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How to prove that: $$\lim_{(x,y) \rightarrow (0,0)} (x+y)\sin(\frac{1}{x})\sin(\frac{1}{y})=0$$

I want to know how to do it. I'd appreciate it a lot.

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  • $\begingroup$ You should write the limit properly. $\endgroup$
    – user
    Commented Nov 7, 2018 at 13:52

2 Answers 2

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So we have, $$ f(x,y)=(x+y)\sin(\frac{1}{x})\sin(\frac{1}{y})$$ then, using the fact that $\forall u\in\Bbb{R},|\sin(u)|<1$, $$ 0<|f(x,y)|=|(x+y)\sin(\frac{1}{x})\sin(\frac{1}{y})|\leq|x+y|$$ Thus, $$ 0<|f(x,y)|\leq |x+y|\xrightarrow[(x,y)\rightarrow(0,0)]{}0$$ By the sandwich theorem we have the answer requested.

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    $\begingroup$ In general, in such cases, a good hint would be preferable to a full solution. $\endgroup$
    – user
    Commented Nov 7, 2018 at 14:33
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\lim_{x,y\to 0,0}(x+y)\sin1/\xsin1/y=0 \lim_{x\to 0}(x+y)[cos(x-y)/cosxcosy]=0 appling limit:- \lim_{x\to 0}(x+y)][\lim: \lim_{x\to 0}[\cos(x-y)/\cosxcosy]]=0 [0+0][cos(0-0)/cos0cos0]=0 0*[cos0/1*1]=0 0*[1/1]=0 0=0 proved

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  • $\begingroup$ iIt would be better if you use MathJax to write math formulas. $\endgroup$
    – Varazda
    Commented Nov 7, 2018 at 14:34
  • $\begingroup$ Visit following link for MathJax tutorial math.meta.stackexchange.com/questions/5020/… $\endgroup$
    – Mayuresh L
    Commented Nov 7, 2018 at 14:37
  • $\begingroup$ Thank for reminder $\endgroup$
    – Kizaru
    Commented Nov 7, 2018 at 14:52

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