Continuity, partial derivatives, differentiability questions for $f(x, y):= \frac{x^2 \sin y^2}{x^2+y^4}$

analysiscontinuityderivativespartial derivativesolution-verification

I would like to know if you think my answers are correct, I am not so sure about $1$ and $3$ and I don't know how to answer $4$.

Define $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ by
\begin{align*}
f(x, y):= \frac{x^2 \sin y^2}{x^2+y^4}, & (x, y) \neq(0,0) & \\ f(x,y) := 0, & (x, y)=(0,0)
\end{align*}

  1. Is the function continuous in (0,0)?
  2. Do the partial derivatives $(D_1f)(0,0)$ en $(D_2f)(0,0)$ exist?
  3. For which $u \in \mathbb{R}^2 \backslash \{0\}$ does the directional derivative $(D_uf)(0,0)$ exist?
  4. Is $f$ differentiable in $(0,0)$?
  5. Is $f$ a $C^1$ function?
  1. To show that $f$ is continuous at $(0,0)$, we need to show that $\lim_{(x,y)\to(0,0)}f(x,y)=f(0,0)=0$. We can do this by using the epsilon-delta definition of a limit. To do this, we need to show that for every $\epsilon > 0$, there exists a $\delta > 0$ such that if $||(x,y) – (0,0)|| < \delta$, then $|f(x,y) – f(0,0)| < \epsilon$.

Let $\epsilon > 0$ and take $\delta = \sqrt{\epsilon}$. If $||(x,y) – (0,0)|| < \delta$, then we have:
\begin{align*}
|f(x,y) – f(0,0)| &= ||\frac{x^2\sin y^2}{x^2+y^4} – 0|| \
&= |\frac{x^2\sin y^2}{x^2+y^4}| \
& \leq |\frac{x^2\sin y^2}{x^2}| \
& \leq |\frac{x^2y^2}{x^2}| = y^2 \
& \leq x^2 + y^2 = \delta^2 = \epsilon.
\end{align*}

Here, we used that $x^2 \leq x^2 + y^4$, since $y^4 \geq 0$ for all $y$ and observed that $0 \leq |\sin y^2| \leq 1$ for all $y$. Thus, the limit of the function exists at 0 and $f$ is continuous at 0.

  1. We prove this using the definition of partial derivative. We have $(D_jf)(a) := \lim_{t \to 0} \frac{f(a + te_j) – f(a)}{t}$ with $a = (0 \quad 0)^T$. Thus:
    \begin{align*}
    (D_1f)(0,0) & = \lim_{t \to 0} \frac{f\left(\left(\begin{array}{c} 0 \ 0 \ \end{array}\right) + t\left(\begin{array}{c} 1 \ 0 \ \end{array}\right)\right) – 0}{t} \
    & = \lim_{t \to 0} \frac{f(t,0) – 0}{t} \
    & = \lim_{t \to 0} \frac{\frac{t^2\sin(0^2)}{t^2 + 0^4} – 0}{t} \
    & = \lim_{t \to 0} \frac{\frac{t^2 \cdot 0}{t^2}}{t}\
    & = \lim_{t \to 0} \frac{0}{t^3} \
    & = \lim_{t \to 0} 0 \
    & = 0
    \end{align*}
    and
    \begin{align*}
    (D_2f)(0,0) & = \lim_{t \to 0} \frac{f\left(\left(\begin{array}{c} 0 \ 0 \ \end{array}\right) + t\left(\begin{array}{c} 0 \ 1 \ \end{array}\right)\right) – 0}{t} \
    & = \lim_{t \to 0} \frac{f(0,t) – 0}{t} \
    & = \lim_{t \to 0} \frac{\frac{0^2\sin(t^2)}{0^2 + t^4} – 0}{t}\
    & = \lim_{t \to 0} \frac{\frac{0}{t^4}}{t}\
    & = \lim_{t \to 0} \frac{0}{t^5} \
    & = \lim_{t \to 0} 0 \
    & = 0.
    \end{align*}
    So the partial derivatives $(D_1f)(0,0)$ and $(D_2f)(0,0)$ exist.

  2. We look at the definition of the directional derivative $(D_uf)(0,0) = \lim_{t \to 0} \frac{f(a + tu) – f(a)}{t}$. For $a = (0 \quad 0)^T$, this gives (with $u := (u_1, u_2)$):
    \begin{align*}
    \lim_{t \to 0} \frac{f(a + tu) – f(a)}{t} &= \lim_{t \to 0} \frac{f(tu)}{t} \
    & = \lim_{t \to 0} \frac{f(t(u_1, u_2))}{t} \
    &= \lim_{t \to 0} \frac{f(tu_1, tu_2)}{t} \
    & = \lim_{t \to 0} \frac{\frac{t^2u_1^2 \cdot \sin(t^2u_2^2)}{t^2u_1^2 + t^4u_2^4}}{t} \
    & = \lim_{t \to 0} \frac{t^2u_1^2 \cdot \sin(t^2u_2^2)}{t^3u_1^2 + t^5u_2^4} \
    & = \lim_{t \to 0} \frac{u_1^2\sin(t^2u_2^2)}{tu_1^2 + t^3u_2^4}\
    & = \frac{u_1^2\sin(0^2 u_2^2)}{0u_1^2 + 0^3u_2^4} \
    & = \frac{0}{0}.
    \end{align*}
    So for no $u \in \mathbb{R}^2 \backslash {0}$, the directional derivative $(D_uf)(0,0)$ exists.

  3. We use the theorem that says when $f$ is differentiable in $a$, then the partial derivatives $(D_jf_i)(a)$ exist and we have the total derivative $Df(a)$. ??

  4. To show that $f$ is a $C^1$ function, we need to show that the partial derivatives exist and are continuous on $\mathbb{R}^2$.

The partial derivatives of $f$ are:
\begin{align*}
\frac{\partial f}{\partial x}(x,y) &= \frac{2x\sin y^2 (x^2+y^4) – x^2\sin y^2 \cdot 2x}{(x^2+y^4)^2} \
&= \frac{2x\sin y^2 (y^4)}{(x^2+y^4)^2}, \\
\frac{\partial f}{\partial y}(x,y) &= \frac{x^2 \cdot 2y\cos y^2 (x^2+y^4) – x^2\sin y^2 \cdot 4y^3}{(x^2+y^4)^2} \
\end{align*}
Both partial derivatives are continuous on $\mathbb{R}^2$ since they are defined as quotients of continuous functions with denominators that do not become zero. Therefore, $f$ is a $C^1$ function on $\mathbb{R}^2$.

Best Answer

The third part.

$$(D_uf)(0,0)=\lim_{t\rightarrow 0}\frac{tu_1^2}{u_1^2+t^2u_2^2}\frac{\sin(t^2u_2^2)}{t^2}=0. u_2^2=0.$$

The fourth part. I am not sure about the solution yet. WolframAlpha confused me. It says "SlowLarge".

By using the definition $$\lim_{(x,y)\rightarrow (x_0,y_0)}\frac{f(x,y)-f(x_0,y_0)-D_1(x_0,y_0)(x-x_0)-D_2(x_0,y_0)(y-y_0)}{\sqrt{(x-x_0)^2+(y-y_0)^2}}$$ of differentiablity for functions of two variables and letting $f(x,y)=\frac{x^2\sin y^2}{x^2+y^4}$ and $(x_0,y_0)=(0,0)$, the origin, we get $$\lim_{(x,y)\rightarrow (0,0)}\frac{x^2\sin y^2}{(x^2+y^4)\sqrt{x^2+y^2}}.$$ Due to Maclaurin series of sine function it is enough to consider $$\lim_{(x,y)\rightarrow (0,0)}\frac{x^2y^2}{(x^2+y^4)\sqrt{x^2+y^2}}.$$ By passing to polar coordinates with $x=r\cos\theta$ and $y=r\sin\theta$, $r>0$, we obtain the limit $$\lim_{r\rightarrow 0}\frac{r\sin^2\theta}{1+r^2\sin^2\theta\tan^2\theta}=0$$ where $\theta=\theta(r).$ So the function is differentiable at the origin.

The fifth part. Although $\frac{\partial f}{\partial x}(0,0)=0$, the limit $$\lim_{(x,y)\rightarrow (0,0)}\frac{\partial f}{\partial x}=\lim_{(x,y)\rightarrow (0,0)}\frac{2xy^4\sin y^2}{(x^2+y^4)^2}$$ does not exist. So, $\frac{\partial f}{\partial x}$ is not continuous and $f$ is not a $C^1$-function. To show the non-existence of the limit above, choose the path $x=y^2$. Then, the limit along this path is $\lim_{y\rightarrow 0}\frac{\sin y^2}{2y^2}=\frac12\neq 0.$

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