# Math Help - Limit and differentiability

1. ## Limit and differentiability

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?

2. 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

3. 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!

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

5. 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

6. 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?

7. 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?
.

8. 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}}
$

9. 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?

10. 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?)

11. 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}}
$
.