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How to find the following limit without using l'Hospital rule $$\lim_{x\rightarrow0}\frac{\tan x-\sin x}{x^3}$$ Using l'Hospital I got $1\over2$. Thanks for your help.

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Hints:

$$\frac{\tan x-\sin x}{x^3}=\frac{\sin x-\sin x\cos x}{x^3\cos x}=\frac1{\cos x}\frac{\sin x}x\frac{1-\cos x}{x^2}$$

Now, use arithmetic of limits and also

$$\frac{1-\cos x}{x^2}=\frac{\sin^2x}{x^2(1+\cos x)}=\left(\frac{\sin x}x\right)^2\frac1{1+\cos x}$$

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    $\begingroup$ thanks for your answer, really help me $\endgroup$ – andrey Jun 1 '14 at 13:15
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Alternatively, write $\tan x = x + \frac13 x^3 + \cdots$ and $\sin x = x - \frac16 x^3 + \cdots$. Their difference has the form $\frac12 x^3 + \cdots$, which divided by $x^3$ takes the value $\frac12$ in $x=0$.

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  • $\begingroup$ thanks for your answer, really help me $\endgroup$ – andrey Jun 1 '14 at 13:15
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$$ \begin{aligned} \lim _{x\to 0}\left(\frac{\tan \left(x\right)-\sin \left(x\right)}{x^3}\right) & = \lim _{x\to 0}\left(\frac{\left(x+\frac{x^3}{3}+o\left(x^3\right)\right)-\left(x-\frac{x^3}{3!}+o\left(x^3\right)\right)}{x^3}\right) \\& = \lim _{x\to 0}\left(\frac{x+\frac{x^3}{3}-x+\frac{x^3}{3!}+o\left(x^3\right)}{x^3}\right) \\& = \color{red}{\frac{1}{2}} \end{aligned} $$ Solved with Taylor expansion

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