# Finding the limit, extraneous angle in denominator

$$\lim_{\theta \to 0} \frac{\sin7\theta}{\theta + \tan7\theta}$$

What do you do with the theta in the denominator, the one that isn't with tan? If it weren't in there, $\tan\theta$ would just be $\frac{\sin\theta}{\cos\theta}$, but with the other angle in the denominator. What do you do? How does that other values presence change the question?

(Sorry if I'm not asking right, I'm trying to find homework help for my roommate on courses I'm not taking)

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HINT

$$\dfrac{\sin(7\theta)}{\theta + \tan(7 \theta)} = \dfrac{\dfrac{\sin(7\theta)}{7 \theta}}{\dfrac{\theta}{7 \theta} + \dfrac{\tan(7 \theta)}{7 \theta}}$$ Hence, $$\lim_{\theta \to 0} \dfrac{\sin(7\theta)}{\theta + \tan(7 \theta)} = \dfrac{\lim_{\theta \to 0} \dfrac{\sin(7\theta)}{7 \theta}}{\lim_{\theta \to 0} \dfrac{\theta}{7 \theta} + \lim_{\theta \to 0} \dfrac{\tan(7 \theta)}{7 \theta}}$$

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Or, if you prefer, you could use $$\dfrac{1}{\dfrac{1}{7}\dfrac{7\theta}{\sin(7\theta)} + \dfrac{\tan(7\theta)}{\sin(7\theta)}}$$ –  Robert Israel Oct 28 '12 at 19:21
...if those three limits exist =) –  Pedro Tamaroff Oct 28 '12 at 20:22

The problem yields in a simple way to L'Hospital's Rule.

If you don't want to use L'Hospital's Rule, divide top and bottom by $7\theta$.

The limit of the new top is $$\lim_{\theta\to 0}\frac{\sin 7\theta}{7\theta},$$ which is $1$.

As for the new bottom, it is $$\frac{1}{7}+\frac{\tan7\theta}{7\theta}.$$ You should not have much trouble finding the limit of this as $\theta\to 0$. Express $\tan7\theta$ in terms of sines and cosines.

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See for $\theta\to 0$, $\sin\theta\approx \theta$ and $\tan\theta\approx \theta$

hence $\lim_{\theta\to 0}\frac{\sin7\theta}{\theta+\tan7\theta}=\lim_{\theta\to 0}\frac{7\theta}{\theta+7\theta}=\frac{7}{8}$

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For $\theta$ sufficiently close to (but not equal to) $0$, neither the numerator nor the denominator will be zero. In this case, it's useful to consider the reciprocal $$\frac{\theta+\tan 7\theta}{\sin 7\theta}=\frac{\theta}{\sin 7\theta}+\frac1{\cos 7\theta}=\frac17\cdot\frac{7\theta}{\sin 7\theta}+\sec 7\theta.$$ Making the substitution $\phi=7\theta$ (so that $\theta\to 0$ precisely as $\phi\to 0$), using continuity of the secant function at $0$, and using the basic fact that $$\lim_{\phi\to 0}\frac{\phi}{\sin\phi}=1,$$ we have $$\lim_{\theta\to 0}\frac{\theta+\tan 7\theta}{\sin 7\theta}=\lim_{\phi\to 0}\left[\frac17\cdot\frac{\phi}{\sin\phi}+\sec\phi\right]=\frac17\lim_{\phi\to 0}\frac{\phi}{\sin\phi}+\sec 0=\frac17+1=\frac87.$$ Thus, $$\lim_{\theta\to 0}\frac{\sin 7\theta}{\theta+\tan 7\theta}=\frac78.$$

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We can use L'Hospitals rule as follows:

$\lim_{x\to 0}(\frac{\sin(7x)}{x+\tan(7x)})=\frac{7\cos(x)}{1+7\sec^2(x)}$ (using chain rule as well to find the derivative of $\sin(7x)$. Put $x=0$ into this equation to give us $\frac{7}{1+7}=\boxed{\frac78}$.

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