how do I calculate this limit?

Question

Lim             (sin(x^2 + y^2))/[1  -  cos((x^2 + y^2)^(1/2))]
(x,y)->(0,0)

I ran it through WolframAlpha and i know it's 2, but I just can't figure out why

No Mythology · Accepted Answer

I'd recommend using polar coordinates. If you take x = rcosΘ and y = rsinΘ, and don't place any restrictions on Θ or how r may depend on Θ, then all possible paths (x, y) -> (0, 0) correspond to r -> 0. 

lim sin(x²+y²)/(1 - cos(√(x² + y²)) = 
(x,y)->(0,0)

lim sin(r²)/(1 - cos(r)) = 
r->0

lim  2r cos(r²)/sin(r) = lim 2cos(r²) (r/sin(r)) = 2(1)(1) = 2.
r->0 . . . . . . . . . . . . . r-> 0

The third line follows by applying L'Hopital's rule, and the last equality follows from the observation that sin(t)/t tends to 1 as t -> 0.

how do I calculate this limit?

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