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$f:\mathbb{R}^2 \rightarrow \mathbb{R}$

$f(x,y)=\begin{cases} \frac{\sin{y}+2yx^2}{y}, & y \gt0 \\ \frac{y+3x}{1-y}, & y\le 0 \end{cases}$

I have to check

a) What points is $f$ continuous at?

b) Is $f$ class $C^1$ at $(1,0)$

My idea for a)

Function $f$ is continuous at all points $(x_0,y_0)$ where $y_0 \neq 0$ as sum, difference and quotient of continuous functions. Now for $y_0=0$, I have to check if $\lim_{x \to x_0} f(x,0) = f(x_0,0)$. I don't know what to plug in instead of $f(x,0)$ in limit.

b) I can find differential $Df(1,0)$ and check if it is continuous in $(1,0)$? I am not sure about b)

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    I just saw that. So I can just say that it is continuous as sum,difference, product and quotient of continuous functions? – 2017-02-10
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    No, my error, I misread it. – 2017-02-10
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    No, I think you are right – 2017-02-10
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    For $y=0$ I plug in down function – 2017-02-10
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    To check for continuity at $(x_0,0)$, what is the limit, leaving $x_0$ fixed, as $y$ approaches $0$ from the above? What about from below? – 2017-02-10

1 Answers 1

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As you say, the function is continuous for all $y \neq 0$, and so we need to consider $y=0$.

$$\lim_{y \to 0^+} \left(\frac{\sin y +2x^2y}{y}\right) \ = \ \lim_{y\to 0^+}\left(\frac{\sin y}{y} + \frac{2x^2y}{y}\right) \ = \ 1+\lim_{y \to 0^+}\left(\frac{2x^2y}{y}\right) \ = \ 1+2x^2$$

$$\lim_{y\to 0^-}\left(\frac{3x+y}{1-y}\right) \ = \ 3x$$

For the function to be continuous, we need the $x$- and $y$-limits to agree.

We need (at least) $1+2x^2=3x$, i.e. $2x^2-3x+1=0$, i.e. $x=\frac{1}{2}$ or $x=1$.

The function is discontinuous at the the points $y=0$ and $x \neq \frac{1}{2},1$.

We need to consider $x=\frac{1}{2}$ and $x=1$ separately. $$f(\tfrac{1}{2},y) = \left\{ \begin{array}{ccc} \frac{2\sin y + y}{2y} & : & y > 0 \\ \frac{2y+3}{2-2y} & : & y \le 0 \end{array}\right.$$

As $y \to 0^-$, $f \to \frac{3}{2}$. As $y \to 0^+$, $f \to \frac{3}{2}$ and so $f(\frac{1}{2},y)$ is continuous at $y=0$.

$$f(1,y) = \left\{ \begin{array}{ccc} \frac{\sin y + 2y}{y} & : & y > 0 \\ \frac{3+y}{1-y} & : & y \le 0 \end{array}\right.$$

As $y \to 0^-$, $f \to 3$. As $y \to 0^+$, $f \to 3$. It follows that $f$ is continuous at $(x,y)=(1,0)$.

Although the function is continuous as $(1,0)$, it needs to be continuous in a neighbourhood of $(1,0)$ for it to be differentiable. The function is not continuous at $(1+\varepsilon,0)$ for arbitrarily small $\varepsilon>0$.

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    Thank you, can you help me with b) question? – 2017-02-10
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    @RushThaMan See additions – 2017-02-10
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    How do you know that its not continuous in a neighbourhood of $(1,0)$? – 2017-02-10
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    Oh, I get it, its not, but does it need to be continuous in neighbourhood to be differentiable? – 2017-02-10
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    @RushThaMan How do you make sense of the limit definition of a derivative if the function is not continous? Even in one variable, what does $$\lim_{h \to 0} \left( \frac{f(x+h)-f(x)}{h}\right)$$ mean of $\lim_{h \to 0} f(x+h)$ is not well-defined? – 2017-02-10
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    You wrote: "The function is not continuous at $(1,\varepsilon))$ for arbitrarily small $\varepsilon >0$". How can that be right? – 2017-02-10
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    Isn't $f(\frac{1}{2},y) = \frac{2y+3}{2-2y}$? Why do you need to reconsider $x=\frac{1}{2}, 1$? Your first analysis works for all $x$. – 2017-02-10
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    @Fly by Night: Jonathan Hahn's point is correct. The function $f$ is continuous at $(1/2,0)$ and $(1,0)$. – 2017-02-10
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    @quasi Thanks for spotting my mistake, I meant to move a bit along the $x$-axis and not the $y$-axis, so $(1,\varepsilon)$ ought to have been $(1+\varepsilon,0)$. JH's point was valid too - I'd made a sign error which lead me to believe it was not continuous. I've fixed both errors. – 2017-02-10
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    @JonathanHahn Thanks for spotting the sign error. My mistake made me think it was not continuous ($-\frac{3}{2} \neq \frac{3}{2}$). But of course, you were right. Thanks again. – 2017-02-10