<s>[INST] 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?
 [/INST] 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.
</s>