A function:
\begin{equation}
f(x,y) =
\begin{cases}
\frac{\sqrt[3]{x^2+y^2}}{\ln(x^2+y^2)}, & \text{(x,y) $\neq$ (0,0)}\\
0, & \text{(x,y) = (0,0)}\\
\end{cases}
\end{equation}
is given. Task is to check:
$1.$ Continuity of $f$
$2.$ Existence of directional derivative at any direction at point $(0,0)$
$3.$ Existence of partial derivatives at point $(0,0)$, and if they exist to calculate the value
$4.$ Differentiability of $f$
This is how I tried solving it:
$1.$ When I replace x and y with zeroes I get $\frac{0}{-\infty}$ which is $0$, so limit is zero and function is continuous
$2.$ I chose direction $u=(\cos\alpha,\sin\alpha)$ and tried solving limit:
Also when I am free to chose direction which one would you chose, my professor usually goes with this one..
$$
\lim_{t\rightarrow0}\frac{f(0+t\cos\alpha,0+t\sin\alpha)-f(0,0)}{t}=..=\lim_{t\rightarrow0}\frac{1}{\sqrt[3]{t}\ln(t^2)}
$$
here I got stuck..
$3.$ When checking existence of partial derivatives , precisely limits I get the same limit expression as in $2.$ so also got stuck there
$4.$ since I haven't shown existence of partials I couldn't continue to differentiability …
Appreciate the help!
Best Answer
Using L'Hôpital's rule, one obtains $$ \lim_{t\rightarrow0} \sqrt[3]{t}\ln(t^2)= \lim_{t\rightarrow0} \frac{\ln(t^2)}{1/\sqrt[3]{t}}= \lim_{t\rightarrow0} \frac{2t/t^2}{-\frac13{t}^{-4/3}}= \lim_{t\rightarrow0} \frac{-6t}{{t}^{2/3}}= \lim_{t\rightarrow0} (-6t^{1/3})= 0; $$ thus, $$ \lim_{t\rightarrow0}\frac{1}{\sqrt[3]{t}\ln(t^2)}= \infty. $$ ($\infty$ is not $+\infty$ here, see the comments below)
Hence, the directional and partial derivatives are infinite and the function is not differentiable at $(0,0)$.