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I am working on a practice question, and I am not sure if what I have done would be considered, 'complete justification'. I would greatly appreciate some feedback or helpful advice on how it could be better etc. The question is here:

Let $f: \mathbb {R}^2 \to \mathbb{R} $ be a function defined by:

$$ \ f(x,y) = \begin{cases} \frac{\sin(x^2 + y^2)}{x^2 + y^2} & \text{if } (x,y) \ne (0,0) \\ 1 & \text{if } (x,y) = (0,0) \end{cases} $$

Is $f$ diff'ble at $(x,y) = (0,0)$?

Here is what I have:

By definition, $f(x,y)$ is diff'ble at $(0,0)$ if

$$ \lim_{(x,y) \to (0,0)} f(x,y)= \frac{f(x,y) - [f(0,0) + f_{x}(0,0)(x-0) + f_{y}(0,0)(y-0)]}{\sqrt{x^2 + y^2}} = 0\tag{*}$$

Since $f(x,y)$ is piecewise, $f_{x}(0,0)$ and $f_{y}(0,0)$ is derived from 1st principles:

So, $$\begin{align}f_{x}(0,0) &= \lim_{h \to 0} \frac{\frac{\sin((0+h)^2 + (0)^2)}{(0+h)^2 + (0)^2} - f_{x}(0,0)}{h} \\ &= \lim_{h \to 0} \frac{\frac{\sin(h^2)}{h^2} - 1}{h}\\ &= \lim_{h \to 0} \frac{\sin(h^2) - h^2}{h^3} \\ &= \lim_{h \to 0} \frac{2h \cos(h^2) - 2h}{3h^2}\end{align} $$ by L'Hopital's rule, apply this twice more I can see that the limit is $0$. Similarly, $f_y(0,0)$ is derived the same way.

Then from $(*)$, I have:

$$f_{x}(0,0) = \lim_{h \to 0} \frac{\frac{\sin(x^2 + y^2)}{x^2 + y^2} - 1}{\sqrt{x^2 + y^2}} $$

From here I haven't had luck trying to get this to $0$. So instead, I try to make it easier.

Since $f_x(0,0)$ and $f_y(0,0)$ exist we must evaluate whether $f_x(x,y)$ is continuous at $(x,y)=(0,0)$ i.e. if

$$\lim_{(x,y) \to (0,0)} \frac{\sin(x^2 + y^2)}{x^2 + y^2} = 1 $$ then $f(x,y)$ is differentiable at $(0,0)$. So letting $u = x^2 + y^2$,

$$\lim_{(x,y) \to (0,0)} \frac{\sin(x^2 + y^2)}{x^2 + y^2} = \lim_{ (x,y) \to (0,0)} \frac{\sin(u)}{u} = 1 $$ by L'Hopital once more.

Therefore, $f$ is differentiable at (0,0).

I feel like I am perhaps not completely justifying this, because I take a short cut - but is it valid? Should I perhaps use a epsilon-delta proof? Or is there a way to work with that tricky limit I have in $(*)$?

Many thanks for the help in advance!

JackReacher
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  • If you want to avoid messy calculations, you can show that for any function $f$ such that $|f(x)| \le |x|^2$, $f$ is differentiable at $0$ (and $f'(x)$ is the zero transformation) where $x\in \Bbb R^n$. – user10444 Mar 20 '14 at 23:59

1 Answers1

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You've already found

$$\lim \limits_{(x,y)\to (0,0)}\left(\dfrac{f(x,y) - [f(0,0) + f_{x}(0,0)(x-0) + f_{y}(0,0)(y-0)]}{\sqrt{x^2 + y^2}}\right)=\lim \limits_{(x,y)\to (0,0)}\left( \dfrac{\frac{\sin(x^2 + y^2)}{x^2 + y^2} - 1}{\sqrt{x^2 + y^2}}\right).$$

To evaluate $\lim \limits_{(x,y)\to (0,0)}\left( \dfrac{\frac{\sin(x^2 + y^2)}{x^2 + y^2} - 1}{\sqrt{x^2 + y^2}}\right)$ simplify the denominator and change the variables with $x=\rho \cos(\theta)$ and $y=\rho\sin(\theta)$ to get $$\begin{align}\lim \limits_{(x,y)\to (0,0)}\left( \dfrac{\frac{\sin(x^2 + y^2)}{x^2 + y^2} - 1}{\sqrt{x^2 + y^2}}\right)&=\lim \limits_{(x,y) \to (0,0)}\left(\dfrac{\sin(x^2+y^2)-(x^2+y^2)}{(x^2+y^2)^{3/2}}\right)\\ &=\lim \limits_{\rho \to 0}\left(\dfrac{\sin\left(\rho^2\right)-\rho ^2}{|\rho| ^3}\right)\\ &=\lim \limits_{\rho \to 0}\left(\dfrac{\rho ^2+o\left(\rho^2\right)-\rho^2}{|\rho|^3}\right)\\ &=0.\end{align}$$

Git Gud
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