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Find the limit algebraically :

$$\lim_{x\to 0} \left( \frac{x}{x-\sin x}-\frac{6}{x^2}\right)$$

My Try :

$$\lim_{x\to 0} \frac{x^2}{x^2}\left( \frac{x}{x-\sin x}-\frac{6}{x^2}\right)$$

$$\lim_{x\to 0} \frac{1}{x^2}\left( \frac{x^3}{x-\sin x}-6\right)$$

now:

$$\lim_{x\to 0} \left( \frac{x^3}{x-\sin x}\right)=6$$

So :

$$\lim_{x\to 0} \frac{1}{x^2}\left( \frac{x^3}{x-\sin x}-6\right)=0$$

is it right ?

Community
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Almot1960
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    No. $\displaystyle \lim_{x \to 0} \frac1{x^2} = +\infty$ and $\displaystyle \lim_{x \to 0} \left(\frac{x^3}{x-\sin x} - 6\right) = 0$, so the limit of their product is the indeterminate form $+\infty \cdot 0$, which isn't necessarily $0$. Also, how'd you get $\displaystyle \lim_{x \to 0} \frac{x^3}{x - \sin x} = 6$ without using l'Hopital? I ask because this is currently tagged [tag:limits-without-lhopital]. –  Mar 21 '17 at 12:18
  • Indeed the list of tools at your disposal and not at your disposal is most unclear at present, although this very much conditions the type of answer you are expecting. – Did Mar 21 '17 at 14:12
  • I now see that this exact remark was made several times about some previous recent questions of yours. What makes that you are not hearing this message? – Did Mar 21 '17 at 14:15

1 Answers1

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HINT: use $$\sin x =x-\frac{x^3}{3!}+\frac{x^5}{5!}-...$$so $$\lim_{x\to 0} \left( \frac{x}{x-\sin x}-\frac{6}{x^2}\right)=\\ \lim_{x\to 0} \left( \frac{x}{x-\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}\right)}-\frac{6}{x^2}\right)=\\ \lim_{x\to 0} \left( \frac{x}{\frac{x^3}{3!}-\frac{x^5}{5!}}-\frac{6}{x^2}\right)=\\ \lim_{x\to 0} \left( \frac{1}{\frac{x^2}{3!}-\frac{x^4}{5!}}-\frac{6}{x^2}\right)=\\ \lim_{x\to 0} \frac{6}{x^2}\left( \frac{1}{1-\frac{x^2}{20}}-1\right)=\\$$can you go on ?

Michael Rozenberg
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Khosrotash
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