Why $\displaystyle\lim_{(x,y)\to(0,0)}\frac{\sin(x^2+y^2)}{x^2+y^2}=\lim_{t\to 0}\frac{\sin t}{t}$( and hence equals to $1$)?
Any rigorous reason? (i.e. not just say by letting $t=x^2+y^2$.)
Why $\lim\limits_{(x,y)\to(0,0)}\frac{\sin(x^2+y^2)}{x^2+y^2}=\lim\limits_{t\to 0}\frac{\sin t}{t}$?
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use polar coordinates... – tired Oct 29 '16 at 10:39
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1@Eric In fact $t=x^2+y^2$ seems to me pretty rigorous, doesn't it? – Miguel Oct 29 '16 at 10:43
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Why the polar coordinates method is consistent with the original $\epsilon-\delta$ definition? – Eric Oct 29 '16 at 10:44
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@MiguelAtencia So do you think $\displaystyle\lim_{(x,y)\to(0,0)}\frac{\sin(x^\sqrt{2}+\log y^4)}{x^\sqrt{2}+\log y^4}=\lim_{t\to 0}\frac{\sin t}{t}=1$? – Eric Oct 29 '16 at 10:46
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1@Eric No, because $\log y^4$ is not defined at $y=0$. Well, ok, let us be rigourous: $t(x,y)=x^2+y^2$ is continuous, so it swaps with the limit; and $t(0,0)=0$. – Miguel Oct 29 '16 at 10:50
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Here's a point of view I like. The fact that $$\lim_{t\to0}\frac{\sin t}{t}=1$$ means precisely that the function $\phi:\mathbb R\to\mathbb R$ defined by $$\phi(t)=\begin{cases}\frac{\sin t}t;&t\neq0,\\1;&t=0,\end{cases}$$ is continuous. Therefore, we have $$\lim_{(x,y)\to(0,0)}\frac{\sin(x^2+y^2)}{x^2+y^2}=\lim_{(x,y)\to(0,0)}\phi(x^2+y^2)=\phi(0)=1.$$ The first equality holds because the definition of limit doesn't involve the value of the function at $(0,0)$ and the second equality holds because $\phi(x^2+y^2)$ is continuous (since it is the composition of two continuous functions).
Dejan Govc
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Thanks! I love your argument!! Can you also help me check why the change to polar coordinates method to compute the limits of some several variables functions that calculus textbooks often used also work? – Eric Oct 29 '16 at 11:02
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Can't it just be thought as the compositions of some functions? Just like the example of this post. Let $\Psi:\mathbb{R}^+\times[0,2\pi)\to\mathbb{R}^2$ defined by $\Psi(r,\phi)=(r\cos\phi,r\sin\phi)$, and regard $\lim f(x,y)$ as $f(r\cos\phi,r\sin\phi)$, but I can't figure out the left process.. – Eric Oct 29 '16 at 14:48
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(I have re-elaborated my comments in my answer to this question, so I have removed them here.) – Dejan Govc Oct 30 '16 at 20:18
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Using $x = t \cos \alpha $ and $y = t \sin \alpha$, then:
$$ \frac{ \sin (x^2 + y^2 ) }{x^2+y^2} = \frac{ \sin (t^2(\sin^2 \alpha + \cos^2 \alpha)) }{t^2(\sin^2 \alpha + \cos^2 \alpha)} = \frac{ \sin (t^2) }{t^2}$$ .
Note as $x,y \to 0,0$, then $t \to 0 $.
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2Can you show rigorously that $x,y\to 0$ then $t\to 0$? Not just say by "intuition". – Eric Oct 29 '16 at 10:48
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@Eric Just solve the first two equations for t and then take the limit seperately – Miksu Oct 29 '16 at 11:26