What is $\lim_\limits{x\to 0}\frac{x}{1-\cos{x}}$ equal to?

Question

At 29:30 in lecture 8 of UMKC's Calculus I course, the instructor makes the claim that its limit is equal to zero mentioning that he proved this result earlier in the lecture. The thing is that I watched the entire lecture and yet never actually saw him do that. Here's my result:

\begin{align} \lim_{x\to 0}\frac{x}{1-\cos{x}} &=\lim_{x\to 0}\frac{x}{1-\cos{x}}\\ &=\lim_{x\to 0}\left(\frac{x}{1-\cos{x}}\cdot\frac{1+\cos{x}}{1+\cos{x}}\right)\\ &=\lim_{x\to 0}\frac{x}{1-\cos^2{x}}\cdot\lim_{x\to 0}(1+\cos{x})\\ &=2\cdot\lim_{x\to 0}\frac{x}{\sin^2{x}}\\ &=2\cdot\lim_{x\to 0}\frac{x}{\sin{x}}\cdot\lim_{x\to 0}\frac{1}{\sin{x}}\\ &=2\cdot\lim_{x\to 0}\frac{1}{\frac{\sin{x}}{x}}\cdot\lim_{x\to 0}\frac{1}{\sin{x}}\\ &=2\cdot\frac{\lim_\limits{x\to 0}1}{\lim_\limits{x\to 0}\frac{\sin{x}}{x}}\cdot\lim_{x\to 0}\frac{1}{\sin{x}}\\ &=2\cdot\frac{1}{1}\cdot\lim_{x\to 0}\frac{1}{\sin{x}}\\ &=2\cdot\lim_{x\to 0}\frac{1}{\sin{x}} \end{align}

$$ \lim_{x\to 0^{-}}\frac{1}{\sin{x}}=-\infty \text{ and} \lim_{x\to 0^{+}}\frac{1}{\sin{x}}=\infty $$

Thus, the limit does not exit. But the instructor clearly says that it does. What's wrong with him?

Answer 1

Write $\dfrac{x}{1-\cos x} = \dfrac{x(1+\cos x)}{1-\cos^2 x}= \dfrac{x^2(1+\cos x)}{x\sin^2 x}= \left(\dfrac{x}{\sin x}\right)^2\cdot \dfrac{1+\cos x}{x}$. The limit does not exist because $\dfrac{1+\cos x}{x} \to +\infty$ and $-\infty$ when $x \to 0^{+}, 0^{-}$ respectively.

Answer 2

Note that $$\lim_{x \to 0} \frac{x}{1-\cos(x)}=\lim_{x \to 0} \left[\frac{x}{1-\cos(x)}\cdot \frac{1+\cos(x)}{1+\cos(x)} \right]=\lim_{x \to 0} \frac{x[1+\cos(x)]}{1-\cos^2(x)}=\lim_{x \to 0} \frac{x[1+\cos(x)]}{\sin^2(x)}.$$ Now, using the fact that $$\lim_{x \to 0} \frac{\sin(x)}{x}=1=\lim_{x \to 0} \frac{x}{\sin(x)}$$ tehn the limit is equal to $$\lim_{x \to 0} \frac{1+\cos(x)}{\sin(x)}.$$ Note that $\displaystyle \lim_{x \to 0^-} \dfrac{1+\cos(x)}{\sin(x)}=-\infty$ and $\displaystyle \lim_{x \to 0^+} \dfrac{1+\cos(x)}{\sin(x)}=+\infty$. Hence the two sided-limits are distinct and therefore the limit doesn't exist.

Answer 3

$\forall x>0: \\\qquad\displaystyle 0\le 1-\cos(x)=\int_0^x \sin(t)\, dt\le \int_0^x t\,dt\le \frac {x^2}2\implies\dfrac{x}{1-\cos(x)}\ge \dfrac{2x}{x^2}\ge \dfrac 2x\to +\infty$

And since $\dfrac{x}{1-\cos(x)}$ is an odd function, the limit for $x<0$ is $-\infty$.

Thus there is no limit in $0.\quad$ [ since the limits in $0^-$ and $0^+$ disagree ]

Answer 4

• If $f(x)>0$ to the right of $a\in\mathbb R$ and $f(x)\overset{x\to a^+}\longrightarrow 0$, then $$\lim_{x\to a^+}\frac{1}{f(x)}=\infty.$$

Proof: Given $M>0$ there exists $\delta>0$ such that $$\begin{align*}a<x<a+\delta\quad &\Longrightarrow \quad |f(x)-0|<\frac{1}{M}\\ \\ &\Longrightarrow \quad \frac{1}{f(x)}=\frac{1}{|f(x)-0|}>M. \qquad\square \end{align*}$$

• If $f(x)<0$ to the left of $a\in\mathbb R$ and $f(x)\overset{x\to a^-}\longrightarrow 0$, then $$\lim_{x\to a^-}\frac{1}{f(x)}=-\infty.$$

Proof: Given $M>0$ there exists $\delta>0$ such that $$\begin{align*}a-\delta<x<a\quad &\Longrightarrow \quad |f(x)-0|<\frac{1}{M}\\ \\ &\Longrightarrow \quad \frac{1}{f(x)}=-\frac{1}{|f(x)-0|}<-M. \qquad \square \end{align*}$$

• The limit $\lim\limits_{x \to 0} \frac{x}{1 - \cos x}$ does not exist.

Proof: Recall that $\frac{1-\cos(x)}{x}\overset{x\to0}{\longrightarrow} 0$. Thus, taking $f(x)=\frac{1-\cos(x)}{x}$ and $a=0$ we conclude from the above properties that $$\lim\limits_{x \to 0^+} \frac{x}{1 - \cos x}=\lim\limits_{x \to 0^+} \frac{1}{f(x)}=\infty\neq -\infty=\lim\limits_{x \to 0^-} \frac{1}{f(x)}=\lim\limits_{x \to 0^-} \frac{x}{1 - \cos x}.\qquad \square$$