 2012, 09:56 AM
mrmackey
Re: Proving the limit of a multivariable function
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 $\displaystyle \delta$ 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 $\displaystyle \left|\ln[x]\right|$ is unwise since that means the $\displaystyle 0 < \sqrt{(x-1)^2+y^2}<\delta$ becomes unused in the proof.

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

Maybe I should restart the process all together? Any advice on how to approach this?
• Oct 11th 2012, 12:00 AM
hollywood
Re: Proving the limit of a multivariable function
To show that the limit is zero:

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

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

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

I think that's a sound argument.

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

- Hollywood
• Oct 11th 2012, 02:39 AM
mrmackey
Re: Proving the limit of a multivariable function
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!