I've seen a the argument that $\sin x\approx x$ when $x\to0$ on this site many times, Thinking about this, would the following be true, and how would it be proved?
$$\lim_{x\to0}f(x, \sin x)=\lim_{x\to0}f(x, x)$$
Where $f(x,y)$ is some function of two variables.
What if $f(x,y)$ is continuous in $(0,0)$?
In general no. For example, take $$f(u,v) = \frac{u-v}{u^3}$$
then $$\lim_{x \to 0} f(x,x) = 0 \neq \frac{1}{6} = \lim_{x \to 0} f(x,\sin x) $$
What you need is that $f$ is continuous in $(0,0)$ to ensure that the two limits are equal. Infact, suppose $f$ is continuous in $(0,0)$ and call $f(0,0) = a$.
Since the functions $g,h : \mathbb{R} \longrightarrow \mathbb{R}^2$ defined by
$$g(x) = (x,x)$$ and $$h(x) = (x, \sin x)$$ are continuous, then $ f \circ g$ and $f \circ h$ are continuous, so $$\lim_{x \to 0} f(h(x)) = f(0, \sin 0 ) = f(0,0) = \lim_{x \to 0} f(g(x))$$
If $f$ is continuous in $(0,0)$, $\forall (x_n,y_n)\to (0,0),f(x_n,y_n)\to f(0,0)$, hence the two limits are equal.
