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Here is the limit:

$$\lim_{\theta\rightarrow0}\frac{\tan(5\theta)}{\tan(10\theta)}$$

how to use calculate this limit without using a graphic calculator??

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  • $\begingroup$ do you mean $\theta\to 0$? So you know how the tangent behaves for small arguments? $\endgroup$ – Marcel Oct 11 '17 at 14:31
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    $\begingroup$ Write everything as sines and cosines, introduce some extra $\theta$s, and do an algebraic dance using what you know about limits of $\frac{\sin \theta}{\theta}$, etc. $\endgroup$ – Randall Oct 11 '17 at 14:33
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$$\lim_{\theta\rightarrow0}\frac{\tan5\theta}{\tan10\theta}=\lim_{\theta\rightarrow0}\left(\frac{\sin5\theta}{5\theta}\cdot\frac{10\theta}{\sin10\theta}\cdot\frac{5\theta}{10\theta}\cdot\frac{\cos10\theta}{\cos5\theta}\right)=\frac{1}{2}$$

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Hint: Use the double-angle formula for tangent: $$ \tan(10 \theta) = \frac{2 \tan(5 \theta)}{1 - \tan^2(5 \theta)} \text{.} $$

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    $\begingroup$ This one is the simplest because it relies on the simpler limit $\lim_{x\to 0}\tan x =0$ than the slightly complicated limit $\lim_{x\to 0}\dfrac{\sin x} {x} =1$. +1 $\endgroup$ – Paramanand Singh Oct 11 '17 at 16:33
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Call $x = 5\theta$ and note that

\begin{eqnarray} \lim_{\theta \to 0} \frac{\tan 5\theta}{\tan 10\theta} &=& \lim_{x\to 0}\frac{\tan x}{\tan 2x} = \lim_{x\to 0}\frac{\tan x}{2\tan x/ (1 - \tan^2x)} \\ &=&\frac{1}{2}\lim_{x\to 0} (1-\tan^2x) = \dots \end{eqnarray}

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  • $\begingroup$ And this is almost the same as another answer here. +1 for the reasons stated in comments to another answer. $\endgroup$ – Paramanand Singh Oct 11 '17 at 16:34
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$y:= 5\theta.$

$f(y) := \dfrac{\tan y}{\tan 2y}=$

$= \dfrac{\tan y}{\dfrac{2\tan y}{1-\tan^2 y}}=$

$=\dfrac{(1-\tan^2 y) \tan y}{2\tan y}$.

$= \dfrac{1-\tan^2 y}{2}$.

$z := \tan y;$

$F(z): = \dfrac{(1-z^2)}{2}$.

$\lim_{z \rightarrow 0} F(z) = 1/2$

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using $$\frac{\tan(5\theta)}{5\theta}\cdot \frac{1}{\frac{\tan(10\theta)}{(10\theta)}}\cdot \frac{1}{2}$$

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