Calculating the limit of $\frac{\cos(x)-1}{x}$ as $x \rightarrow 0$

Question

Show that the $\lim \limits_{x \to 0}\frac{\cos(x)-1}{x}=0$

My attempt

$$\begin{align} \lim \limits_{x \to 0}\frac{\cos(x)-1}{x}*\frac{\cos(x)+1}{\cos(x)+1}&=\lim \limits_{x \to 0}\frac{1-\cos^2(x)}{x(1+\cos(x))}\\\\ &=\lim \limits_{x \to 0}\frac{\sin^2(x)}{x(1+\cos(x))}\\\\ &=\lim \limits_{x \to 0}\frac{\sin(x)}{x}*\frac{\sin(x)}{1+\cos(x)}\\\\ &=1*\frac{0}{2}\\\\ &=0 \end{align}$$ $$QED$$

Since $\lim \limits_{x \to 0}\frac{\sin(x)}{x}=1$ by the sandwich theorem.

i know this is correct however i would like it if anyone could show me the natural argument of using power series instead.

Recall the definition of the cosine function by the power series:

$$\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}x^{2n}, \forall x \in \mathbb R$$

Also can anyone show me how to show that the radius of convergence of the cosine power series is $$\infty$$

Answer 1

If you insist in power series:

$$\cos x=1-\frac{x^2}2+\frac{x^4}{24}-\ldots\implies \cos x-1=x\left(-\frac x{2}+\frac{x^3}{24}-\ldots\right)\implies$$

$$\frac{\cos x-1}x=\left(-\frac x{2}+\frac{x^3}{24}-\ldots\right)=-\frac x2+\mathcal O(x^3)\xrightarrow[x\to0]{}0$$

Answer 2

Note that $\cos(x)=\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n)!}$. Therefore, we can write

$$\frac{\cos(x)-1}{x}=\sum_{n=1}^\infty\frac{(-1)^n x^{2n-1}}{(2n)!} \tag 1$$

Hence, inasmuch as the lowest order power in the series is $1$, all terms approach $0$ as $x\to 0$.

Interestingly, we immediately have from $(1)$ that

$$\lim_{x\to 0}\frac{1-\cos(x)}{x^2}=\frac12$$

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Second Question:

The radius of convergence for the power series for the cosine function can be found using the ratio test. Note that the series converges when

$$\lim_{n\to \infty}\left|\frac{a_{n+1}}{a_n}\right|\le1$$

where $a_n=\frac{(-1)^nx^{2n}}{(2n)!}$. Proceeding we see that

$$\lim_{n\to \infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to \infty}\frac{x^2}{(2n+2)(2n+1)}=0$$

for all $x$. Hence, the radius of convergence is indeed $\infty$.

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ASIDE:

Note that $1-\cos(x)=2\sin^2(x/2)$. Therefore, we can write

$$\frac{\cos(x)-1}{x}=-\underbrace{\left(\frac{\sin(x/2)}{x/2}\right)}_{\to1}\,\underbrace{\sin(x/2)}_{\to0}$$

Answer 3

Hint:

using the series of $\cos x$ you have: $$ \cos x-1 =-1 +\cos x= -1+1-\frac{x^2}{2!}+o(x^3)= -\frac{x^2}{2}+o(x^3) $$