## Thomas' Calculus 13th Edition

We know $$\lim\limits_{\theta \to 0} \dfrac{\mathrm{sin}\left(\theta \right)}{\theta }=1.$$ Thus for any nonzero real number $a$, it follows that $$\underset{\theta \to 0}{\lim}\dfrac{\mathrm{sin}\left(a\theta \right)}{a\theta }=1.$$ We will use this fact to compute the limit below. So to see that this is true, we let $x=a\theta.$ Then since $x \to 0$ as $\theta \to 0$, we have $$\underset{\theta \to 0}{\lim}\dfrac{\mathrm{sin}\left(a\theta \right)}{a\theta }=\underset{x\to 0}{\lim}\dfrac{\mathrm{sin}\left(x\right)}{x}=1.$$ Now, we want to find $$\underset{x\to 0}{\lim}\frac{\mathrm{tan}\left(2x\right)}{x}.$$ Note that $$\frac{\mathrm{tan}\left(2x\right)}{x}=\frac{\left(\frac{\mathrm{sin}\left(2x\right)}{\mathrm{cos}\left(2x\right)}\right)}{x}=\frac{\mathrm{sin}\left(2x\right)}{x\mathrm{cos}\left(2x\right)}=\frac{2\mathrm{sin}\left(2x\right)}{2x\mathrm{cos}\left(2x\right)}=\left(\frac{\mathrm{sin}\left(2x\right)}{2x}\right)\left(\frac{2}{\mathrm{cos}\left(2x\right)}\right).$$ Thus $$\underset{x\to 0}{\lim}\frac{\mathrm{tan}\left(2x\right)}{x}=\underset{x\to 0}{\lim}\left[\left(\frac{\mathrm{sin}\left(2x\right)}{2x}\right)\left(\frac{2}{\mathrm{cos}\left(2x\right)}\right)\right]=\left(1\right)\left(\frac{2}{1}\right)=2.$$