Limit of multivariable function

**Tascja** · Apr 7th 2009, 12:15 PM

The question asks to find the limit (x,y) -> (0,0) and this is the function:
$\frac{x^2 sin(y)^2}{x^2 + 5y^2}$
so i tried approaching (0,0) from y=cx and i get this:
$\frac{x^2 sin(cx)^2}{x^2(1 + 5c)}$
but im not quite sure where to go from here??
an explanation would be greatly appreciated

**thelostchild** · Apr 7th 2009, 01:23 PM

This may not be correct in the slightest so please don't take my word for it

First of all you made a slight mistake it should be

$\frac{x^2 \sin^2 (cx)}{x^2(1+5c^2)}$

so

$=\frac{\sin^2 (cx)}{1+5c^2}$ (you can just cancel the x^2's

$=\frac{(cx-\frac{cx^3}{6}+ \ldots)^2}{1+5c^2}$ (the taylor expansion of sin is probably unnecessary!)

hence as x goes to 0 (and so y goes to 0 along the line y=cx) the limit approaches 0 for $c \neq 0$

**Tascja** · Apr 7th 2009, 02:01 PM

thank you for catching my mistake

just one question: if i had just assumed that since as x approaches 0, sin(0)=0 .. so the limit =0, would that be alright??

**thelostchild** · Apr 8th 2009, 01:30 AM

yea it would be fine!

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