I was preparing for a calculus test tomorrow on limits and continuity and I came upon this problem while I was practicing. I have no idea how to attempt it, so any help would be appreciated.
The only thing I really got is cos(0) = 1
Ah, thanks so much!! Looks like I need to review my trig identities more.
Best Answer
The expression is equivalent to $$\lim_{\theta \rightarrow 0}\frac{\tan \theta \tan{2\theta}}{\theta^2}.$$ Using double angle identities simplies the expression to $$\lim_{\theta \rightarrow 0}\frac{2\tan^2\theta}{\theta^2(1-\tan^2 \theta)} = \lim_{\theta \rightarrow 0}\frac{2\frac{\sin^2\theta}{\cos^2\theta}}{\theta^2(\frac{\cos^2\theta - \sin^2\theta}{\cos^2\theta})} = \lim_{\theta \rightarrow 0}\frac{\sin^2\theta}{\theta^2}\cdot\frac{2}{\cos^2\theta - \sin^2\theta}$$
$$=\lim_{\theta \rightarrow 0}\frac{\sin^2\theta}{\theta^2}\cdot\lim_{\theta \rightarrow 0}\frac{2}{\cos^2\theta - \sin^2\theta} = 1\cdot 2 = 2.$$
The separation into product of limits is rigorous because both limits exist. Note that $\frac{\sin^2\theta}{\theta^2} = \frac{\sin\theta}{\theta}\cdot\frac{\sin\theta}{\theta}$.