# 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!

• Seems like a classical application of l'Hospital. Also you could consider $$\ln\left( \frac{\sin x}x\right) = \ln\sin x - \ln x$$ – AlexR Jan 26 '14 at 23:52
• @AlexR, I agree, tough I am not looking for a solution for the limit which used only for demonstration. The main issue is: When you choose evaluating a limit using Taylor expansion, what order would be preferred? – SuperStamp Jan 26 '14 at 23:58
• The order which requires the less calculations, I'd say.. – Berci Jan 27 '14 at 0:00
• Since the functions are even, only an even-order expansion makes sense, because this gifts us the first (odd-order) error term as $0$. In other words: you only need to calculate even-order derivatives up to a certain degree, say $2n$, and your error is automatically $\mathcal O(h^{2n+2})$ – AlexR Jan 27 '14 at 0:03
• (As for the specific example, note that $\frac{\sin x}{x}$ gives you $1 + \text{small}_1$, which the $\ln$ transforms into $\text{small}_1$, so you need to know what the $\text{small}_1$ term is. Similarly $\cos x = 1 - \text{small}_2$ but because of the $1 - \cos x$ you will need to know what $\text{small}_2$ is.) – TMM Jan 27 '14 at 0:07

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.

$$\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{{1 - \cos x}}\ln \left( {\frac{{\sin x}}{x}} \right)} \right)=\mathop {\lim }\limits_{x \to 0}\ln \left( {\frac{{\sin x}}{x}} \right)^{\frac{1}{1-\cos x}}=\ln \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin x}}{x}} \right)^{\frac{1}{1-\cos x}}=$$$$=\ln e^{\mathop{\lim }\limits_{x \to 0}(\frac{\sin x}{x}-1)\frac{1}{1-\cos x}}= {\mathop{\lim }\limits_{x \to 0}(\frac{\sin x}{x}-1)\frac{1}{1-\cos x}} = {\mathop{\lim }\limits_{x \to 0}\frac{\sin x -x}{x}\cdot\frac{1}{1-\cos x}}=$$ $$={\mathop{\lim }\limits_{x \to 0}\frac{\sin x -x}{x^3}\cdot\frac{x^2}{1-\cos x}}= ={\mathop{\lim }\limits_{x \to 0}\frac{(\sin x -x)'}{(x^3)'}\cdot\frac{x^2}{1-\cos x}}=$$ $$={\mathop{\lim }\limits_{x \to 0}\frac{\cos x -1}{3x^2}\cdot\frac{x^2}{1-\cos x}}=-\frac{1}{3}$$ We applied "the shortcut" for $1^\infty$: $\lim_{x \to a } (f(x))^{g(x)}= e^{\lim_{x \to a }(f(x)-1)\cdot g(x)}$

• There is problem in the step where you create $x^{3}$ factor in denominator and $x^{2}$ in numerator and then apply LHR. Suppose that we don't create such factors and then apply LHR. Then $\dfrac{\sin x - x}{x}$ becomes $\dfrac{\cos x - 1}{1}$ and then we get the limit as $\dfrac{\cos x - 1}{1}\cdot\dfrac{1}{1 - \cos x} = -1$ – Paramanand Singh Jan 27 '14 at 8:16

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 $-\frac{1}{3}$