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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?

  • 0
    $\dfrac{\tan x}{x}$ cannot reduce to $1$. And $\tan^3 x = (\tan x)^3$ – 2017-01-27
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    See this answer http://math.stackexchange.com/a/1783818/72031 which discusses the flaw in your approach. – 2017-01-27
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    @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. – 2017-01-27

5 Answers 5

5

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

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    Before putting $\dfrac 1x=\infty$ can we cancel it? – 2017-01-27
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    @LOL., You can't cancel as $$\dfrac{\sin x}x$$ becomes $1$ only when you set $$x\to0$$ – 2017-01-27
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    So as $x \to 0$ then $\dfrac 1x \to \infty$ so we can cancel? – 2017-01-27
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    @LOL., You will reach $$\dfrac1{x^2}-\dfrac1{x^2}$$ only when you have set $x\to0$ – 2017-01-27
4

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

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    So final answer is $\infty$ ? – 2017-01-27
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    No, the final answer is not ∞. There is nothing in MyGlasses' answer that looks anything like that... – 2017-01-27
3

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

2

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

2

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

Now use the fundamental limits:

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