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Find $$\lim\limits_{x\to0}{\dfrac{\tan x-\sin x}{x^3}}$$

So my first thought was to simplify the numerator: $$\tan x-\sin x=\dfrac{\sin x}{\cos x}-\sin{x}=\dfrac{\sin x-\sin x\cos x}{\cos x}=\dfrac{\sin x(1-\cos x)}{\cos x}$$ Now let's plug it back: $$\lim_{x\to0}{\dfrac{\tan x-\sin x}{x^3}}=\lim_{x\to0}{\dfrac{x^3\sin x(1-\cos x)}{\cos x}}$$ I don't see how to use the limit $$\lim_{x\to0}{\dfrac{\sin x}{x}}=1$$

Rodrigo de Azevedo
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kormoran
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    Note that your $x^3$ shouldn't have jumped to the numerator. You can do $1-\cos(x)=\cos(0)-\cos(x)=-2\sin((0+x)/2)\sin((0-x)/2)$. Then pair each sine with one of the $x$ from $x^3$. Since some sines have $x/2$ inside, instead of $x$, we can pair them with an $x/2$, where the $/2$ we will need to make them appear, by multiplying both numeratorn and denominator by $1/2$ as needed. – plop Aug 17 '22 at 20:13
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    @user85667 That's right. Might be clearer if you posted an Answer where you did this, instead of just saying what should happen... – David C. Ullrich Aug 17 '22 at 20:17
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    Hint: $\frac1{x^3}\ne x^3$. – David C. Ullrich Aug 17 '22 at 20:17

3 Answers3

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Hint: $$\lim_{x\to0}{\dfrac{\tan x-\sin x}{x^3}}=\lim_{x\to0}{\dfrac{\sin x(1-\cos x)}{x^3\cos x}}=\lim_{x\to0}{\dfrac{\sin x(1-\cos^2 x)}{x^3\cos x(1+\cos x)}}=\lim_{x\to0}{\dfrac{\sin^3 x}{x^3\cos x(1+\cos x)}}$$

Vasili
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Note that you should be looking at (referring to your next-to-last line)

$$\lim_{x\to0}{\dfrac{\sin x(1-\cos x)}{x^3 \cos x}}$$

and we have

$$\lim_{x\to0}{\dfrac{\sin x(1-\cos x)}{x^3\cos x}} = \lim_{x \to 0} \frac{\sin x}{x} \frac{1-\cos x}{x^2} \frac{1}{\cos x}$$

The only limit in question here would be the middle of the three factors. Note that

$$\begin{align*} \frac{1-\cos x}{x^2} &= \frac{1 - \cos x}{x^2} \frac{1+ \cos x}{1+\cos x} \\ &= \frac{1-\cos^2 x}{x^2} \frac{1}{1+\cos x}\\ &=\frac{\sin^2 x}{x^2} \frac{1}{1+\cos x}\\ &=\left( \frac{\sin x}{x} \right)^2 \frac{1}{1+\cos x} \end{align*}$$

This gives you the middle factor as well when knowing $\sin(x)/x \to 1$.

(Ideas essentially as seen here and here.)

PrincessEev
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$$\lim_{x\to0}{\dfrac{\tan x-\sin x}{x^3}}=\lim_{x\to0}{\dfrac{\sin x(1-\cos x)}{x^3\cos x}}=\lim_{x\to0}{\dfrac{\sin x(1-\cos x)}{x^3}}=\lim_{x\to0}{\dfrac{2\sin x\sin^2\frac{x}{2}}{x^3}}$$

Zhang
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