What is $\lim_\limits{x\to 0}\frac{x}{1-\cos{x}}$ equal to? 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?
 A: 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.
A: 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.
A: $\forall x>0: \\\qquad\displaystyle 0\le 1-\cos(x)=\int_0^x \sin(t)\, dt\le \int_0^x t\,dt\le \frac {x^2}2\implies\dfrac{x}{1-\cos(x)}\ge \dfrac{2x}{x^2}\ge \dfrac 2x\to +\infty$
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 ]
A: 
  
*
  
*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
$$\lim\limits_{x \to 0^+} \frac{x}{1 - \cos x}=\lim\limits_{x \to 0^+} \frac{1}{f(x)}=\infty\neq -\infty=\lim\limits_{x \to 0^-} \frac{1}{f(x)}=\lim\limits_{x \to 0^-} \frac{x}{1 - \cos x}.\qquad \square$$
