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$$f(x,y) = \left\{ \begin{array}{ll} \frac{xy}{x^2+y^3} & \mbox{if $(x,y) \not = (0,0)$};\\ 0 & \mbox{if $(x,y) = (0,0)$}.\end{array} \right. $$

Here is what I got

Note that

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

take limit of the above inequality, we have

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

By the squeeze theorem

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

So $f(x,y)$ is continuous at $(0,0)$.

My professor told me this is wrong without telling where is the mistake. I don't know where did I do wrong and how to fix it.

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1  
why is $|y/x| \le |y|$? –  user10444 Apr 7 at 14:16
    
because $|y/x| \leq |y|/|x| \leq |y|$ is this wrong ? –  Diane Vanderwaif Apr 7 at 14:21
    
what if $|x|<1$ –  user10444 Apr 7 at 14:22

5 Answers 5

up vote 0 down vote accepted

Hello Diane,

1) The limit of this function as it approaches the origin does not exist.

2) This is because you seem to have made an error in your inequality-- precisely at the point where you convert [xy/(x^2)] to [y/x] and then further deduce that this is less than or equal to abs(y). The error lies in the fact that x could be zero and hence, you would essentially be dividing by 0, not something you would want to do here. I hope that makes it clear.

3) Note, however, that the limit MAY exist if you choose a path in the complex plane. I have not checked it, but it may be so. But if you are only dealing with the reals, then there is no limit convergence for this function at the origin.

EDIT: Note that the limit does not exist even in the complex plane.

All the best.

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How could choosing a path ever show that the limit exists? If the limit does not exist, then that means there is a path along which the limit does not exist, and then that path will still be an offending path even if you view the function as a complex function. It's the other way around: a limit that exists when viewing the function as real might not exist when you view the function as complex. –  alex.jordan Apr 8 at 0:30

Hint: make the change of variables $$ x = r\cos\theta\\ y = (r\sin\theta)^{2/3} $$

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Hint: Take the sublimits along the lines $y=kx$.

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Comment, down voter? –  Git Gud Apr 7 at 21:42

Another approach would be to take the sequence $x_n = ( \frac{1}{n}, \frac{1}{n}) $. we know $x_n \to (0,0) $. So, if $f$ were continuous, then we would have $f(x_n) \to f(0,0) = 0 $. But,

$$f\left(\frac{1}{n}, \frac{1}{n}\right) = \frac{ \frac{1}{n^2}}{\frac{1}{n^2}+ \frac{1}{n^3}} = \frac{n}{n+1} \to 1 \neq 0 = f(0,0) $$

Hence, $f$ cannot be continuous at the origin. Hope this helps.

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There's a much simpler reason that this limit does not exist. There is a path leading to $(0,0)$ along which the function is not defined at any step. The path is given by $x^2+y^3=0$, or perhaps more familiarly as $y={-x^{2/3}}$. This is always the first thing I would look for in problems like this.

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