Evaluating $ \lim_{x\to 0} ({\frac{1}{1-\cos(x)}} - {\frac{2}{x^2}})$ without L'hopital's rule?

Question

$$ \lim_{x\to 0} ({\frac{1}{1-\cos(x)}} - {\frac{2}{x^2}})$$

I know I can so some L'hopital's rule and get the answer, but is there another way, more clever? Like with Taylor expansion? I tried it and got $\infty*0$

Answer 1

By replacing $x$ with $2z$, we are left with:

$$\begin{eqnarray*} \frac{1}{2}\cdot\lim_{z\to 0}\left(\frac{1}{\sin^2 z}-\frac{1}{z^2}\right)&=&\frac{1}{2}\lim_{z\to 0}\frac{(z-\sin z)(z+\sin z)}{z^4+o(z^4)}\\&=&\frac{1}{2}\lim_{z\to 0}\frac{\left(\frac{z^3}{6}+o(z^3)\right)(2z+o(z))}{z^4+o(z^4)}\\&=&\frac{1}{2}\cdot\frac{1}{3}=\color{red}{\frac{1}{6}}.\end{eqnarray*}$$

Answer 2

A convenient notation that I will use here: $o(f(x))$ is a symbol for an unspecified function such that $\lim_{x \to 0} \frac{o(f(x))}{f(x)}=0$. For example, $o(x^2)$ could stand for $x^3$. This is called "little oh notation".

With this included, let's start with some algebra:

$$\frac{1}{1-\cos(x)}-\frac{2}{x^2}=\frac{x^2-2+2\cos(x)}{x^2(1-\cos(x))}.$$

The idea: determine what order the denominator vanishes to, and then calculate the numerator to that order. The denominator turns out to vanish to order $4$, since both $x^2$ and $1-\cos(x)$ vanish to order $2$. So the numerator, computed to order $4$, is

$$x^2-2+2(1-x^2/2+x^4/24+o(x^4))=x^2-2+2-x^2+x^4/12+o(x^4)=x^4/12+o(x^4).$$

The denominator, computed to order $4$ (i.e. to leading order) is $\frac{x^4}{2}+o(x^4)$. So the ratio is

$$\frac{x^4/12+o(x^4)}{x^4/2+o(x^4)}.$$

Can you see how to finish the argument from here? Hint: the way you finish it is the same as it would be if you replaced the $o(x^4)$ with $x^5$.

What we're doing here is actually very similar to L'Hopital's rule; in fact this produces a proof that if you apply L'Hopital's rule to that function 4 times, you'll get the limit.

Answer 3

$$ \lim_{x\to 0} {\frac{1}{1-\cos(x)}} - {\frac{2}{x^2}}=\lim_{x\to 0} {{x^2-2+2 \cdot \cos(x)} \over {(1-\cos(x)) \cdot x^2}}$$

Take advantage of the Taylor/Maclurian expansion for cosine and substitute for each instance of cosine. Note that some of the terms cancel.

$$\lim_{x\to 0} {{x^4/12+O(x^6)} \over {x^4/2+O(x^6)}}={1 \over 6}$$

Answer 4

An alternative view is the following.

\begin{align} \frac{1}{1- \cos x} - \frac{2}{x^{2}} &= \frac{1}{1 - \left(1 - \frac{x^{2}}{2!} - \frac{x^{4}}{4!} + \cdots \right)} - \frac{2}{x^{2}} \\ &= \frac{2}{x^{2}} \, \left[ \frac{1}{1 - \frac{2 \, x^{2}}{4!} + \frac{2 \, x^{4}}{6!} - \cdots } - 1 \right]. \end{align} Now, by "long division" it is seen that \begin{align} \frac{1}{1 - \frac{2 \, x^{2}}{4!} + \frac{2 \, x^{4}}{6!} - \cdots } = 1 + \frac{2 \, x^{2}}{4!} + \frac{3 \, x^{4}}{6!} + \mathcal{O}(x^{6}) \end{align} for which \begin{align} \frac{1}{1 - \cos x} - \frac{2}{x^{2}} &= \frac{2}{x^{2}} \, \left[ \left( 1 + \frac{2 \, x^{2}}{4!} + \frac{3 \, x^{4}}{6!} + \mathcal{O}(x^{6}) \right) - 1 \right] \\ &= \frac{1}{3!} \, \left[ 1 + \frac{x^{2}}{20} + \mathcal{O}(x^{4}) \right]. \end{align} By taking the limit as $x \to 0$ leads to \begin{align} \lim_{x \to 0} \left\{ \frac{1}{1- \cos x} - \frac{2}{x^{2}} \right\} = \frac{1}{3!}. \end{align}