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I know this $$\lim_{x\to0}\frac{3x^2}{\tan(x)\sin(x)}$$ But I have no idea how make a result different of: $$\lim_{x\to0}\frac{3x}{\tan(x)}$$ I would like understand this calculation without using derivation or L'hôpital's rule. Thank you.

Américo Tavares
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user73276
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    This is kinda related to your earlier question, http://math.stackexchange.com/questions/421473/how-can-evaluate-lim-x-to0-frac-sin3x2-tanx-sinx , and can be calculated by similar means... – colormegone Jun 17 '13 at 20:22
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    I can't understand how come you ask this new question after you asked, and were answered galore, another very similar question: can't you compare?! – DonAntonio Jun 17 '13 at 20:48
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    So, I would put this remark in the header, but as I am still beginner as a user, I am trying to learn what is relevant or not, but I tried to learn how to do without the sine, but I could not see, but I'm learning new techniques, basically was by this. – user73276 Jun 17 '13 at 22:54

5 Answers5

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By Taylor series we have $\tan x\sim_0 x$ hence $$\lim_{x\to0}\frac{3x}{\tan(x)}=\lim_{x\to0}\frac{3x}{x}=3$$

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    You cannot always make substitutions like this, right? For example, if you have the limit as x approaches 0 of $\frac {tan(x)-x}{x^3}$ this is equal to $1/3$, but if you were to make your substitution, you would get $0$. – Ovi Jun 17 '13 at 21:49
  • Only when it is a multiplier on the whole limit. – NightRa Sep 05 '13 at 08:33
  • Very nice answer! – amWhy May 29 '14 at 16:27
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Since $\tan x = \dfrac{\sin x}{\cos x}$, we have the identity:

$$\frac{3x^2}{\tan x\sin x} = \frac{3x^2\cos x}{\sin^2 x}$$

valid in at least an open, punctured neighbourhood around $0$.

Now using the product rule for limits, and that $\lim\limits_{x\to 0}\dfrac{\sin x}x = 1$ (proofs here, also some without differentiation):

$$\lim_{x\to0}\frac{3x^2}{\tan x\sin x} = \lim_{x\to0}\cos x\cdot\lim_{x\to 0}\frac{3x^2}{\sin^2 x} = 3$$

Community
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Lord_Farin
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Replace $\tan x$ by $\frac{\sin x}{\cos x}$. Our expression becomes $$3(\cos x)\left(\frac{x}{\sin x}\right)^2. $$ You are undoubtedly familiar with the fact that $\lim_{x\to 0}\frac{\sin x}{x}=1$.

André Nicolas
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Hints

$$\frac{3x^{2}}{\tan x\sin x}=3\times \frac{x}{\tan x}\times \frac{x}{\sin x},$$

$$$$

$$0<x<\frac{\pi }{2}\Rightarrow \sin x<x<\tan x\Leftrightarrow 1<\frac{x}{ \sin x}<\frac{1}{\cos x}.$$

Américo Tavares
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We have, $$\lim_{x\to0}\frac{ x^2 }{\tan(x)\sin(x)} =3\lim_{x\to0}\frac{x}{\sin(x)}\cdot\frac{x}{\tan(x)} =3$$

Given that $$\lim_{h\to0}\frac{\sin h}{h} =\lim_{h\to 0}\frac{\tan h}{h} =1$$

Guy Fsone
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