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So, I have to prove that the limit on $(0,0)$ for the following function exists (or not!), with the (ε,δ) limit proof :

if for every $ϵ > 0$ there exists a $δ$ such that, for all $x ∈ D$ , if $0 < | x − c | < δ$, then $| f ( x ) − L | < ϵ $

The function is:

$$f(x,y)=\begin{cases} \frac {\sin(xy)}{x} & \text{ if } x \neq 0 \\ y & \text{ if } x = 0\end{cases}$$

I already know that the limit, if it exists, should be $0$ by doing reiterated and directional limits, but I am stuck in the proof of the limit itself when trying to get a similar norm, which is how we prove it. This is how far I got:

$||f(x)|| = \frac {|\sin(xy)|}{|x|} \leq \frac {1}{|x|} \leq ??? $

Any help is welcome

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  • $\begingroup$ you are using wrong estimate. Instead note that around $0$, $|sin(x)| \le |x|$ $\endgroup$ – user160738 Apr 17 '19 at 9:05
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For each $t \in \mathbb R$ we have $|\sin(t) \le |t|.$ Hence

$|f(x,y)| \le |y|$ for all $(x,y)$.

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$\delta =\epsilon$ will do because $|f(x,y)| \leq \max \{|x|,|y|\}$ (since $|sin\, t| \leq |t|$).

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