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I have this limit: $$\lim_{x \to 0}\frac{\sin x-x}{x^3}$$ How can i resolve it without l'Hopital's rule?

agha
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egarro
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  • Use LaTeX to show math properly. Also, where is $x$ going? $\pm\infty,,0,\ldots$? – cjferes Aug 19 '14 at 21:00
  • x going to 0, sorry i don't know how to use LaTeX.Thank you. – egarro Aug 19 '14 at 21:03
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    People are down voting because you have not provided any working or context. Please consider adding your thoughts. – Ali Caglayan Aug 19 '14 at 21:11
  • I tried use L'Hopital's rule for get the result of the limit, it works, but i want to know how resolve it without this rule, that's what i thought. – egarro Aug 19 '14 at 21:15

1 Answers1

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You can use the power series for $\sin x$. Then $\sin x=x-\frac{x^3}{3!}+o(x^3)$, where $o(x^3)$ stands for the higher order terms, in particular $\lim_{x \to 0}\frac{o(x^3)}{x^3}=0$. Subtracting $x$ on both sides, dividing by $x^3$ and taking the limit we get $-\frac1{3!}$.

Conifold
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