Can we always replace $\sin x$ with $x$ in limit as $x\to 0$ Let $D\subset \mathbb{R}^2$, $0$ be a limit point of $D$, and $f:D\to \mathbb{R}$ be a function.
How to prove or disprove that $\displaystyle\lim_{x\to 0}f(x,\sin x)=\lim_{x\to 0}f(x, x)$ ?
If it is not true, what are the sufficient and necessary conditions such that $\displaystyle\lim_{x\to 0}f(x,\sin x)=\lim_{x\to 0}f(x, x)$ ?
Thanks in advances.
 A: Idea is to analyze the difference $f(x, \sin x) -  f(x, x) $ which by mean value theorem is equal to $$(\sin x - x) f_{y} (x, \xi) $$ for some $\xi$ between $x$ and $\sin x$. Now you can see that an obvious condition is that $f_{y} (x, x) $ should not diverge too much. More specifically our job is done if $f_{y} (x, x)$ is continuous and satisfies the condition $x^{k} f_{y} (x, x) \to 0$ for some $k\in(-\infty, 3)$.

In general one should resist the temptation to replace $\sin x$ by $x$ while evaluating limit of an expression as $x\to 0$. This is simply because the meaning of statement $$\lim_{x\to 0}\frac{\sin x} {x} = 1\tag{1}$$ is not that you can replace $\sin x$ by $x$ when $x\to 0$ but rather that whenever you see the expression $\lim_{x\to 0}\dfrac{\sin x} {x} $ you can replace it with the symbol $1$. It is much simpler to evaluate limits by a direct use of the limit formula $(1)$ instead of trying to figure out the conditions under which $\sin x$ can be replaced by $x$. 
A: I wonder if the function $$ f(x,y) = \frac{e^{x -y}-1}{x-y}$$ provides a counterexample.
$f$ is not defined for $(x,x)$, so it does not even make sense to talk about the limit $$lim_{x \to 0} f(x,x) $$
Yet, $$lim_{x \to 0} f(x, \sin x) = 1 $$ seems perfectly defined.
