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I need to prove that the limit of $\frac{\sin(xy)}{xy}$ exists using the $\epsilon$-$\delta$ limit definition as $(x,y)\to (0,0)$

I know that the limit exist and is equal to $1$. I worked on it using the squeeze theorem, but been trying on how to prove it using the limit definition but I don't really know how to start.

Did
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Jrgs
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    $f(x,y)=xy$ is continuous, $g(z)=\dfrac{\sin z}z$ is continuous (at least it has a continuous extension, so $g \circ f$ is also continuous. – Kenny Lau Sep 01 '17 at 14:38
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    Furthermore, you can turn that comment into a proper answer to your question; if you are allowed to use that $\sin(z)/z$ has a continuous extension, then the proof that multiplication is continuous and that a composition of continuous functions is continuous are both easy, and they finish your argument. – Mees de Vries Sep 01 '17 at 14:41
  • unfortunately I am not allowed to use that \sin(z)/z has a continuous extension.... even though how can I prove that the composition of two multivariable continuous function is continuous? – Jrgs Sep 01 '17 at 15:10
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    If you are not allowed to use $\lim_{t\to 0}{\sin t\over t}=1$ you have to tell us what your definition of $\sin$ is. – Christian Blatter Sep 01 '17 at 15:20
  • Related : https://math.stackexchange.com/questions/777272/how-to-find-the-following-limit-lim-x-y-to0-0-frac-sinxyxy – Arnaud D. Dec 02 '19 at 14:38

2 Answers2

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If you used the squeeze theorem you have walked half of the way. However, a proof using the definition could go like this. If we assume that $\sin z = \sum_{n=0}^{\infty} (-1)^n\frac{z^{2n+1}} {(2n+1)!}$ then $|{\frac{\sin z}{z}-1}| \le z^2$ if $|z|<1$. So, if $\epsilon>0$ then pick $\delta= \sqrt{\epsilon}$. It follows that if $|x| < \delta$ and $|y|<\delta$ then $|\frac{\sin(xy)}{xy}-1| < |xy| < \delta^2=\epsilon$

Arundo Donax
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(2017-09-01 21:00: This answer just received a revenge downvote. Oh well...)

For every real number $t$,

$$\left|\sin t-t\right|\leqslant\tfrac16|t|^3\tag{$\ast$}$$

hence $$f(x,y)=\frac{\sin(xy)}{xy}$$ is such that, for every $(x,y)$ such that $xy\ne0$ (that is, where $f(x,y)$ exists), $$|f(x,y)-1|\leqslant\tfrac16|xy|^2$$ which should be enough to conclude that, indeed,

$$\lim_{(x,y)\to(0,0)}f(x,y)=1$$

To prove $(\ast)$, start from $$|\cos t|\leqslant1\tag{$\circ$}$$ and integrate thrice, that is, note that this upper bound yields $$|\sin t|=\left|\int_0^t\cos s\,ds\right|\leqslant|t|$$ which implies $$|\cos t-1|=\left|\int_0^t\sin s\,ds\right|\leqslant\tfrac12t^2$$ which implies $$|\sin t-t|=\left|\int_0^t(\cos s-1)\,ds\right|\leqslant\tfrac16|t|^3$$ as desired.

To conclude, $(\circ)$ is the only prerequisite we used, and $(\circ)$ itself follows readily, for example, from the fact that $$\cos^2t+\sin^2t=1$$

Did
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