I want to know when can I replace $\sin(x)$ by $x$ when $x$ approaches zero. My teacher taught me that when $x$ approaches zero, $\sin(x)/x = 1$. Now, in large expressions of the form $\sin(x)\cdot f(x)$ or $\sin(x) + f(x)$, when can I replace $\sin(x)$ by $x$.
If possible, please provide me more material on where can I learn about these shortcuts.
On
No it is not possible in general make this kind of substitution without take the all limit at once. We only can when the first order term is the one which really determine the limit and in any case always using little-o or Big-O notation in order to handle properly the remainders for higher order terms.
Take also a look here for an example and some references Problem with limit solving.
On
You can always do this:
$$\lim_{x\rightarrow 0} (\sin{(x)} \cdot f(x))= \lim_{x\rightarrow 0} (\frac{\sin{(x)}}{x}\cdot x \cdot f(x))$$ If $$\lim_{x\rightarrow 0} ( x \cdot f(x)) = a \Rightarrow \lim_{x\rightarrow 0} (\sin{(x)} \cdot f(x))= \lim_{x\rightarrow 0} (\frac{\sin{(x)}}{x}\cdot x \cdot f(x))= 1\cdot a = a$$
On
If $f\sim g$, i.e. $\lim\limits_{x\to somewhere}\frac fg=1$ then for any $\alpha\in \mathbb R$ we have $ f^\alpha h= \left(\frac{f}{g}\right)^\alpha\cdot g^\alpha h$ and so $ \lim f^\alpha h= \lim g^\alpha h$ whenever one of the limits exists.
So we can replase $f$ by $g$ in such expressions without changing the result. Such replacement in the sums there is no justification except the desire and may lead to wrong results.
Let $f=x+1,\,\, g=x$ then $f\sim g$ when $x\to \infty$. Try to replase $f$ by $g$ in the $\lim\limits_{x\to\infty} (f-g).$ :)
For small $x$, $\sin x$ is indeed "of order" $x$. So you can replace it by $x$ unless it is canceled by another term of order $x$ (when subtracted).
For instance, $e^x-1$ is also of order $x$ and
$$\frac{\sin x}{e^x-1}\to\frac xx=1$$ is valid.
And
$$\sin x-(e^x-1)\to x-x=0$$ is not because the cancellation exposes residual lower order terms.