Limit of $\lim_{x \rightarrow 0} \frac{\sin xy^2}{x}$ Limit of $$\lim_{x \rightarrow 0} \frac{\sin xy^2}{x}$$
I know (thanks to wolfram) it is equal to $y^2$, but i do not know how to show that.
 A: For fixed $y\ne 0$, change the variable to $u=xy^2$. This transforms your limit to
$$ \lim_{u\to 0}\frac{\sin u}{uy^{-2}} = \frac{1}{y^{-2}}\lim_{u\to 0}\frac{\sin u}{u} = y^2 \lim_{u\to 0}\frac{\sin u}{u}$$

Alternatively, if you know the standard calculus toolkit, you can brute-force it using l'Hospital, which amounts to recognizing the limit as the derivative of $f(x)=\sin xy^2$ at $x=0$. You can then differentiate $f$ symbolically using the chain rule, getting $f'(x) = y^2 \cos xy^2$ -- which, when you plug in $x=0$, yields $f'(0)=y^2$.
A: The simplest (and shortest) is with equivalents:
$$\sin ax\sim_0 ax,\enspace\text{hence}\quad \frac{\sin ax}x\sim_0\frac{ax}x=a.$$
A: Provided you are aware of continuity of polynomials in $\mathbb R^n$ you can plug $t:=xy^2$
The limit then becomes $\lim_{t \rightarrow 0} \dfrac {\sin t}{t}y^2$
A: Hint:
\begin{align*}
\frac{\sin xy^2}{x}=y^2\frac{\sin xy^2}{xy^2}
\end{align*}
now take the limit!
A: for $f(x,y)=\sin xy^2$, it is $f_x(0,y)=y^2 cos 0y^2=y^2$
A: Just remember that
$$\lim_{\text{a number}\to0} \frac{\sin(\text{that number})}{\text{that very same number}} = 1$$
