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#### GreenGoblin

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- Feb 22, 2012

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- Feb 22, 2012

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- Jan 11, 2013

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You have to show that **Any** sequence $X_n \in \mathbb{R} \times \mathbb{R}$ s.t that $X_n \to (0,0)$, $f(X_n) \to L$ and L = 0.

As for directional derivatives, you have to show that $lim_{h \to 0} \frac{f(x+hv)-f(x)}{h}$ exists along any 2 dimensional vector v at point (0,0) = x

so $\displaystyle \lim_{h \to 0} \frac{f(hv)}{h}$ has a limit for a non zero vector v.

As for directional derivatives, you have to show that $lim_{h \to 0} \frac{f(x+hv)-f(x)}{h}$ exists along any 2 dimensional vector v at point (0,0) = x

so $\displaystyle \lim_{h \to 0} \frac{f(hv)}{h}$ has a limit for a non zero vector v.

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- #3

- Feb 22, 2012

- 68

huh?i posted a pica.of the problem. that cant be the answer because this is not x, this is (x,y) so the answer has to be as much to do with y as x.

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- Jan 11, 2013

- 68

i used x in the sense that $x \in \mathbb{R} \times \mathbb{R}$, so x = ($x_1,x_2$)huh?i posted a pica. of the problem. that cant be the answer because this is not x, this is (x,y) so the answer has to be as much to do with y as x.

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- Jan 29, 2012

- 1,151

The "difficult" part of showing continuity is showing that the limit **is** 0. You cannot, for example, just note that if x= 0 the fraction is 0 for any y and so approaching (0, 0) along the y-axis, the limit is 0. You have to show that if you are "close" to (0, 0), in any direction, the value of the function is "close" to 0. I think the simplest way to handle this is to change to polar coordinates. That way, the single variable, r, measures the distance from (0, 0). [tex]x= r cos(\theta)[/tex], [tex]y= r sin(\theta)[/tex], and [tex]x^2+ y^2= r^2[/tex] so this fraction becomes

[tex]\frac{r^3cos(\theta)sin^2(\theta)}{r^2}= r cos(\theta)sin^2(\theta)[/tex] which clearly is close to 0 for r close to 0.

No, "directional derivatives" are not, strictly speaking, "partial derivatives". The "directional derivative" is the rate of change as you move in the given direction, not necessarily just in the direction of the x and y axis. The directional derivative of f(x,y) in the direction of**unit** vector, v, is [tex]\nabla f\cdot v[/tex]. Equivalently, the directional derivative of f(x, y) in direction [tex]\theta[/tex] to the positive x-axis is [tex]cos(\theta)\frac{\partial f}{\partial x}+ sin(\theta)\frac{\partial f}{\partial y}[/tex]. To determine whether or not the directional derivatives exist at (0, 0), replace y with mx and take the ordinary derivative of the resulting function of x. That works for all "directions" except along the y-axis where, of course, the directional derivative is just the partial derivative with respect to y.

As for showing that the function is not "differentiable" at (0, 0), what definition of "differentiable" are you using for a function of two variables?

(Note that the function, f(x,y), defined to be 0 if xy= 0, 1 if xy is not 0, has both partial derivatives at (0, 0) but is not even continuous there.)

[tex]\frac{r^3cos(\theta)sin^2(\theta)}{r^2}= r cos(\theta)sin^2(\theta)[/tex] which clearly is close to 0 for r close to 0.

No, "directional derivatives" are not, strictly speaking, "partial derivatives". The "directional derivative" is the rate of change as you move in the given direction, not necessarily just in the direction of the x and y axis. The directional derivative of f(x,y) in the direction of

As for showing that the function is not "differentiable" at (0, 0), what definition of "differentiable" are you using for a function of two variables?

(Note that the function, f(x,y), defined to be 0 if xy= 0, 1 if xy is not 0, has both partial derivatives at (0, 0) but is not even continuous there.)

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- Feb 22, 2012

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- Jan 11, 2013

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$$f(x,y) = \left\{

\begin{array}{lr}

(x+y+1,\frac{x^2-y^2}{x^2+y^2}) & : (x,y) \not = (0,0)\\

(1,0) & : (x,y) = (0,0)

\end{array}

\right.

$$

Then f(0,0) = (1,0)

But Take the sequence $X_j$ = ($\frac{1}{j}$,0) where $j \in \mathbb{N}$

then $X_j \to (0, 0)$ as $j \to \infty$

but $lim_{j \to \infty}f(0,\frac{1}{j}) = (\frac{1}{j} + 1, 1) \to (1,1)$

but (1,1) $\not = f(0,0) = (1,0)$.

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- #8

- Feb 22, 2012

- 68

what has that got to do with anything? i already dont like you.

$$f(x,y) = \left\{

\begin{array}{lr}

(x+y+1,\frac{x^2-y^2}{x^2+y^2}) & : (x,y) \not = (0,0)\\

(1,0) & : (x,y) = (0,0)

\end{array}

\right.

$$

Then f(0,0) = (1,0)

But Take the sequence $X_j$ = ($\frac{1}{j}$,0) where $j \in \mathbb{N}$

then $X_j \to (0, 0)$ as $j \to \infty$

but $lim_{j \to \infty}f(0,\frac{1}{j}) = (\frac{1}{j} + 1, 1) \to (1,1)$

but (1,1) $\not = f(0,0) = (1,0)$.

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- #9

- Jan 26, 2012

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Those were the magic words. Thread closed.what has that got to do with anything? i already dont like you.

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