I'm learning how to evaluate limits of trigonometric functions. This is my attempt to solve $$\lim\limits_{\theta\to0} \dfrac{\theta^2}{1-\cos\theta}$$
The answer should be $2$. Can anyone confirm if I did it right?
If not, can you please explain me how to do it?
\begin{align*}
\lim_{\theta\to0}\frac{\theta^2}{1-\cos\theta}&=\lim_{\theta\to0}\frac{\theta^2}{1-\cos\theta} \\
&=\lim_{\theta\to0}\frac{\theta^2}{1-\cos\theta}\frac{1+\cos\theta}{1+\cos\theta} \\
&=\lim_{\theta\to0}\frac{\theta^2(1+\cos\theta)}{1-\cos^2\theta} \\
&=\lim_{\theta\to0}\frac{\theta^2(1+\cos\theta)}{\sin^2\theta} \\
&=\lim_{\theta\to0}\frac{\theta}{\sin\theta}\cdot\frac{\theta}{\sin\theta}\cdot\frac{(1+\cos\theta)}{1} \\
&=\lim_{\theta\to0}\csc^2\theta\cdot(1+\cos\theta) \\
&=\lim_{\theta\to0}\csc^20\cdot(1+\cos0) \\
&=\lim_{\theta\to0}1+1=2 \\
\end{align*}
Best Answer
Your work is correct up to the step $$\lim_{\theta \to 0} \frac{\theta^2}{1 - \cos\theta} = \lim_{\theta \to 0} \frac{\theta}{\sin\theta} \cdot \frac{\theta}{\sin\theta} \cdot \frac{1 + \cos\theta}{1}$$ However, as Gary pointed out in the comments, $\csc\theta = \dfrac{1}{\sin\theta}$ rather than $\dfrac{\theta}{\sin\theta}$.
As Gary also pointed out in the comments, $$\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$$ from which it follows that $$\lim_{\theta \to 0} \frac{\theta}{\sin\theta} = 1$$ Therefore, \begin{align*} \lim_{\theta \to 0} \frac{\theta^2}{1 - \cos\theta} & = \lim_{\theta \to 0} \frac{\theta}{\sin\theta} \cdot \frac{\theta}{\sin\theta} \cdot \frac{1 + \cos\theta}{1}\\ & = \lim_{\theta \to 0} \frac{\theta}{\sin\theta} \cdot \lim_{\theta \to 0} \frac{\theta}{\sin\theta} \cdot \lim_{\theta \to 0} (1 + \cos\theta)\\ & = 1 \cdot 1 \cdot (1 + 1)\\ & = 2 \end{align*} Since $$\lim_{\theta \to 0} \sin\theta = 0,$$ your work is also incorrect because $$\lim_{\theta \to 0} \csc^2\theta = \lim_{\theta \to 0} \frac{1}{\sin^2\theta} = \infty$$