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$$\lim_{x\to0}\left(\frac{\tan (x)+4\tan(2x)-3\tan(3x)}{x^2\tan(x)}\right)$$

Martin Sleziak
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SutharAryan
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    Your thoughts, please? As this is not a homework-solving site, all efforts are appreciated.. –  Jan 01 '18 at 13:40
  • Actually, I had evaluated this by myself but I didn't quite like the method (Already mentioned in the answers) with which I did it. So, I was looking for a new good method. – SutharAryan Jan 01 '18 at 16:10
  • After cancelling common factor $ \tan x$ what do you get? – Narasimham Jan 01 '18 at 18:41
  • @SutharAryan If you are ok, you can accept the answer and set as solved. Thanks! – user Jan 03 '18 at 17:25

6 Answers6

5

Hint:

$$\tan x+\tan2x+3(\tan2x-\tan3x)=\dfrac{\sin3x}{\cos x\cos2x}-\dfrac{3\sin x}{\cos2x\cos3x}$$

$$=\dfrac{\sin3x\cos3x-3\sin x\cos x}{\cos x\cos2x\cos3x}$$

$$=\dfrac{\sin6x-3\sin2x}{2\cos x\cos2x\cos3x}$$

$$=\dfrac{(3\sin2x-4\sin^32x)-3\sin2x}{2\cos x\cos2x\cos3x}$$

Now use $\lim_{h\to0}\dfrac{\sin h}h=1$

lab bhattacharjee
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4

We have $\tan x\sim x+x^3/3$ near $0$, so we get $$\lim _{x\to 0}\frac{\tan x+4\tan 2x-3\tan3x}{x^2\tan x}=\lim_{x\to 0}\frac{x+\frac{x^3}{3}+4\left(2x+\frac{8x^3}{3}\right)-3\left(3x+\frac{27x^3}{3}\right)}{x^2\left(x+\frac{x^3}{3}\right)}=\color{red}{-16}.$$

rae306
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4

Without Taylor's expansions (briefly):

  1. Denote $\tan x=t$ then $$ \tan 2x=\frac{2t}{1-t^2},\quad\tan 3x=\tan(x+2x)=\frac{t(3-t^2)}{1-3t^2}. $$
  2. Substitute $$ \frac{\tan x+4\tan 2x-3\tan 3x}{\tan x}=[\text{simplify}]=-\frac{16 t^2}{(1-t^2)(1-3t^2)}. $$
  3. Calculate the limit, using $\frac{\tan x}{x}\to 1$ as $x\to 0$.
A.Γ.
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3

We know that the Taylor expansions of the tangent are: $$\tan(x)=x+\frac{x^3}{3}+\frac{2x^5}{15}+o(x^6)$$ $$\tan(2x)=2x+\frac{8x^3}{3}+\frac{64x^5}{15}+o(x^6)$$ $$\tan(3x)=3x+9x^3+\frac{162x^5}{5}+o(x^6)$$ Now we substitute the expansions into the limit, and it becomes: $$\lim_{x\to 0}\frac{\left(x+\frac{x^3}{3}+\frac{2x^5}{15}+o(x^6)\right)+4\cdot\left(2x+\frac{8x^3}{3}+\frac{64x^5}{15}+o(x^6)\right)-3\cdot\left(3x+9x^3+\frac{162x^5}{5}+o(x^6)\right)}{x^3\cdot\left(1+\frac{x^2}{3}+\frac{2x^4}{15}+o(x^6)\right)}=\\=\lim_{x\to 0}\frac{-16x^3-80x^5+o(x^6)}{x^3\cdot\left(1+\frac{x^2}{3}+\frac{2x^4}{15}+o(x^6)\right)}=\lim_{x\to 0}\frac{x^3\cdot(-16-80x^2+o(x^6))}{x^3\cdot\left(1+\frac{x^2}{3}+\frac{2x^4}{15}+o(x^6)\right)}=\\=\lim_{x\to 0}\frac{-16-80x^2+o(x^6)}{1+\frac{x^2}{3}+\frac{2x^4}{15}+o(x^6)}=\color{red}{-16}$$

Bacause all the other terms of the limit goes to 0.

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    But isn't using Taylor expansion the same as L'Hospital, in a way? By using the expansion you concede that both nominator and denominator approach zero at the limit, and so you examine the derivatives... – Itai Jan 01 '18 at 14:25
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$\tan x =x+\frac{x^3}{3}+o(x^3)$

$$\frac{\tan\left(x\right)+4\tan\left(2x\right)-3\tan\left(3x\right)}{x^2\tan\left(x\right)} =\frac{x+\frac{x^3}{3}+4x+\frac{32x^3}{3}-3x-\frac{81x^3}{3}+o(x^3)}{x^3 +o(x^3)}=\frac{-16x^3+o(x^3)}{x^3 +o(x^3)}=\frac{-16+o(1)}{1 +o(1)}\to-16$$

user
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0

$$\dfrac{\tan x+4\tan2x-3\tan3x}{x^2\tan x}=\dfrac{\tan x-x+4(\tan2x-2x)-3(\tan3x-3x)}{x^3}\cdot\dfrac x{\tan x}$$

$$F=\dfrac{\tan x-x+4(\tan2x-2x)-3(\tan3x-3x)}{x^3}$$

$$=\dfrac{\tan x-x}{x^3}+4\cdot2^3\cdot\dfrac{\tan2x-2x}{(2x)^3}-3\cdot3^3\cdot\dfrac{\tan3x-3x}{(3x)^3}$$

Using Are all limits solvable without L'Hôpital Rule or Series Expansion, $$\lim_{y\to0}\dfrac{\tan y-y}{y^3}=\dfrac13$$

$$\lim_{x\to0}F=\dfrac13\left(1+4\cdot2^3-3\cdot3^3\right)=-16$$

lab bhattacharjee
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