find lim, (x,y)->(0,0), of (x^4+y^4)/(x^2+y^2) w/ polar coor

[jackieu]

I thought I was good at math. Turns out I'm an idiot. So, help on any or all questions would be appreciated. I thought I knew what I was doing, but these threw me for a loop. So, I ask you, dear stranger, teach me to swim. I keep sinking!

1. Use polar coordinates to find the following limits. Clearly show steps and explain your reasoning as necessary.

a.) lim (x,y) -> (0,0) (x^4+y^4)/(x^2+y^2)

I simplified this to get an answer of zero. Is this correct. Btw, relevant equations x=rcos(theta) y=rsin(theta) r^2=X^2+y^2

b.) lim (x,y) -> (0,0) (x^2+y^2+1) ^ (1/(x^2+y^2))

Of course, this is equal to lim r -> 0 (r+1) ^ (1/r). I thought this would head to 1, but the answer is e. I have no idea how!

c.) lim (x,y) -> (0,0) (X^2 + Y^2) ln (x^2+y^2)

Again, simplified to lim r->0 r^2 ln (r^2)
I have no idea how to do this. I thought since r is going to zero, the whole thing would. But, that doesn't seem right for some reason.


2.) IF z^3-xz-y=0 prove that (partial^2 z)/(partial x partial y) = - (3z^2 +x)/(3z^2 -x)^3


Alright, for this I changed the first part (the partial nonsense) to partial with respect to x of ( - partial f/ partial y)/(partial f/partial z)

where f = z^3 -xz-y
partial f/partial y = -1
partial f/partial z = 3z^2-x

which means the whole partial nonsense equals
(3z^2 -x)^-2
but when i work it all out, i get a negative on one side but not the other.

What gives?

Thank you in advance! =)

 

[mmm4444bot]

Classic error in limit reasoning ...

... this is equal to lim r -> 0 [(r+1) ^ (1/r)]. I thought this would head to 1, but the answer is e ...

Hello Jackieu:

I'm not qualified to discuss multi-variable calculus right now, but I have a comment to make on the limit quoted above.

You thought its value should be 1.

The following is a very common misconception when trying to analyze what's happening to the value of a function in some limit. (I know because I've been revisiting limits in one variable over the past two weeks, trying to refresh my memory, and I've been commiting the same type of logic errors myself.)

I'm going to assume that you did something like the following.

First, you looked at 1/r as r approaches zero from the right. (In other words, r is always a tiny POSITIVE value.)

So, you reasoned that 1/r ? ? as r ? 0+ .

That's good reasoning.

But then, you may have concluded that (r + 1) BECOMES 1 when r ? 0+ .

This type of direct substitution (0 + 1) is often fatal when working with expressions not covered by the direct-substitution principle.

We need to remember that, in the limit, r never becomes zero! Therefore, we are not even interested in what happens when r is zero.

Here is how I look at it.

The SUM of (an infinitely small positive value + 1) is being multiplied by itself an infinite number of times ...

In other words, this one-sided limit equals some HUGE power of some base that is an extremely TINY FRACTION more than 1.

Now it becomes obvious to me that there's no way that this power can equal 1. I really have no way of recognizing in my head what this power equals. I need to use mathematical notation, definitions, and paper and pencil to reason it out.

There are clearly some limits that can be evaluated by direct substitution. Those involving polynomial expressions are one type. Your limit is not a polynomial because the exponent 1/r is not an integer.

Remember the Limit Laws; better yet, try to understand the implications of what they're telling you.

I wish you good fortune.

Cheers,

~ Mark

 

[galactus]

Re: I thought I was good at math. Turns out I'm an idiot. Please

\(\displaystyle \lim_{(x,y)\to (0,0)}\left(x^{2}+y^{2}+1\right)^{\frac{1}{x^{2}+y^{2}}}\)

If we make the substitution \(\displaystyle r^{2}=x^{2}+y^{2}\), we get:

\(\displaystyle \large{\displaystyle}\lim_{r\to 0}\left(r^{2}+1\right)^{\frac{1}{r^{2}}}\)

Now, this is a famous limit. Do you, by chance, recognize it?. This is the 'e' limit, \(\displaystyle \lim_{x\to 0}\left(1+h\right)^{\frac{1}{h}}\)

Only in this case, we have r^2 instead of h.

We can prove this by building on the differentiability of ln(x), more specifically, at x=1.

\(\displaystyle \frac{d}{dx}[ln(x)]=1\)

Using the definition of a derivative:

\(\displaystyle 1=\lim_{h\to 0}\frac{ln(1+h)-ln(1)}{h}=\lim_{h\to 0}\frac{ln(1+h)}{h}=\lim_{h\to 0}ln(1+h)^{\frac{1}{h}}\)

Then , it follows that:

\(\displaystyle \large{\displaystyle}e=e^{\lim_{h\to 0}ln(1+h)^{\frac{1}{h}}}\)

From the continuity of e:

\(\displaystyle e=\lim_{h\to 0}e^{ln(1+h)^{\frac{1}{h}}}=\lim_{h\to 0}(1+h)^{\frac{1}{h}}\)

In this case, \(\displaystyle h=r^{2}\) and we're done.

Pretty cool, huh?. Follow all that?.