I have been trying to calculate this limit using Taylor polynomials:
$$ \lim_{x \to 0} \frac{e^{x}-e^{-x}-2x}{\sin(x)-x}.$$
I understand that the main idea is to use the remainders so I'm only left with the remainders and some polynomials, then from this point I execute some algebra manipulations and it becomes clear. However, it feels tricky and technical to me which makes it a little hard. I would like to get some ideas or general tips which can help me to determine where to start, for example, at the case above.
Thanks
One way to present the method is to use increasingly precise Taylor expansions, and continue as long as the limit you obtain is undefined. Let's work it out.
First, let's say you perform Taylor expansions of order $0$. You write $e^x = 1 + O(x)$, $e^{-x} = 1 + O(x)$ and hence $e^x - e^{-x} - 2x = 1 - 1 + O(x) = 0 + O(x)$. For the denominator, $\sin(x) = 0 + O(x)$, so $\sin(x) - x = O(x)$. Hence, you end up with $$ \lim_{x\to0} \frac{0+O(x)}{0+O(x)} = ?$$ This limit is undefined, it could be anything. Your expansions were not precise enough to determine the relative behavior of both terms.
So we move to order $1$ Taylor expansions. Working it out, $$\begin{split} e^{x} - e^{-x} - 2x & = (1+x+O(x^2)) - (1-x+O(x^2)) - 2x \\ & = (1-1) + (x-x) + O(x^2) \\ & = 0 + O(x^2). \end{split} $$ And $\sin(x) - x = (x+O(x^2)) - x = 0 + O(x^2)$. Hence you still have an undetermined form $0/0$.
Similarly, at order $2$, we still have $0/0$ (try it out).
Eventually, with order $3$ Taylor-expansion, $e^{x} - e^{-x} - 2x = 2 \frac{x^3}{3!} + O(x^4)$ and $\sin(x) - x = -\frac{x^3}{3!} + O(x^4)$. Hence we obtain $$ \lim_{x \to 0} \frac{2 \frac{x^3}{3!} + O(x^4)}{-\frac{x^3}{3!} + O(x^4)} = \lim_{x \to 0} \frac{2+O(x)}{-1 + O(x)} = - 2. $$