I need to use Taylor's theorem to compute: $\lim\limits_{(x,y)\to (0,0)}{\sin(xy)-xy\over x^2y}$ Using the theorem we have that: $\sin(xy)=xy-{(xy)^3\over6}+R_6(x,y)$ where $\lim\limits_{(x,y)\to (0,0)}{R_6(x,y)\over (x^2+y^2)^3}=0$

So $$\lim\limits_{(x,y)\to (0,0)}{\sin(xy)-xy\over x^2y}=\lim\limits_{(x,y)\to (0,0)}-{{(xy)^3\over6}+R_6(x,y)\over x^2y} \\[2em] =\lim_{(x,y)\to (0,0)}-xy^2/6+\lim\limits_{(x,y)\to (0,0)}{R_6(x,y)\over x^2y}$$

But my problem is that I don't know how to compute $\lim\limits_{(x,y)\to (0,0)}{R_6(x,y)\over x^2y}$

Any hints, suggestions or ideas would be highly appreciated.


$\sin A=A-\frac{1}{3!}A^3+\frac{1}{5!}A^5+\cdots$, so $$\sin (xy)=xy-\frac{1}{3!}x^3y^3+\frac{1}{5!}x^5y^5+\cdots$$ The powers are all high enough to cancel the $x^2y$ on the bottom, so you get $$\frac{\sin(xy)-xy}{x^2y}=\frac{-\frac{1}{3!}x^3y^3+\frac{1}{5!}x^5y^5+\cdots}{x^2y}=-\frac{1}{3!}xy^2+\frac{1}{5!}x^3y^4-\cdots,$$ which tends to $0$ as $(x,y)\to(0,0)$.


Here's an alternative proof that may be more intuitive for others who do not necessarily need to use Taylor's theorem.

Note that $\sin(-x)= -\sin(x),$ therefore the above function $f(x,y) = \frac{\sin(xy) - xy}{x^2y}$ has the following property $$f(x, y) = - f(-x, y)$$ therefore for any fixed value of $y$ if the limit of $f$ exists at all as $x$ approaches $0$ then that limit for $\lim_{x \to 0} f(x, y)$ must be zero. Since $y$ can be taken to be arbitrarily small therefore $\lim_{y \to 0} \lim_{x \to 0}f(x,y) = 0$.

  • $\begingroup$ $f$ is not defined on the $y$-axis $\endgroup$ – zhw. Dec 4 '17 at 20:48
  • $\begingroup$ Sorry yes, I mixed up the limit with the value of the function in my explanation. Fixed it now. $\endgroup$ – Pushpendre Dec 4 '17 at 20:54

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