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I have $$\lim \limits_{x\to 0} \frac {\tan(2x)}{\sin(x)}$$ and in my case the result is $\frac{2}{1}$ =2 not whether it is right.

This is my procedure.

$$\lim \limits_{x\to 0} \frac{\frac {\sin(2x)}{\cos(2x)}}{\frac{\sin(x)}{1}}= \lim \limits_{x\to 0} {\dfrac {\sin(2x)}{(\cos(2x))(\sin(x))}}=\dfrac{2x\frac {\sin(2x)}{2x}}{\cos(2x)\frac{x\sin(x)}{x}}$$

I separate the limit.

$$\frac{\left(\lim \limits_{x\to 0}2x\right) \cdot \left(\lim \limits_{x\to 0}\frac {\sin(2x)}{2x}\right)}{\lim \limits_{x\to 0}\left(\cos(2x)\right)\cdot\left(\lim \limits_{x\to 0}\frac{x\sin(x)}{x}\right)} = \lim \limits_{x\to 0} \dfrac{2x}{x}=\frac{2}{1} =2$$

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Ah, thanks for clarification @GitGud. –  AWertheim Aug 14 '13 at 21:34
@user2568086 \$\tan\$ to get $\tan$, \$\lim\$ to get $\lim$, \$\sin\$ to get $\sin$... –  Git Gud Aug 14 '13 at 21:37
Yes,sorry in spanish is sen(x). –  Leonardo Aug 14 '13 at 21:38
You could use that $\sin(2x) = 2\sin(x) \cos(x) $ –  Zaid Alyafeai Aug 14 '13 at 21:40
What is your question? –  Cameron Buie Aug 14 '13 at 21:42
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3 Answers

up vote 14 down vote accepted

That does give you the correct answer, but it takes a bit more work than necessary.

An alternative method is to use the double-angle formula for $\sin$. That is: $$\sin(2x) = 2\cos(x)\sin(x)$$

Thus: $$\begin{align}\require{cancel} \lim_{x\to0} \frac{\tan(2x)}{\sin(x)} &= \lim_{x\to0} \frac{\sin(2x)}{\cos(2x)\sin(x)} \\ &= \lim_{x\to0} \frac{2\cos(x)\cancel{\sin(x)}}{\cos(2x)\cancel{\sin(x)}} \\ &= \lim_{x\to0} \frac{2\color{blue}{\cos(x)}}{\color{red}{\cos(2x)}} \\ &= \frac{2\cdot \color{blue}{1}}{\color{red}{1}} \\ &= \boxed2 \end{align}$$

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What you've written is not entirely correct. One thing is probably essentially a typo: You omitted $\lim\limits_{x\to0}$ in your third expression. But you can't separate the limits as fully as you did when it leads to the numerator and denominator both being $0$. Instead of $$ \frac{(\lim_{x\to 0}2x) \cdot \left(\lim_{x\to 0}\frac {\sin(2x)}{2x}\right)}{(\lim_{x\to 0}\cos(2x))\cdot\left(\lim_{x\to 0}\frac{x\sin(x)}{x}\right)}, $$ you need $$ \lim_{x\to 0}\frac{2x}{x} \cdot \frac{\left(\lim_{x\to 0}\frac {\sin(2x)}{2x}\right)}{(\lim_{x\to 0}\cos(2x))\cdot\left(\lim_{x\to 0}\frac{\sin(x)}{x}\right)}. $$

You can't separate $\displaystyle\lim_{x\to 0}\frac{2x}{x}$ into $\displaystyle\frac{\lim_{x\to0} 2x}{\lim_{x\to0} x}$ because both limits are $0$. You need to cancel the $x$ from the numerator and denominator before taking the limit.

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In the privacy of my mind, the way I do this is what Euler did in the 18th century: remember that when $x$ is infinitely small, then $\sin x$ and $\tan x$ are the same as $x$ so that $\tan(2x)/\sin x$ is $2x/x=2$. –  Michael Hardy Aug 14 '13 at 21:52
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What are you allowed to use? An easier way of solving it is by expanding both numerator and denominator in Maclaurin series for $ x \to 0$: $$ \lim_{x \to 0}\frac{2x+O(x^3)}{x+O(x^3)}=2 $$

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