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I need to solve $$\lim_{x\to 0} \dfrac{\tan ^3 x - \sin ^3 x}{x^5}$$

I did like this:

$\lim \limits_{x\to 0} \dfrac{\tan ^3 x - \sin ^3 x}{x^5} = \lim \limits_{x\to 0} \dfrac{\tan ^3 x}{x^5} - \dfrac{\sin ^3 x}{x^5}$

$=\dfrac 1{x^2} - \dfrac 1{x^2} =0$

But it's wrong. Where I have gone wrong and how to do it?

  • $\dfrac{\tan x}{x}$ cannot reduce to $1$. And $\tan^3 x = (\tan x)^3$ – Brian J Jan 27 '17 at 13:14
  • See this answer http://math.stackexchange.com/a/1783818/72031 which discusses the flaw in your approach. – Paramanand Singh Jan 27 '17 at 13:20
  • @LOL Do you know how to accept an answer to your question? At the left side of every answer, there is that check mark for you to click it. Doing so, that indicates that an answer is accepted. – Juniven Acapulco Jan 27 '17 at 14:46

5 Answers5

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HINT:

$\lim_{x\to0}\left(\left(\dfrac{\tan x}x\right)^3\cdot\dfrac1{x^2}-\left(\dfrac{\sin x}x\right)^3\cdot\dfrac1{x^2}\right)$ is of the form $\infty-\infty$

See List of indeterminate forms

Use $$\dfrac1{\cos^3x}\cdot\left(\dfrac{\sin x}x\right)^3\cdot\dfrac{1-\cos x}{x^2}\cdot(1+\cos x+\cos^2x)$$

lab bhattacharjee
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$$\lim_{x\to 0} \dfrac{\tan ^3 x - \sin ^3 x}{x^5}=\lim_{x\to 0} \dfrac{\frac{\sin ^3 x}{\cos^3x} - \sin ^3 x}{x^5}=\lim_{x\to 0} \dfrac{\sin ^3 x(1-\cos^3x)}{x^5\cos ^3 x}$$ with $\cos x\approx 1-\frac12x^2$ as $x\to0$ $$\lim_{x\to 0} \dfrac{\sin ^3 x}{x^3}\times\dfrac{1-(1-\frac12x^2)^3}{x^2\times1}=\lim_{x\to 0}\frac32-\frac34x^2+\frac18x^4=\frac32$$

Nosrati
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$$\begin{align}\lim_{x\to 0} \dfrac{\tan ^3 x - \sin ^3 x}{x^5}&=\lim_{x\to 0} \dfrac{\frac{\sin ^3 x}{\cos^3x} - \sin ^3 x}{x^5}\\ &=\lim_{x\to 0} \dfrac{\sin ^3 x(1-\cos^3x)}{x^5\cos ^3 x}\\ &=\lim_{x\to 0} \dfrac{\sin ^3 x\Big[(1-\cos x)(1+\cos x+\cos^2x)\Big]}{x^5\cos ^3 x}\\ &=\lim_{x\to 0} \left(\dfrac{\sin ^3 x\Big[(1-\cos x)(1+\cos x+\cos^2x)\Big]}{x^5\cos ^3 x}\cdot\frac{1+\cos x}{1+\cos x}\right), \quad\text{note that }(1-\cos x)(1+\cos x)=\sin^2x\\ &=\lim_{x\to 0} \dfrac{\sin ^5 x\quad(1+\cos x+\cos^2x)}{x^5\cos ^3 x\quad(1+\cos x)}\\ &=\lim_{x\to 0}\left[\left(\frac{\sin x}{x}\right)^5\cdot \frac{1}{\cos^3x}\cdot\frac{1+\cos x+\cos^2x}{1+\cos x}\right]\\ &=1^5\cdot\frac{1}{1^3}\cdot\frac{1+1+1^2}{1+1}=\frac{3}{2}. \end{align}$$

Juniven Acapulco
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$$\dfrac{\tan ^3 x - \sin ^3 x}{x^5}=\ldots=\frac{\sin^3x\left(1-\cos^3x\right)}{x^5\cos^3x}=\tan^3x \cdot\frac{1-\cos^3x}{x^5}$$ $$\sim x^3 \cdot\frac{1-\cos^3x}{x^5}=\frac{1-\cos^3x}{x^2}\sim \frac{1-\left(1-x^2/2+o(x^2)\right)^3}{x^2}$$ $$=\frac{3x^2/2+o(x^2)}{x^2}=\frac{3}{2}+\frac{o(x^2)}{x^2}\underbrace{\to}_{\text{ as }x\to 0}\frac{3}{2}+0=\frac{3}{2}$$

Fernando Revilla
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$$\frac{\tan ^3 x - \sin ^3 x}{x^5}=\tan^3x\dfrac{1 - \cos ^3 x}{x^5}=\left(\dfrac{\tan x}{x}\right)^3\left(\dfrac{(1-\cos x)(1+\cos x+\cos^2x)}{x^2}\right)=\\ =\left(\dfrac{\tan x}{x}\right)^3\left(\dfrac{1-\cos x}{x^2}\right)(1+\cos x+\cos^2x)=\\ =\left(\dfrac{\tan x}{x}\right)^3\left(\dfrac{\sin x}{x}\right)^2\left(\dfrac{1}{1+\cos x}\right)(1+\cos x+\cos^2x)$$

Now use the fundamental limits:

$$\lim_{x\to 0}\frac{\sin x }{x}=1=\lim_{x\to 0}\frac{\tan x }{x}$$

NotADeveloper
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Arnaldo
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