$lim_{(x,y)to(0,0)}frac{x^2y^2}{sin(x)cos(y)}$ is this done correctly?
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$lim_{(x,y)to(0,0)}frac{x^2y^2}{sin(x)cos(y)}$ is it allowed to split a multi-variable limit into its component variables as in the next step? $= (lim_{xto0}frac{x^2}{sin(x)})(lim_{xto0}frac{y^2}{sin(y)})$ this is an indeterminate form and now I use L'Hopital $=(lim_{xto0}frac{2x}{cos(x)})(lim_{xto0}frac{2y}{cos(y)})$ $=(frac{0}{1})(frac{0}{1})=0$
calculus limits multivariable-calculus
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edited Nov 23 at 15:27
Martin Sleziak
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asked Feb 7 '16 at 8:52