## Thomas' Calculus 13th Edition

We know $\lim\limits_{\theta \to 0} \frac{\mathrm{sin}\left(\theta \right)}{\theta }=1$. Thus for any nonzero real number $a$, it follows that $\underset{\theta \to 0}{\lim}\frac{\mathrm{sin}\left(a\theta \right)}{a\theta }=1$. To see this, we let $x=a\theta$. Then since $x \to 0$ as $\theta \to 0$, we have $\underset{\theta \to 0}{\lim}\frac{\mathrm{sin}\left(a\theta \right)}{a\theta }=\underset{x\to 0}{\lim}\frac{\mathrm{sin}\left(x\right)}{x}=1$. So, to find $\underset{\theta \to 0}{\lim}\frac{\mathrm{sin}\left(\sqrt{2}\theta \right)}{\sqrt{2}\theta }$, we let $x=\sqrt{2}\theta$. Then $\underset{\theta \to 0}{\lim}\frac{\mathrm{sin}\left(\sqrt{2}\theta \right)}{\sqrt{2}\theta }=\underset{x\to 0}{\lim}\frac{\mathrm{sin}\left(x\right)}{x}=1$.