I am trying to come up with an epsilon delta proof of $$\lim_{(x,y)\to (0,0)}sin(x^2+y^2)=0.$$
I know I have to find the form square root of $x^2+y^2 < \delta$, but the only thing I know is that the absolute value of $\sin (x^2+y^2) <= 1$
Thanks!
2026-03-26 17:53:38.1774547618
Epsilon-Delta proof of limit as $(x,y)\to 0,0)$ of $\sin(x^2+y^2)=0$
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We can make use of the fact that for $0 < z < 1$, $0 < \sin z < z$.
For $0 < x, y < 1$, then, $0 < x^2 + y^2 < 1$, and hence $0 < \sin (x^2 + y^2) < x^2 + y^2$.
From this point, we can then use the squeeze theorem and state that since $x^2 + y^2 \rightarrow 0$ in the given limit, $\sin (x^2 + y^2)$ must also approach $0$. You can dress that up in $\epsilon-\delta$ notation as required.