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$$\lim\limits_{(x,y)\rightarrow (0,0)} \dfrac{(x^3-y^3) \sin (2x^2+3y^2)}{x^2+2y^2}$$

I don't know what to do doesn't it give always $0$?

Whether $x=0$ $x=y$ or $y=0$ or $x=y^2$ it always give $0$ since it goes to $(0,0)$

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But other similar-looking ones might not. Let $x=r\cos\theta$, $y=r\sin\theta$. That will give you a tight argument. Here everything is under control, we don't even care that the sine part goes fast to $0$, because $x^3-y^3$ goes to $0$ faster than the denominator. –  André Nicolas Dec 5 '12 at 23:00
    
are you suggesting i should evaluate (x^3-y^3)/(x^2+2y^2) only? –  question Dec 5 '12 at 23:11
    
@question use @ and the name of the user you want to talk to, if you want him to be notified of your comment –  clark Dec 5 '12 at 23:18
    
Not really. With substitution I suggested (but for brevity I will not do it fully) we get $\frac{(r^3\cos^3\theta-r^3\sin^3\theta)\sin w}{r^2\cos^2\theta+2r^2\sin^2\theta}$. Cancel $r^2$ from top and bottom. The bottom is now $\cos^2\theta+2\sin^2\theta$, bigger than $1$. The top has an $r$ in it, times stuff that can't get big in absolute value. As $(x,y)\to(0,0)$, $r\to 0$, so the limit is $0$. Note: The substitution is often useful. –  André Nicolas Dec 5 '12 at 23:19
    
isn't there a better way? like calculating 3 limits when 2 of them gives 0? –  question Dec 5 '12 at 23:31

2 Answers 2

up vote 2 down vote accepted

HINT

Note that $$-2 \leq \dfrac{\sin(2x^2 + 3y^2)}{x^2 + 2y^2} \leq 2$$ Hence, $$(x^3 - y^3)\dfrac{\sin(2x^2 + 3y^2)}{x^2 + 2y^2} \in \left[-2\vert(x^3-y^3) \vert, 2\vert(x^3-y^3)\vert \right]$$

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isn't there any other way? –  question Dec 5 '12 at 23:09
    
It doesn't seem to be very intuitive... Is there a way that is much simpler? –  question Dec 5 '12 at 23:29

$\left | \dfrac{(x^3-y^3) \sin (2x^2+3y^2)}{x^2+2y^2} \right |\leq \dfrac{|x^3-y^3| (2x^2+3y^2)}{x^2+2y^2} \leq \dfrac{|x^3-y^3| (2x^2+4y^2)}{x^2+2y^2}\leq 2|x^3-y^3|$

Where we used the inequality $|\sin (x)| \leq |x|$

ADDED Let $f(x,y)= x^3-y^3$, now knowing that $ \lim _{ (x,y) \rightarrow (0,0)} f(x,y)= 0$

It means that for $\epsilon >0$ there is a $\delta >0$ such that $ \forall (x,y) \in \{ (x,y ) | \sqrt{ x^2+y^2} < \delta \}$ we have $ | x^3-y^3| < \frac{ \epsilon }{2}$. Therefore for that $\delta >0$ we have $\left |\dfrac{(x^3-y^3) \sin (2x^2+3y^2)}{x^2+2y^2} \right | < \epsilon$, whenever $ (x,y) \in \{ (x,y ) | \sqrt{ x^2+y^2} < \delta \}$. Since $\epsilon $ was arbitary we conclude be definition of the limit that

$ \lim _{ (x,y) \rightarrow (0,0)} \dfrac{(x^3-y^3) \sin (2x^2+3y^2)}{x^2+2y^2}=0$.

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isn't there an easy way to see this? –  question Dec 5 '12 at 23:10
    
are you suggesting to calculate the limit of all 3 expressions? –  question Dec 5 '12 at 23:31
    
you only need that $2|x^3-y^3| \rightarrow 0$ then $ \left | \dfrac{(x^3-y^3) \sin (2x^2+3y^2)}{x^2+2y^2} \right | \rightarrow 0$ from squeeze theorem –  clark Dec 5 '12 at 23:35
    
can you do it the typical way? where you replace x or y and turn it into a single variable limit? it doesn't seem to be working here. –  question Dec 5 '12 at 23:48

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