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Find the limit $$\lim_{(x,y) \to (0,0)}\frac{x^3\sin(x+y)}{x^2+y^2}.$$ How exactly can I do this? Thanks.

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2 Answers

The first thing is to say that $|\mathrm{sin}(x + y)| \leq 1$.

The second thing is to use polar coordinates. The expression will reduce to $r \cos(\theta)^3$ with $r \to 0$ and so the limit is 0.

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What if sin(x+y) was ln(x+y), for instance? –  user59917 Feb 3 '13 at 14:17
    
@user59917: Then it would certainly diverge because the denominator goes to $0$ and the numerator goes to $-\infty$, so it diverges very quickly. –  Clayton Feb 3 '13 at 14:18
    
@user59917: Sounds like you might need to ask another question as opposed to putting it in a comment... –  Clayton Feb 3 '13 at 14:34
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Using $|\sin z|\leq 1$ we find that the absolute value of your function is not greater than $$ \frac{|x^3|}{x^2+y^2}\leq \frac{|x^3|}{x^2}=|x|. $$ This is first when $x\neq 0$. Then observe that $|f(x,y)|\leq |x|$ also holds when $x=0$.

Can you conclude?

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