Taylor expansion - what order would be preferred? Let say you want to calculate the following limit:  
$$\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{{1 - \cos x}}\ln \left( {\frac{{\sin x}}{x}} \right)} \right)$$
Obviously, Taylor Expansion can comes in handy here.
But how do you decide the order of the expansion (error)? For this question, and for a general case.  
Thanks!
 A: The order of magnitude must of course be chosen so as not to end up where you started, i.e., with an indeterminate expression on your hands.
$\qquad\sin x\simeq x-\dfrac{x^3}6\iff\dfrac{\sin x}x\simeq1-\dfrac{x^2}6\iff\ln\dfrac{\sin x}x\simeq\ln\bigg(1-\dfrac{x^2}6\bigg)\simeq-\dfrac{x^2}6$
$\qquad\cos x\simeq1-\dfrac{x^2}2\iff1-\cos x\simeq\dfrac{x^2}2$ . I think you can take it from here.
A: $$\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{{1 - \cos x}}\ln \left( {\frac{{\sin x}}{x}} \right)} \right)=\mathop {\lim }\limits_{x \to 0}\ln \left( {\frac{{\sin x}}{x}} \right)^{\frac{1}{1-\cos x}}=\ln \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin x}}{x}} \right)^{\frac{1}{1-\cos x}}=$$$$ =\ln e^{\mathop{\lim }\limits_{x \to 0}(\frac{\sin x}{x}-1)\frac{1}{1-\cos x}}= {\mathop{\lim }\limits_{x \to 0}(\frac{\sin x}{x}-1)\frac{1}{1-\cos x}} = {\mathop{\lim }\limits_{x \to 0}\frac{\sin x -x}{x}\cdot\frac{1}{1-\cos x}}=$$
$$={\mathop{\lim }\limits_{x \to 0}\frac{\sin x -x}{x^3}\cdot\frac{x^2}{1-\cos x}}=
={\mathop{\lim }\limits_{x \to 0}\frac{(\sin x -x)'}{(x^3)'}\cdot\frac{x^2}{1-\cos x}}=$$
$$={\mathop{\lim }\limits_{x \to 0}\frac{\cos x  -1}{3x^2}\cdot\frac{x^2}{1-\cos x}}=-\frac{1}{3}$$
We applied "the shortcut" for $1^\infty$: $\lim_{x \to a } (f(x))^{g(x)}= e^{\lim_{x \to a }(f(x)-1)\cdot g(x)}$
See and Finding the limit $\displaystyle\lim_{x\to 0+} \left(\frac{\sin x}x\right)^{1/{x^2}}$
A: All that you need for this one is the second order expansion.
Start with $\frac{1}{1-\cos x}$
This is the geometric series of $\cos x$, so you can expand $\cos x$ out into it's usual series and plug that into the geometric series:
$$\frac{1}{1-\cos x}=\frac{1}{6} + \frac{2}{x^2} + HOT$$
Then, start work at $\ln(\frac{\sin x}{x})$:
$$\ln \frac{\sin x}{x}= -\frac{1}{3!}x^2 + HOT$$
Multiplying these two together yields:
$$-\frac{2}{x^2}\cdot\frac{x^2}{3!}$$
Which is simply $-\frac{1}{3}$
