If $$\frac{x-\sin{x}} {x^3}+\frac {4(\sin {x})^3}{3x^3}=\frac {3x-\sin3x}{3x^3}$$ How do we solve $$\lim_{x\to 0}\frac{x-\sin{x}} {x^3}$$ without L'hospital's

Ans: $\frac{1}{6}$

  • $\begingroup$ Taylor expansion $\endgroup$ – Alec B-G Feb 17 at 7:06
  • $\begingroup$ What is Taylor expansion $\endgroup$ – KillErHawK X Feb 17 at 7:08
  • $\begingroup$ Can you pls solve it as an answer $\endgroup$ – KillErHawK X Feb 17 at 7:08
  • $\begingroup$ I would recommend reading this en.wikipedia.org/wiki/Taylor_series $\endgroup$ – Alec B-G Feb 17 at 7:18
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    $\begingroup$ Taylor series are clearly not the spirit of the question. Here, the goal is to compute the limit 1/6 by relying on a "clever trick" (the first equation) and limit manipulations. $\endgroup$ – Clement C. Feb 17 at 7:38

You could use Taylor expansions, and that would be a simple and quick way to do it, but that's clearly not the spirit of the question. So let's do it the expected way.

I am assuming you have at your disposal the fact that $$ \lim_{x\to 0} \frac{\sin x}{x} = 1 \tag{1} $$ which is standard, and equivalent to knowing that $\sin'(0)=1$. Now, based on that, we get that $$ \lim_{x\to 0} \frac{4(\sin x)^3}{3x^3} = \frac{4}{3}\lim_{x\to 0} \left(\frac{\sin x}{x}\right)^3 = \frac{4}{3} \left(\lim_{x\to 0}\frac{\sin x}{x}\right)^3= \frac{4}{3} \tag{2} $$(can you see why?)

Great. Now, let's define $$ \ell \stackrel{\rm def}{=} \lim_{x\to 0}\frac{x-\sin x}{x^3} \tag{3} $$ the limit we seek. Setting $u=3x$, we see that $$ \ell = \lim_{x\to 0}\frac{3x-\sin 3x}{27x^3} \tag{4} $$ as well. Therefore, from the identity we started with, by taking limits on both sides, we get $$ \lim_{x\to 0}\frac{x-\sin x}{x^3} + \lim_{x\to 0}\frac{4(\sin x)^3}{3x^3} = \lim_{x\to 0}\frac{3x-\sin 3x}{3x^3}\tag{5} $$ and, by the above, this is equivalent to $$ \ell + \frac{4}{3} = 9\ell\tag{6} $$ (can you see why?) Solving (6) for $\ell$ gives $8\ell = 4/3$, that is, $\boxed{\ell = \frac{1}{6}}$, as expected.

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    $\begingroup$ Out of curiosity, to the downvoter: why the downvote? Anything wrong with my answer? Anything I can do to improve it? $\endgroup$ – Clement C. Feb 17 at 7:36
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    $\begingroup$ Someone just downvoted all the answers.even the ones which actually helped me out.who ever that is just undo it $\endgroup$ – KillErHawK X Feb 17 at 7:42
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    $\begingroup$ Malevolence has no boundaries. I have all voted yes questions and answers. $\endgroup$ – Sebastiano Feb 17 at 8:00
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    $\begingroup$ @Sebastiano yep, we can't fight them. $\endgroup$ – manooooh Feb 17 at 11:44
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    $\begingroup$ @manooooh Very kind friend naughtiness must not have the victory. Positive votes are also the fruit of our unity and strength. $\endgroup$ – Sebastiano Feb 17 at 12:46

We have $$\frac{x-\sin{x}} {x^3}+\frac {4(\sin {x})^3}{3x^3}=\frac {3x-\sin3x}{3x^3}$$ Let $$\lim_{x\to 0}\frac{x-\sin{x}} {x^3}=l$$ We compute $\lim_{x\to 0}\frac {3x-\sin3x}{3x^3}$. Let $u=3x$, then as $x\to 0$, $u\to 0$ too and $x=u/3$. Substituting, we have $$\lim_{x\to o}\frac {3x-\sin3x}{3x^3}=\lim_{u\to 0}\frac{u-\sin u}{u^3/9}\\=9\lim_{u\to 0}\frac{u-\sin u}{u^3}=9l.$$ On the other hand, $$\lim_{x\to 0}\frac{4(\sin x)^3}{3x^3}=\frac{4}{3}\lim_{x\to 0}\left(\frac{\sin x}{x}\right)^3=\frac{4}{3}$$ Therefore $$l+\frac{4}{3}=9l$$ and $l=\frac{1}{6}$.


Starting off with what you're given:$$\frac{x-\sin(x)}{x^3} + \frac {4\sin^3 (x)}{3x^3} = \color{blue}{\frac{3x-\sin(3x)}{3x^3}}$$

Notice that the blue part can be re-written to use $3x$ as what's being cubed:

$$\frac {3x-\sin(3x)}{3x^3} = 9\cdot\dfrac{3x-\sin(3x)}{27x^3} = 9\cdot\dfrac{3x-\sin(3x)}{(3x)^3}$$

Taking the limit as $x \to 0$ with the RHS re-written gives

$$\lim_{x \to 0}\frac{x-\sin(x)}{x^3} + \lim_{x \to 0}\frac{4\sin^3(x)}{3x^3} = 9\color{blue}{\lim_{x \to 0}{\dfrac{3x-\sin(3x)}{(3x)^3}}}$$

If $\lim_\limits{x \to 0}\dfrac{x-\sin(x)} {x^3} = L$, then $\lim_\limits{x \to 0}\dfrac{3x-\sin(3x)}{(3x)^3} = \lim_\limits{\color{blue}{3x} \to 0}\dfrac{\color{blue}{3x}-\sin\color{blue}{(3x)}}{(\color{blue}{3x})^3} = L$ as well, which was the whole point of manipulating the RHS in the first place, so

$$L+\lim_{x \to 0}\frac {4\sin^3(x)}{3x^3} = 9L$$

The second limit can be re-written slightly:

$$L+\frac{4}{3}\lim_{x \to 0}\left[\frac {\sin(x)}{x}\right]^3 = 9L$$

Finally, $\lim_\limits{x \to 0} \dfrac{\sin(x)}{x} = 1$, so $L+\dfrac{4}{3} = 9L \iff 8L = \dfrac{4}{3} \iff \boxed{L = \dfrac{1}{6}}$.

  • $\begingroup$ To the downvoter: care to explain why the downvote? All answers have been downvoted, with no explanation. $\endgroup$ – Clement C. Feb 17 at 7:38
  • $\begingroup$ down vote, because you just repeated the answers of Qurultay and Clement C. $\endgroup$ – Khayyam Feb 17 at 7:39
  • $\begingroup$ @Khayyam The 3 were nearly simultaneous. I am Clement C. My own answer was downvoted: is it because "I repeated the answer of KM101 (who was downvoted because they repeated my answer)"? $\endgroup$ – Clement C. Feb 17 at 7:40
  • $\begingroup$ To whoever is doing this: this makes no sense... I'm upvoting this answer, to try to counteract whatever is going on. (This answer is perfectly fine.) $\endgroup$ – Clement C. Feb 17 at 7:42
  • $\begingroup$ I was just trying to emphasize the limit manipulation. Guess you can't post a similar answer even if you're emphasizing something else since it has be "repeating." $\endgroup$ – KM101 Feb 17 at 7:44

Using $\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \mathcal{O}(x^7)$ then \begin{align} \lim_{x\to 0}\frac{x-\sin{x}} {x^3} &= \lim_{x \to 0} \frac{1}{x^3} \, \left( \frac{x^3}{3!} - \frac{x^5}{5!} + \mathcal{O}(x^7) \right) = \lim_{x \to 0} \left( \frac{1}{3!} - \frac{x^2}{5!} + \mathcal{O}(x^4) \right) = \frac{1}{6}. \end{align}

Based on the "If" statement then the problem appears to be more along the lines of \begin{align} \lim_{x \to 0} \left( \frac{x-\sin{x}} {x^3}+\frac {4(\sin {x})^3}{3x^3} \right) &= \lim_{x \to 0} \frac {3x-\sin3x}{3x^3} = 9 \, \lim_{x \to 0} \frac {3x-\sin3x}{(3x)^3} = \frac{9}{6} = \frac{3}{2}. \end{align}

  • $\begingroup$ The last paragraph you added after your edit is confusing, at best. What are you trying to do with it? If you already ahve computed the limit $1/6$ by a Taylor-based argument, what is the point of the last paragraph which uses this limit? $\endgroup$ – Clement C. Feb 17 at 7:25
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    $\begingroup$ Dang, someone around here really likes downvoting without explanation. $\endgroup$ – Clement C. Feb 17 at 7:45

$$L=\lim_{x \rightarrow 0} \frac{x-\sin x}{x^3}=\lim_{x \rightarrow 0} \frac{3x-\sin 3x}{27x^3}=\lim_{x \rightarrow 0} \frac{3x-3\sin x+4 \sin^3 x}{27x^3}$$ $$\implies L=\lim_{x \rightarrow 0}\frac{1}{9} \frac{x-\sin x}{x^3} +\frac{4}{27} \lim_{x \rightarrow 0} \left(\frac{\sin x}{x} \right)^3 \implies L=\frac{L}{9}+\frac{4}{27} \implies L=\frac{1}{6}.$$


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