In trying to find this limit, all the (x,y) paths I check approaching (0,0) produce the same result, namely, the function approaches 0. consequently, I came up with the simplest proof I could that the limit indeed exists and is zero. However wolframalpha tells me that the limit doesnt exist. Assuming I'm wrong, can anyone shed light on where my proof is flawed, and possibly provide a path on which the function doesn't approach 0.

\begin{equation} \lim_{(x,y) \rightarrow (0,0)} \frac { x \cdot \sin(\frac{y}{\sqrt{x}}) } {\sqrt{ x^2 + y^2 }} \end{equation}

provided $x>0$

my proof:

let's take $\varepsilon > 0$ and prove the existence of some $\delta > 0$ such that
$\forall (x,y) \in B(0, \delta ) : |f(x,y)| < \varepsilon$
(where f is our function above, and as stated $x>0$)

taking the x to the denominator, we get:

\begin{equation} \frac { \sin(\frac{y}{\sqrt{x}}) } {\sqrt{ 1 + (\frac{y}{x})^2 }} \end{equation}

sin is bounded between -1 and 1, so from here i intend from here to show that i can choose a relevant $\delta$ where either the denominator is large enough to make the expression smaller than our $\varepsilon$, and if not, the numerator is smaller than our $\varepsilon$.

We find the relevant denominator bound by:

\begin{equation} \frac {1} {\sqrt{ 1 + (\frac{y}{x})^2 }} < \varepsilon \end{equation}

which is simply:

\begin{equation} \frac {y}{x} > \sqrt{\frac{1}{\varepsilon^2} - 1} \end{equation}

clearly, if this inequality holds, then the absolute value of the function is indeed bound by our $\varepsilon$ as required.

so lets deal with the seconds case:

\begin{equation} \frac {y}{x} \leq \sqrt{\frac{1}{\varepsilon^2} - 1} \end{equation}

multiplying through by $\sqrt {x}$ we get

\begin{equation} \frac {y}{\sqrt{x}} \leq \sqrt{x} \cdot \sqrt{\frac{1}{\varepsilon^2} - 1} \leq \sqrt{\delta} \cdot \sqrt{\frac{1}{\varepsilon^2} - 1} \end{equation}

the last inequality is because $x$ is obviously bound by the $\delta$ we choose
it should be clear that the right hand side of the inequality is arbitrarily small, although we will formalise somewhat:

take $\delta < \varepsilon^4$: \begin{equation} \frac {y}{\sqrt{x}} \leq \sqrt{\delta} \cdot \sqrt{\frac{1}{\varepsilon^2} - 1} < \sqrt{\varepsilon^4} \cdot \sqrt{\frac{1}{\varepsilon^2}} = \varepsilon^2 \cdot \frac{1}{\varepsilon} = \varepsilon \end{equation}

without too much formality, we say that for sufficiently small $\delta$, $\sin x \approx x$ and therefore

\begin{equation} {\sin{ \frac {y}{\sqrt{x}}} \approx \frac {y}{\sqrt{x}} \leq \varepsilon} \end{equation}

Of course, I was lazy and didn't add in the necessary stages to prove the approximation will hold in the inequality, although it does, because I can take the $\delta$ to be such that the difference between the sin and the value itself is less than $\frac{\varepsilon}{3}$ and also such that $\frac{y}{\sqrt {x}} < \frac{\varepsilon}{3}$ which means the sin is indeed less than $\varepsilon$ as required. this takes care of the second case.


(I didn't take much care with whether y is positive or negative, although the function is odd, so we can work with positives and then just prove the result for the negatives)

  • 1
    $\begingroup$ By the way, it would be nice if you could make your title a little more descriptive. $\endgroup$
    – user856
    Nov 15, 2010 at 1:38

3 Answers 3


Your proof is correct, although somewhat long-winded. It's a good practice, once you have a proof, to go back and see if you can simplify it as much as possible, aiming to find the most direct path to the desired result. In the process, one often gains a better understanding of the problem.

Since you ended up using the fact that $\sin \theta \le \theta$ for all $\theta > 0$, let's see if we can use this from the outset. Since the function is odd with respect to $y$, it suffices to only consider $y \ge 0$ as you noticed. Then we have $$\frac{x\sin\frac{y}{\sqrt{x}}}{\sqrt{x^2+y^2}} \le \frac{y\sqrt{x}}{\sqrt{x^2+y^2}} \le \sqrt{x}$$ because $y \le \sqrt{x^2+y^2}$. In other words, the function is positive and bounded above by $\sqrt{x}$. From here, you can quite easily get to where you want.


Wolframalpha is not infallible. Your proof looks OK to me.

One trick that sometimes helps for problems like this is to write $y = ax$ and then examine your expression. In this case you get ${1 \over \sqrt{1 + a^2}} \sin (a \sqrt{x})$. Then you can use that $|\sin(\theta)| \leq |\theta|$ for all $\theta$, to get that it is bounded in absolute value by ${a \over \sqrt{1 + a^2}} \sqrt{x}$. Since $\big|{a \over \sqrt{1 + a^2}}\big|$ is at most 1, your expression's absolute value is at most $\sqrt{x}$, which goes to zero as $(x,y)$ approaches the origin.

  • 1
    $\begingroup$ Does this trick always work? A contrived counterexample is $f(x,y) = 1$ if $y = x^2 \ne 0$ and $0$ otherwise, which goes to zero on every $y = ax$. $\endgroup$
    – user856
    Nov 15, 2010 at 1:25
  • $\begingroup$ by problems like this, I mean examples with radial symmetry in the expression. In this case the $x^2 + y^2$ in the denominator led me to try that. $\endgroup$
    – Zarrax
    Nov 15, 2010 at 1:28

I think Wolfram-Alpha is right in this case. But technically you're asking it the wrong question. In order for you to state $\forall(x,y)\in B(\delta,0)$, $|f(x,y)|<\delta$, your function $f(x,y)$ has to be defined on a local neighbourhood around the origin. In this case it isn't, as the function is undefined on $x=0$. I think what you need to input into Wolfram Alpha is something along the lines of a multivariable one-sided limit.
$$ \lim_{x\rightarrow 0^+,y\rightarrow 0} \frac{x\sin\frac{y}{\sqrt{x}}}{x^2 + y^2}$$ Because on the face-of it, I think Wolfram alpha would check if there's any problem areas in a local neighbourhood of the origin.


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