Prove this limit exists:

$$\lim_{(x,y)\rightarrow (0,3)}\frac{\tan(xy)}{x}=3$$

My work:

Let $\epsilon >0$, $\delta=\cdots$

If $\sqrt{(x-0)^2+(y-3)^3}<\delta$


\begin{align} & \left|\frac{\tan(xy)}{x}-3\right|=\left|\frac{\tan(xy)-3x}{x}\right| = \left|\frac{\sin(xy)-3x\cos(xy)}{x\cos(xy)}\right| \\[10pt] = {} & \frac{|\sin(xy)-3x\cos(xy)|}{|x\cos(xy)|} \leq \frac{|\sin(xy)|+3|x||\cos(xy)|}{|x||\cos(xy)|}\leq \frac{1+3|x|}{|x|} \end{align}

In this step I'm stuck. Can someone help me?

  • $\begingroup$ Your method won't work, since as $x\to\infty$, you just get $$\left|\frac{\tan(xy)}x-3\right|\le\lim_{x\to0}\frac1{|x|}+3=+\infty$$which doesn't help to prove anything. $\endgroup$ – John Doe Oct 28 '17 at 23:38
  • $\begingroup$ Oops: Double check what's the limit of $|\sin(xy)|$ as $(x,y) \rightarrow (0,3)$. $\endgroup$ – aschepler Oct 28 '17 at 23:38

\begin{align}\lim_{(x,y) \to (0,3)}\frac{\tan(xy)}{x}&= \lim_{(x,y) \to (0,3)}\frac{y\sin(xy)}{xy\cos(xy)}\\&= \lim_{(x,y) \to (0,3)}(y)\lim_{(x,y) \to (0,3)}\left(\frac{\sin(xy)}{xy} \right)\lim_{(x,y) \to (0,3)}\frac{1}{\cos(xy)} \end{align}


A not so good solution (which can be made better of course) is

\begin{align*} \lim_{(x,y)\rightarrow(0,3)}\dfrac{\tan(xy)}{x}&=\lim_{(x,y)\rightarrow(0,3)}\dfrac{1}{x}\left(xy+\dfrac{1}{3}(xy)^{3}+\cdots\right)\\ &=\lim_{(x,y)\rightarrow(0,3)}\left(y+\dfrac{1}{3}x^{2}y^{3}+\cdots\right)\\ &=3. \end{align*}

If we decide to go with $\epsilon$-argument, a painstaking way would be:

Observe that $\tan u\leq u+\dfrac{1}{3}u^{3}+\dfrac{1}{15}u^{5}$ for all real numbers $u$, just tackle with the function $\varphi(u)=u+3^{-1}u^{3}+15^{-1}u^{5}-\tan u$ and we know that $\varphi'(u)>0$ for $u\ne 0$ by the inequality follows by Mean Value Theorem.

On the other hand, by the similar fashion, we could have $\tan u\geq u+\dfrac{1}{3}u^{3}$ for $u\geq 0$.

Given $\epsilon>0$ for all $|x|^{2}+|y-3|^{2}<\epsilon^{2}$ and $2<y<4$, one see that for $x>0$ \begin{align*} \frac{\tan(xy)}{x}-3&\leq\frac{xy+3^{-1}(xy)^{3}+5^{-1}(xy)^{5}-3x}{x}\\ &=(y-3)+\frac{1}{3}x^{2}y^{3}+\frac{1}{5}x^{4}y^{5}\\ &\leq\epsilon+\frac{64}{3}\epsilon^{2}+\frac{512}{3}\epsilon^{4}. \end{align*} For $x<0$, one do the trick with $\dfrac{\tan(xy)}{x}=\dfrac{\tan(-xy)}{-x}$ and still get the upper bound. Now we move on to the lower bound. For $x>0$, we have \begin{align*} \frac{\tan(xy)}{x}-3\geq\frac{xy+3^{-1}(xy)^{3}-3x}{x}=(y-3)+\frac{1}{3}x^{2}y^{3}\geq-\epsilon+\frac{8}{3}x^{2}\geq-\epsilon, \end{align*} the rest should be easy.

  • $\begingroup$ The proof need be answer by definitión of épsilon.... $\endgroup$ – Bvss12 Oct 28 '17 at 23:50

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