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Who can help me solving this limit?

$$\lim_{x\to 0 }\frac{\frac{x^3}{6} +\sin x - x }{x^5}$$

I don't need answer Taylor Maclaurin L'Hospital and Derivatives. But I know my answer $\frac 1{120}$.

How can I do?

Michael Rozenberg
  • 194,933
Master Ask
  • 17

5 Answers5

3

Well you can use L'Hospital rule $5$ times because each time we get $0/0$ form:

$$\lim_{x\to 0} \frac{\cos(x)}{5!} = \frac{1}{5!}$$

jonsno
  • 7,521
1

By L'Hospital we obtain: $$\lim_{x\rightarrow0}\frac{\frac{x^3}{6}+\sin{x}-x}{x^5}=\lim_{x\rightarrow0}\frac{\frac{x^2}{2}+\cos{x}-1}{5x^4}=\frac{1}{10}\lim_{x\rightarrow0}\frac{x^2-4\sin^2\frac{x}{2}}{x^4}=$$ $$=\frac{1}{10}\lim_{x\rightarrow0}\left(\frac{x-2\sin\frac{x}{2}}{x^3}\cdot\left(1+\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)\right)=\frac{1}{5}\lim_{x\rightarrow0}\frac{1-\cos\frac{x}{2}}{3x^2}=$$ $$=\frac{1}{15}\lim_{x\rightarrow0}\frac{2\sin^2\frac{x}{4}}{x^2}=\frac{1}{120}\lim_{x\rightarrow0}\frac{\sin^2\frac{x}{4}}{\frac{x^2}{16}}=\frac{1}{120}.$$

Michael Rozenberg
  • 194,933
1

The limit can be proven using derivatives, and without l'Hopital's rule or series expansions as follows:

$$ \begin{aligned} f(x)&=x^{1/2}+6\frac{\sin\left(x^{1/4}\right)}{x^{1/4}} \\ f'(0)&=\lim_{x\to0}\frac{x^{1/2}+6\frac{\sin\left(x^{1/4}\right)}{x^{1/4}}-6}{x}&=6\lim_{x\rightarrow0}\frac{\frac{x^3}{6}+\sin{x}-x}{x^5} \end{aligned} $$

Another way of approaching the problem is noticing is that the numerator contains the Maclaurin expansion of $\sin$.

$$ \begin{aligned} \sin(x)&\in x-\frac{x^3}{6}+\frac{x^5}{120}+\mathcal{O}(x^7) \\ L&=\lim_{x\rightarrow0}\frac{\sin{x}-\left(x-\frac{x^3}{6}\right)}{x^5} \\ &=\lim_{x\rightarrow0}\frac{\left(\color{red}{x-\frac{x^3}{6}}+\frac{x^5}{120}+\ldots\right)-\left(\color{red}{x-\frac{x^3}{6}}\right)}{x^5} &=\frac1{120} \end{aligned} $$

@user45914123's answer shows an algebraic proof without using derivatives.

Jam
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$$\begin{align}L= \lim_{x\to 0 }\frac{\frac{x^3}{6} +\sin x - x }{x^5}&= \lim_{x\to 0 }\frac{\frac{x^3}{6} +3\sin(x/3) - 4 \sin^3(x/3) - x }{x^5} &\\&= \lim_{x\to 0 }\frac{x^3/6 -3(x/3)^3/6 - 4 \sin^3(x/3) }{x^5} &\\&+ \lim_{x \to 0} \dfrac{ 3(x/3)^3/6 +3\sin(x/3)-3(x/3)}{3^5(x/3)^5} &\\&= \dfrac{L}{3^4} + \lim_{x\to 0 }\frac{4(x/3)^3 - 4 \sin^3(x/3) }{x^5}&\\&= \dfrac{L}{3^4} &\\&+ \dfrac{4}{3^5}\lim_{x\to 0 }\frac{((x/3) - \sin(x/3))( (x/3)^2 +\sin^2(x/3)+ (x/3) \sin(x/3)) }{(x/3)^5}&\\&=\dfrac{L}{3^4}+ \dfrac{2}{3^5}\end{align}$$

$$\therefore L = \dfrac1{120}$$

user8277998
  • 2,666
  • Seems pretty slick; but how did you manage to show that $\lim \frac{(x/3\ldots)}{(x/3)^5}=1/2$? – Jam Jan 07 '18 at 22:42
  • Actually, I've answered my own question: $\lim {(x/3\ldots)}=\lim\frac{u-\sin(u)}{u^3}\cdot\left(1+\frac{\sin^2u}{u^2}+\frac{\sin u}{u}\right)$, for $u=x/3$. Then use math.stackexchange.com/questions/134051 . – Jam Jan 07 '18 at 22:47
  • @Jam Yes you are correct; take a look at https://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lh%C3%B4pital-rule-or-series-expansion for other such limits. Thanks for the upvote and mention in your answer. I shall return the favour. – user8277998 Jan 08 '18 at 01:32
  • You use seri Taylor again at sin x - x \approx x^3/6 – Master Ask Jan 08 '18 at 07:06
  • @MasterAsk I don't see how or where I used used taylor. I would be glad if you enlighten me by telling me the step number. If you are asking about $\lim_{x\to 0 }\frac{((x/3) - \sin(x/3))( (x/3)^2 +\sin^2(x/3)+ (x/3) \sin(x/3)) }{(x/3)^5}$, then take a look at second limit of https://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lh%C3%B4pital-rule-or-series-expansion. – user8277998 Jan 08 '18 at 07:18
  • The step number 2 – Master Ask Jan 08 '18 at 07:49
  • The step number 2 – Master Ask Jan 08 '18 at 07:49
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    @MasterAsk I used $\sin 3x = 3\sin x - 4\sin^3 x \iff\sin x = 3\sin (x/3) - 4\sin^3 (x/3)$ – user8277998 Jan 08 '18 at 09:04
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It's $\sin(x)\sim x-\frac{1}{6}x^3+\frac{1}{120}x^5,\, x\to0$. Can you take it from here?

http://mathworld.wolfram.com/MaclaurinSeries.html