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I have to calculate these two limits, and have no idea where to start from. Your guidance for how should I start working with it can help me a lot.

1) $\lim\limits_{(x,y,z)\rightarrow (0,0,0)} (1+xyz)^{(x^2+y^2+z^2)^{-1}}$

2)$\lim\limits_{(x,y)\rightarrow (0,0)} \dfrac{4y^2+3xy^2+2x^2}{x^2+2y^2}$

Thanks in advance.

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actually nothing special. –  adamco Apr 29 '12 at 16:09
    
I am sure you can write the second one as $2+3\cdot$ something. What would be the limit of this something? –  Did Apr 29 '12 at 16:33

1 Answer 1

up vote 2 down vote accepted

Hint:

  • Write $$ \lim_{(x,y,z)\to (0,0,0)} (1+xyz)^{(x^2+y^2+z^2)^{-1}} = \lim_{(x,y,z)\to (0,0,0)} \left[(1+xyz)^{\frac{1}{xyz}}\right]^{\frac{xyz}{x^2+y^2+z^2}} $$ The limit of the base behaves like $\lim\limits_{t\to 0}(1+t)^{\frac{1}{t}}$, while for $\lim\limits_{(x,y,z)\to (0,0,0)} {\frac{xyz}{x^2+y^2+z^2}}$, the top has degree three, while the bottom has only degree two, hence the top should be approaching $0$ in a higher order than the bottom, and the limit $\lim\limits_{(x,y,z)\to (0,0,0)} {\frac{xyz}{x^2+y^2+z^2}}$ should be $0$(this step is left for you to argue in a more mathematical way), and the whole limit should be 1. EDIT: you mention you haven't learned polar/spherical coordinates(that would be the easier way), here we could make use of AMGM inequality to get: $$ \lim\limits_{(x,y,z)\to (0,0,0)} {\frac{|xyz|}{x^2+y^2+z^2}} \leq \lim\limits_{(x,y,z)\to (0,0,0)}{\frac{|xyz|}{3\sqrt[3]{x^2 y^2 z^2}}} $$ and evaluating the limit for the right hand side is not that difficult.

  • Write $$ \lim\limits_{(x,y)\to (0,0)} \dfrac{4y^2+3xy^2+2x^2}{x^2+2y^2} = 2 + \lim\limits_{(x,y)\to (0,0)}\frac{3xy^2}{x^2+2y^2} $$ like Didier said in the comment, now the argument is kinda similar, the top again is a one degree higher polynomial than the bottom(use AMGM inequality again for the bottom).

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Wait, the second limit certainly doesn't exist for the reason explained below. (Approach along $y = 0$ and approach along $y = x$.) I also believe the first limit is 1, but can't prove it. –  user29743 Apr 29 '12 at 16:59
    
@countinghaus the second limit does exist, whenever the fraction type limit in 2D has the same degree top and bottom, it is path dependent, however here if top is approaching 0 much faster than the bottom, it is another story. –  Shuhao Cao Apr 29 '12 at 17:02
    
Actually, I think Jon is right here - somehow in my calculations I missed the extra power of $y$ in the cross term. I should know better than to do arithmetic before coffee... –  Steven Stadnicki Apr 29 '12 at 17:07
    
Oh yeah, I fold! –  user29743 Apr 29 '12 at 17:11

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