2

Compute

$$\lim_{x\to0}\left[-\dfrac{4x}{\sin 2x} + \dfrac{x}{\cos 2x}\right]$$

Obviously I can't plug in 0. I noticed the sin and cos are both 2x. Is there a way to combine them into tan? I don't want to do illegal math, lol.

Martin Sleziak
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    If I am reading your expression correctly, the second part has limit $0$ so we can forget about it. – André Nicolas Aug 28 '14 at 04:46
  • i don't understand what you mean – Hello Aug 28 '14 at 04:48
  • If the limit of $f(x)$ and $g(x)$ both exist, then $\lim_{x\to 0}(f(x)+g(x))=\lim_{x\to 0}f(x)+\lim_{x\to 0}g(x)$. And it is clear that $\lim_{x\to 0}\frac{x}{\cos 2x}=0$, since the top approaches $0$ and the bottom approaches $1$. So we need only find $\lim_{x\to 0}\frac{-4x}{\sin x}$. I will leave this to you, or to someone who writes an answer. – André Nicolas Aug 28 '14 at 04:52
  • -2. Thanks!‏‏‏‏ – Hello Aug 28 '14 at 05:09
  • You are welcome. Yes, the limit is indeed $-2$. – André Nicolas Aug 28 '14 at 05:11

1 Answers1

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It was already explained in the comments that

$$\lim_{x\to0}\left[-\dfrac{4x}{\sin 2x} + \dfrac{x}{\cos 2x}\right] = -\lim_{x\to0}\dfrac{4x}{\sin 2x}$$ since $\lim\limits_{x\to0}\dfrac{x}{\cos 2x}=0$.

To compute $\lim_{x\to0}\dfrac{4x}{\sin 2x}$ le us substitute $t=2x$. We get $$\lim_{x\to0}\dfrac{4x}{\sin 2x} = 2\lim_{t\to0}\dfrac{t}{\sin t}.$$

So we obtained a well-known limit for which there are various approaches: How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?

It is also useful to notice that the limit $\lim\limits_{x\to 0}\frac{\sin x}{x}.$ is precisely the derivative of $\sin x$ at 0, since by definition of derivative we have $$(\sin x)'|_{x=0} = \lim\limits_{x\to 0}\frac{\sin x-\sin 0}{x-0} = \lim\limits_{x\to 0}\frac{\sin x}{x}.$$

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Martin Sleziak
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