# Find the limit (if it exists) $\lim_{(x,y) \to (0,0)} \dfrac{xy}{x^2 + |y|}.$

I want to find the limit $$\lim_{(x,y) \to (0,0)} \dfrac{x \sin(y)}{x^2 + |y|},$$ if it exists. My idea was to use the fundamental trig limit to obtain $$\lim_{(x,y) \to (0,0)} \dfrac{x \sin(y)}{x^2 + |y|} = \lim_{(x,y) \to (0,0)} \dfrac{xy}{x^2 + |y|} \dfrac{\sin{y}}{y},$$ but now, I need to calculate the limit of the first factor, which I can't.

If you solve the limit computationally, you'll know it exists and it equals zero, so you could use polar coordinates to prove it. But the exercise do not say the limit exists, so the use of polar coordinates is not allowed.

Do you know how to find this limit $$\lim_{(x,y) \to (0,0)} \dfrac{xy}{x^2 + |y|}?$$

• No need to consider trigonometry functions like sine or cosine. You may find a bound purely in algebraic way. Apr 14 at 19:56

You can apply the squeeze theorem:

\begin{align*} 0\leq \left|\frac{x\sin(y)}{x^{2} + |y|}\right| & = \frac{|x\sin(y)|}{x^{2} + |y|} \leq \frac{|x\sin(y)|}{|y|} = |x|\left|\frac{\sin(y)}{y}\right|\xrightarrow{(x,y)\to(0,0)} 0\times 1 = 0 \end{align*}

As to the proposed limit in the title of the question, you can proceed as follows: \begin{align*} 0\leq\left|\frac{xy}{x^{2} + |y|}\right| & = \frac{|xy|}{x^{2} + |y|} \leq \frac{|xy|}{|y|} = |x| \xrightarrow{(x,y)\to(0,0)} 0 \end{align*}

• Oh... I see... As $\left|\frac{xy}{x^{2} + |y|}\right| \leq |x|$ we have $-|x| \leq \frac{xy}{x^{2} + |y|} \leq |x|$, so the result follows from the squeeze theorem. Thank you very much! Apr 15 at 12:49
• @Alice you are welcome! I am glad to help. Apr 15 at 17:43

Let $$(x,y)\in\mathbb{R}^{2}\setminus\{(0,0)\}$$. If $$y=0$$, then $$x\neq0$$ and hence $$\frac{xy}{x^{2}+|y|}=0$$. If $$y\neq0$$, we have that $$\begin{eqnarray*} \left|\frac{xy}{x^{2}+|y|}\right| & \leq & \left|\frac{xy}{|y|}\right|\\ & = & |x|. \end{eqnarray*}$$ Therefore, in all cases, we have that $$\left|\frac{xy}{x^{2}+|y|}\right|\leq|x|$$. Note that $$\lim_{(x,y)\rightarrow(0,0)}|x|=0$$, so, by sandwich rule, $$\lim_{(x,y)\rightarrow(0,0)}\frac{xy}{x^{2}+|y|}=0$$.

I thought it might be instructive to present an alternative approach to those in the two posts at the time of this submission.

Proceeding, we first note that since $$(|x|-|y|^{1/2})^2\ge 0$$, then

$$\frac{1}{x^2+|y|}\le \frac1{2|x||y|^{1/2}}\tag 1$$

Then, using $$(1)$$ it is clear that

$$\left|\frac{x\sin(y)}{x^2+|y|}\right|\le \frac12|y|^{1/2}\tag2$$

whereupon application of the squeeze theorem to $$(2)$$ shows that

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

• Thanks for your contribution! It's very nice to learn another ways of thinking. Apr 15 at 13:00
• You're welcome. My pleasure. Apr 15 at 14:54