Can someone explain this trigonometric limit?

Question

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.

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

Are you familiar with the double-angle formula for $\sin(x)$? — abiessu, Aug 14 '13 at 21:36
@user2568086 $\tan$ to get $\tan$, $\lim$ to get $\lim$, $\sin$ to get $\sin$... — Git Gud, Aug 14 '13 at 21:37
Hey, that's pretty ingenious. It's not the easiest way to do it, but it's sound and it gets the right answer. I think you should be pleased with yourself. — MJD, Aug 14 '13 at 21:54
Why not just apply L'Hospital? This seems like a prototypical example of a job for that. — hmakholm left over Monica, Aug 14 '13 at 22:58

score 14 · Accepted Answer · answered Aug 14 '13 at 21:44

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

Michael Hardy · Answer 2 · 2013-08-14T21:53:38.613

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.

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

score 2 · Answer 3 · answered Aug 14 '13 at 23:07

2

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$: $$ \lim_{x \to 0}\frac{2x+O(x^3)}{x+O(x^3)}=2 $$

answered Aug 14 '13 at 23:07

Alex

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Can someone explain this trigonometric limit?

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