Thread: Limit of multivariable function

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

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

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

so

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

$\displaystyle =\frac{(cx-\frac{cx^3}{6}+ \ldots)^2}{1+5c^2}$ (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 $\displaystyle c \neq 0$

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??

yea it would be fine!