# Finding limits for this without L'hospital's rule

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

• Taylor expansion – Alec B-G Feb 17 at 7:06
• What is Taylor expansion – KillErHawK X Feb 17 at 7:08
• Can you pls solve it as an answer – KillErHawK X Feb 17 at 7:08
• I would recommend reading this en.wikipedia.org/wiki/Taylor_series – Alec B-G Feb 17 at 7:18
• 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. – Clement C. Feb 17 at 7:38

## 5 Answers

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.

• Out of curiosity, to the downvoter: why the downvote? Anything wrong with my answer? Anything I can do to improve it? – Clement C. Feb 17 at 7:36
• Someone just downvoted all the answers.even the ones which actually helped me out.who ever that is just undo it – KillErHawK X Feb 17 at 7:42
• Malevolence has no boundaries. I have all voted yes questions and answers. – Sebastiano Feb 17 at 8:00
• @Sebastiano yep, we can't fight them. – manooooh Feb 17 at 11:44
• @manooooh Very kind friend naughtiness must not have the victory. Positive votes are also the fruit of our unity and strength. – 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 $$L+\dfrac{4}{3} = 9L \iff 8L = \dfrac{4}{3} \iff \boxed{L = \dfrac{1}{6}}$$.

• To the downvoter: care to explain why the downvote? All answers have been downvoted, with no explanation. – Clement C. Feb 17 at 7:38
• down vote, because you just repeated the answers of Qurultay and Clement C. – Khayyam Feb 17 at 7:39
• @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)"? – Clement C. Feb 17 at 7:40
• 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.) – Clement C. Feb 17 at 7:42
• 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." – 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}

• 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? – Clement C. Feb 17 at 7:25
• Dang, someone around here really likes downvoting without explanation. – Clement C. Feb 17 at 7:45

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