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I'm learning multivariable calculus, specifically multivariable continuity and differentiability, and need help with the following exercise:

Let

$$f(x,y) = \begin{cases} \frac{2x^3 + 3y^3}{x^2 + y^2}, & (x,y) \neq (0, 0) \\ 0, & (x,y) = (0, 0). \end{cases}$$

$(a)$ Examine if $f$ is continuous at $(0, 0)$.

$(b)$ Examine if $f$ is differentiable at $(0, 0)$.

Since I'm having difficulties for $(b)$, I'm going to show my work for $(a)$.

$(a)$ We note that

$$0 \leq \left| \frac{2x^3 + 3y^3}{x^2 + y^2} \right| \leq \frac{2|x|^3}{x^2 + y^2} + \frac{3|y|^3}{x^2 + y^2} \leq \frac{2|x|^3}{x^2} + \frac{3|y|^3}{y^2} = 2|x| + 3|y|$$

and $$\lim_{(x,y) \to (0, 0)} 2|x| + 3|y| = 0.$$

Therefore, by the Squeeze theorem, we conclude that

$$\lim_{(x, y) \to (0, 0)} f(x, y) = f(0, 0)$$

that is

$$\lim_{(x, y) \to (0, 0)} \frac{2x^3 + 3y^3}{x^2 + y^2} = 0$$

so $f$ is continuous at $(0, 0)$.


Is my work correct for $(a)$? How do I examine differentiability of $f$ at $(0, 0)$? More importantly: what is the method to examine differentiability in this case? What do I need to show?

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  • $\begingroup$ Since the removable discontinuity has been removed, you can simply cancel out the expressions without any harm. $\endgroup$ – Jacob Wakem Aug 14 '16 at 17:10
  • $\begingroup$ Why are we allowed to squeeze the absolute value of our function rather than the function itself? $\endgroup$ – EgoKilla Sep 7 '18 at 11:22
  • $\begingroup$ Never mind, it’s because the absolute value is always larger. $\endgroup$ – EgoKilla Sep 7 '18 at 11:32
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Your part (a) is perfect!

For differentiability a necessary condition is that a directional derivative should exist and depend linearly on the direction: Given $(h,k)$ we calculate $$ Df_{(0,0)} (h,k) = \frac{d}{dt}_{|t=0} f(th,tk) = \frac{d}{dt}_{|t=0} t f(h,k) = f(h,k) = \frac{2h^3+3k^3}{h^2+k^2}$$ The directional derivative exists in any direction but this derivative is not linear as a function of $h$, $k$ (check why!). So the function $f$ is not differentiable at zero.

Note that a sufficient (though not necessary) condition is that partial derivatives exist and are continuous (not the case here).

To complete the story: Suppose (hypothetically) that you had found a directional derivative which is linear in $(h,k)$, thus $Df_{(0,0)} (h,k) =ah+bk$ for some $a$ and $b$. Then there are still some hurdles, as you would have to show that the error term is small (called little 'o') compared to the length of $(h,k)$. More precisely, that $$ \lim_{h^2+k^2\rightarrow 0} \frac{f(h,k)-ah-bk}{\sqrt{h^2+k^2}} = 0$$ Luckily enough you have not arrived at this...

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Your proof about continuity is correct. In order to check differentiability fistly find partial derivatives so that

$$ { f }_{ x }^{ \prime }\left( 0,0 \right) =\lim _{ x\rightarrow 0 }{ \frac { f\left( x,0 \right) -f\left( 0,0 \right) }{ x } } =\lim _{ x\rightarrow 0 }{ \frac { \frac { 2x^{ 3 } }{ x^{ 2 } } }{ x } } =2\\ { f }_{ y }^{ \prime }\left( 0,0 \right) =\lim _{ y\rightarrow 0 }{ \frac { f\left( y,0 \right) -f\left( 0,0 \right) }{ x } } =\lim _{ x\rightarrow 0 }{ \frac { \frac { 3y^{ 3 } }{ y^{ 2 } } }{ y } } =3$$ differentiability should be shown as

$$f\left( x,y \right) -f\left( 0,0 \right) =\frac { 2x^{ 3 }+3y^{ 3 } }{ x^{ 2 }+y^{ 2 } } =2x+3y+\left( \frac { 2x^{ 3 }+3y^{ 3 } }{ x^{ 2 }+y^{ 2 } } -2x-3y \right) =\\={ f }_{ x }^{ \prime }\left( 0,0 \right) x+{ f }_{ y }^{ \prime }\left( 0,0 \right) y+\alpha \left( x,y \right) \sqrt { { x }^{ 2 }+{ y }^{ 2 } } $$ where $$\alpha \left( x,y \right) =\frac { -3{ x }^{ 2 }y-2{ y }^{ 2 }x }{ \left( x^{ 2 }+y^{ 2 } \right) \sqrt { x^{ 2 }+y^{ 2 } } } $$

We should check whether is this infinity small? Let $x=\frac { 1 }{ n } ,y=\frac { 1 }{ n } $ so that

$$\\ \alpha \left( \frac { 1 }{ n } ,\frac { 1 }{ n } \right) =\frac { -\frac { 5 }{ { n }^{ 3 } } }{ \frac { 2\sqrt { 2 } }{ { n }^{ 3 } } } =-\frac { 5 }{ 2\sqrt { 2 } } \neq 0,n\rightarrow \infty ,n\in \ \mathbb{N} $$

it shows $\alpha \left( x,y \right) $ is not infinitly small (when $x\rightarrow 0,y\rightarrow 0$) in other words

$$\alpha \left( x,y \right) \sqrt { { x }^{ 2 }+{ y }^{ 2 } } \neq o\left( \sqrt { { x }^{ 2 }+{ y }^{ 2 } } \right) $$

so it not differentiable

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Differentiability means there is a tangent plane. Let us assume for the moment that there is a tangent plane at $(0,0)$.

First we constrain ourselves to the $x$-axis, i.e. take $y=0$. Then the function is $\dfrac{2x^3}{x^2} = 2x$, and we have a tangent line whose slope is $2$.

Then we constrain ourselves to the $y$-axis, i.e. take $x=0$. Then the function is $\dfrac{3y^3}{y^2} = 3y$, and we have a tangent line whose slope is $3$.

There is only one plane that contains those two lines, so if there were a tangent plane, that would be it.

Next we constrain ourselves to the diagonal line $y=x$. The line in our putative tangent plane lying above the line $y=x$ in the $(x,y)$-plane is $\{(x,x,2x+3x) : x\in\mathbb R\}$. So the directional derivative in that direction, of the function whose graph is that plane, is $5$. So we ask whether that is the directional derivative of the original function in that direction at $(0,0)$. If not, then that plane fails to be a tangent plane.

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Your work for $(a)$ looks good. For $(b)$, you will need to show there exists a linear map $L: \mathbb{R}^2 \rightarrow \mathbb{R}$ such that $$f(h)=Lh+\text{o}(h).$$ Luckily enough there is a theorem that says this is equivalent to saying that the partial derivatives of $f$ exist and are continuous in $(0,0)$. First for existence you calculate the limit with the definition of partial derivatives, that is you calculate $$\frac{\partial f}{\partial x} |_{(0,0)} = \lim_{t \to 0} \frac{f(t,0)}{t}=2$$ and $$\frac{\partial f}{\partial y} |_{(0,0)}=\lim_{t \to 0} \frac{f(0,t)}{t}=3.$$ Also, in $\mathbb{R}^2 - \{(0,0)\}$ you already know what the partials are, since you know $f$ on $\mathbb{R}^2-\{(0,0)\}$. The (partial) derivative of a function at a point is a local property of the function, and since $\mathbb{R}^2-\{(0,0)\}$ is open, for calculating the partial derivatives there we can just use the formula for $f$ on $\mathbb{R}^2-\{(0,0)\}$ and calculate them with the usual rules. Then we know $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y} $ on the whole of $\mathbb{R}^2$, so we can check for their continuity just like you did for $f$. Hope this helps.

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