# Trigonometric limit: Dividing the Numerator and Denominator by $\theta\;?$

I can not see how mathematically one can divide the numerator and denominator by theta. Can some one explain this in a different way. When you divide two fractions you multiply the first fraction by the reciprocal of the second. This is not the case here.

Divide numerator and denominator by $\theta$. ($\sin\theta$ also works.) \eqalign{\lim_{\theta\to0}\dfrac{\sin\theta}{8\theta+\tan\theta}&= \lim_{\theta\to0}\dfrac{\dfrac{\sin\theta}\theta}{8+\dfrac{\sin\theta}\theta\cdot\dfrac1{\cos\theta}}\\&=\dfrac{\lim\limits_{\theta\to0}\dfrac{\sin\theta}\theta}{8+\lim\limits_{\theta\to0}\dfrac{\sin\theta}\theta\lim\limits_{\theta\to0}\dfrac1{\cos\theta}}\\&=\dfrac1{8+1\cdot1}\\&=\dfrac19.}

• Are you asking why $\dfrac a b=\dfrac{a/\theta}{b/\theta}$? Commented Oct 11, 2014 at 16:50
• I am asking someone to describe differently how one can divide by theta. The above solved equations makes no sense to me. Commented Oct 11, 2014 at 16:52
• Sorry, I don't understand your question. I have two interpretations for the question. One is "How does one get the inspiration to divide by $\theta$?" and the other is something different which I still can't understand. You can divide because you can divide... you can also add $2$ to the expression if you like, (but that changes the limit). Commented Oct 11, 2014 at 16:55
• Like, $\frac{a/\theta}{b/\theta} = \frac{a}{b} \cdot \frac{1/\theta}{1/\theta} = \frac{a}{b} \cdot 1 = \frac{a}{b}$? Commented Oct 11, 2014 at 16:59

As $\theta \neq 0$[As $\theta \to 0$]. Division by $\theta$ is valid.
$$\frac{\sin \theta}{8\theta+\tan\theta} = \frac{\sin \theta}{8\theta+\frac{\sin \theta}{\cos\theta}}$$
Now dividing numerator and denominator by $\theta$, we have
\begin{align} \frac{\sin \theta}{\left(8\theta+\dfrac{\sin \theta}{\cos\theta}\right)} \cdot\frac{\dfrac 1\theta}{\dfrac 1\theta} &=\frac{\dfrac{\sin \theta}{\theta}}{\dfrac{8\theta}{\theta} +\dfrac{\sin \theta}{\theta\cos\theta}} \\ \\ & = \frac{\dfrac{\sin \theta}{\theta}}{8 +\dfrac{\sin \theta}{\theta\cos\theta}}\\ \\& = \frac{\dfrac{\sin \theta}{\theta}}{8 +\dfrac{\sin \theta}{\theta}\cdot \dfrac 1{\cos\theta}}\end{align}
So, since $\lim_{\theta \to 0}\dfrac{\sin\theta}{\theta} = 1$
we have $$\lim_{\theta \to 0}\frac{\dfrac{\sin \theta}{\theta}}{8 +\dfrac{\sin \theta}{\theta}\cdot \dfrac 1{\cos\theta}} = \frac{1}{8 +1\cdot \frac 1{1}} = \frac 19$$