Limit and differentiability

**demode** · March 13th 2010, 01:14 AM

Let

$f(x, y) = \sqrt[3]{xy^2 + x^2 y}$ .

By using the definition of differentiability show that f is not differentiable at (0,0).

Definition is:

$ \lim_{(\Delta x,\Delta y) \to (0,0)} \frac{\Delta f - f_x(x_0,y_0)\Delta x -f_y(x_0,y_0)\Delta y}{\sqrt{(\Delta x)^2 + (\Delta y)^2}}=0 $

I already know that $f_x (0,0)=f_y(0,0)=0$ . Also the increment is $\Delta f = (0+\Delta x, 0+ \Delta y) = \sqrt[3]{\Delta x(y)^2 + (\Delta x)^2 \Delta y}$ . So

$\lim_{(\Delta x,\Delta y) \to (0,0)} \frac{\sqrt[3]{\Delta x(y)^2 + (\Delta x)^2 \Delta y}}{\sqrt{(\Delta x)^2+(\Delta y)^2}}$

Now, I need to evaluate this to show that it doesn't =0. I'm stuck here. I don't know if I should use the conjugate OR use polar coordinates to evaluate this limit? Can someone help?

**tonio** · March 13th 2010, 02:51 AM

Originally Posted by demode

Let

$f(x, y) = \sqrt[3]{xy^2 + x^2 y}$ .

By using the definition of differentiability show that f is not differentiable at (0,0).

Definition is:

$ \lim_{(\Delta x,\Delta y) \to (0,0)} \frac{\Delta f - f_x(x_0,y_0)\Delta x -f_y(x_0,y_0)\Delta y}{\sqrt{(\Delta x)^2 + (\Delta y)^2}}=0 $

I already know that $f_x (0,0)=f_y(0,0)=0$ . Also the increment is $\Delta f = (0+\Delta x, 0+ \Delta y) = \sqrt[3]{\Delta x(y)^2 + (\Delta x)^2 \Delta y}$ . So

$\lim_{(\Delta x,\Delta y) \to (0,0)} \frac{\sqrt[3]{\Delta x(y)^2 + (\Delta x)^2 \Delta y}}{\sqrt{(\Delta x)^2+(\Delta y)^2}}$

Now, I need to evaluate this to show that it doesn't =0. I'm stuck here. I don't know if I should use the conjugate OR use polar coordinates to evaluate this limit? Can someone help?

Put $\Delta_y=m\Delta_x$ and you'll get the limit depends on $m$ ... and correct the expression for $f(\Delta_x,\Delta_y)$ , it should be $\sqrt[3]{\Delta_x\Delta^2_y+\Delta^2_x\Delta_y}$ .

Tonio

**demode** · March 13th 2010, 01:00 PM

Originally Posted by tonio

Put $\Delta_y=m\Delta_x$ and you'll get the limit depends on $m$ ... and correct the expression for $f(\Delta_x,\Delta_y)$ , it should be $\sqrt[3]{\Delta_x\Delta^2_y+\Delta^2_x\Delta_y}$ .

Tonio

So

$\lim_{(\Delta x,\Delta y) \to (0,0)} \frac{\sqrt[3]{\Delta x(\Delta y)^2 + (\Delta x)^2 \Delta y}}{\sqrt{(\Delta x)^2+(\Delta y)^2}}$

becomes

$\lim_{(\Delta x,m \Delta x) \to (0,0)} \frac{\sqrt[3]{\Delta x(m \Delta x)^2 + (\Delta x)^2 m \Delta x}}{\sqrt{(\Delta x)^2+(m \Delta x)^2}}$

Again I don't know how to evaluate this!

**demode** · March 13th 2010, 09:20 PM

Would it not have been easier to convert to polar coordinates, so that the limit will be in terms of "r"?

**tonio** · March 14th 2010, 03:08 AM

Originally Posted by demode

So

$\lim_{(\Delta x,\Delta y) \to (0,0)} \frac{\sqrt[3]{\Delta x(\Delta y)^2 + (\Delta x)^2 \Delta y}}{\sqrt{(\Delta x)^2+(\Delta y)^2}}$

becomes

$\lim_{(\Delta x,m \Delta x) \to (0,0)} \frac{\sqrt[3]{\Delta x(m \Delta x)^2 + (\Delta x)^2 m \Delta x}}{\sqrt{(\Delta x)^2+(m \Delta x)^2}}$

Again I don't know how to evaluate this!

You need some high school algebra here to simplify that stuff...for example, $\sqrt[3]{\Delta x(m \Delta x)^2 + (\Delta x)^2 m \Delta x}=\Delta_x\sqrt[3]{m^2+m}$ , etc.

Tonio

**demode** · March 14th 2010, 09:07 PM

Originally Posted by tonio

You need some high school algebra here to simplify that stuff...for example, $\sqrt[3]{\Delta x(m \Delta x)^2 + (\Delta x)^2 m \Delta x}=\Delta_x\sqrt[3]{m^2+m}$ , etc.

Tonio

yes

$\lim_{m \to (0,0)} \frac{\Delta_x\sqrt[3]{m^2+m}}{\Delta x \sqrt{1+m^2}}= \lim_{m \to (0,0)} \frac{\sqrt[3]{m^2+m}}{\sqrt{1+m^2}}$

Since this is a limit involving radicals, I need to rationalize the denominator:

$\lim_{m \to (0,0)} \frac{\sqrt[3]{m^2+m} . \sqrt{1+m^2}}{\sqrt{1+m^2}.\sqrt{1+m^2}}$

This is where I'm stuck. I know the denominator will be 1+m, but how do we simplify the numerator?

**tonio** · March 15th 2010, 04:27 AM

Originally Posted by demode

yes

$\lim_{m \to (0,0)} \frac{\Delta_x\sqrt[3]{m^2+m}}{\Delta x \sqrt{1+m^2}}= \lim_{m \to (0,0)} \frac{\sqrt[3]{m^2+m}}{\sqrt{1+m^2}}$

Why do you take the limit when m --> (0,0) ?? This doesn't even make sense! The limit is when $(\Delta_x,\Delta_y)\rightarrow (0,0)$ ...!

Tonio

Since this is a limit involving radicals, I need to rationalize the denominator:

$\lim_{m \to (0,0)} \frac{\sqrt[3]{m^2+m} . \sqrt{1+m^2}}{\sqrt{1+m^2}.\sqrt{1+m^2}}$

This is where I'm stuck. I know the denominator will be 1+m, but how do we simplify the numerator?

.

**demode** · March 15th 2010, 09:12 PM

Yes but then how are we going to evaluate this limit?

I mean if

$ \lim_{(\Delta_x,\Delta_y)\rightarrow (0,0)} \frac{\Delta_x\sqrt[3]{m^2+m}}{\Delta x \sqrt{1+m^2}}= \lim_{(\Delta_x,\Delta_y)\rightarrow (0,0)} \frac{\sqrt[3]{m^2+m}}{\sqrt{1+m^2}} $

And in the expression on the RHS there are no more $\Delta_x, \Delta_y$ terms left, and we don't know the value of m! So, what do I need to do?

Of course, I could rationalize the denominator but I don't know how to simplify the numerator in the following:

$ \lim_{(\Delta_x,\Delta_y)\rightarrow (0,0)} \frac{\sqrt[3]{m^2+m} . \sqrt{1+m^2}}{\sqrt{1+m^2}.\sqrt{1+m^2}} $

**demode** · March 17th 2010, 03:41 AM

I even tried writing out the powers as a fraction, that is, to let the square roots be ^(1/2) and such. But still I end up with m's and you didn't mention what the value of m was! How am I supposed to show that the limit doesn't equal zero?

**Defunkt** · March 17th 2010, 04:12 AM

You actually do know the value of m - you can set it to be anything you like!
That is why the function is not differentiable around (0,0) - because the limit depends on m (ie. you can get two different values for the limit for two different values of m) and as a result, it does not exist (do you understand why?)

**tonio** · March 17th 2010, 04:22 AM

Originally Posted by demode

Yes but then how are we going to evaluate this limit?

I mean if

$ \lim_{(\Delta_x,\Delta_y)\rightarrow (0,0)} \frac{\Delta_x\sqrt[3]{m^2+m}}{\Delta x \sqrt{1+m^2}}= \lim_{(\Delta_x,\Delta_y)\rightarrow (0,0)} \frac{\sqrt[3]{m^2+m}}{\sqrt{1+m^2}} $

And in the expression on the RHS there are no more $\Delta_x, \Delta_y$ terms left, and we don't know the value of m! So, what do I need to do?

You need to really concentrate a little more: if you put $\Delta_y=m\Delta_x$ , then $(\Delta_x,\Delta_y)\rightarrow (0,0) \Longleftrightarrow \Delta_x\rightarrow 0$ , so THE LIMIT IS $\frac{\sqrt[3]{m^2+m}}{\sqrt{1+m^2}}$ , and then

the limit depends on $m$ (unless you could prove the limit is the same no matter what m we choose...which of course you won't be able to prove since it isn't true

).

Tonio

Of course, I could rationalize the denominator but I don't know how to simplify the numerator in the following:

$ \lim_{(\Delta_x,\Delta_y)\rightarrow (0,0)} \frac{\sqrt[3]{m^2+m} . \sqrt{1+m^2}}{\sqrt{1+m^2}.\sqrt{1+m^2}} $

.

Thread: Limit and differentiability

LinkBack

Thread Tools

Display

Limit and differentiability

Similar Math Help Forum Discussions

[SOLVED] Differentiability.

Help w/ Differentiability

Differentiability

Limit / Differentiability of complex functions

Differentiability

Search Tags