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Consider $$\lim\limits_{x\to 0} \frac{\sin(x)-x+\frac{x^3}{6}}{x^5}$$

I want to evaluate this limit in three different ways.

First method - Hopital's Rule

The limit leads to indetermination $0/0$.

Consider $$\lim_{x\to 0}\frac{\left[\sin(x)-x+\frac{x^3}{6}\right]'}{\left[x^5\right]'}=\lim_{x\to 0}\frac{\cos(x)-1+\frac{x^2}{2}}{5x^4}=\frac{0}{0}$$

that it's still an indetermination. I apply Hopital's rule again:

$$\lim_{x\to 0}\frac{\left[\cos(x)-1+\frac{x^2}{2}\right]'}{\left[5x^4\right]'}=\lim_{x\to 0}\frac{-\sin(x)+x}{20x^3}=\frac{0}{0}$$

that it's still an indetermination. I apply Hopital's rule again:

$$\lim_{x\to 0}\frac{\left[-\sin(x)+x\right]'}{\left[20x^3\right]'}=\lim_{x\to 0}\frac{-\cos(x)+1}{60x^2}=\frac{1}{60}\lim_{x\to 0}\frac{1-\cos(x)}{x^2}=\frac{1}{120}$$

Therefore, the initial limit is:

$$\lim\limits_{x\to 0} \frac{\sin(x)-x+\frac{x^3}{6}}{x^5}=\frac{1}{120}$$

Second method - Taylor's expansion

Using that $\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+o(x^6)$ for $x\to 0$, I can rewrite the initial limit as it follows:

$$\lim_{x\to 0}\frac{\sin(x)-x+\frac{x^3}{6}}{x^5}=\lim_{x\to 0}\frac{x-\frac{x^3}{6}+\frac{x^5}{120}-x+\frac{x^3}{6}}{x^5}=\lim_{x\to 0}\frac{\frac{x^5}{120}}{x^5}=\frac{1}{120}$$

Any suggestions for a third method to solve this limit?

Sigma Algebra
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    why you want a third method? – Sine of the Time Sep 19 '24 at 14:19
  • I wanted an additional method because both methods I used (which are the most standard obviously) involve the use of derivatives and I was wondering if this limit (and others like it) could be calculated using other ways – Sigma Algebra Sep 19 '24 at 14:22
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    Using Taylor expansion is the standard and fastest method for this kind of limits. There's probably a method that does not involve derivatives, but I don't think it's relevant. You can take a look at this question – Sine of the Time Sep 19 '24 at 14:24
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    Any method is going to need to have some quantification of how sin(x) behaves relative to -x + x^3 / 6, for small x, which is determined by the behaviour of the derivatives of sin(x) near zero, which is precisely what the two methods are looking at. (And anyway, in my view, l'Hopital is just implementing the consequence of writing two functions as Taylor series, so method 1 is method 2 in disguise). – MilesB Sep 19 '24 at 14:29
  • Ok, I'm already happy with this two, finding a third method is actually quite useless. – Sigma Algebra Sep 19 '24 at 14:40
  • The usual formal definition of $\sin$ is through the power series, so your second method doesn't use derivatives, just the definition of $\sin$ and basic results about power series and limits. – jjagmath Sep 19 '24 at 14:47

3 Answers3

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COMMENT.-(This is not an answer!) The difficulty in finding a third method for your problem is that your statement uses a partial series expansion of $\sin(x)$ in the numerator and the corresponding power of $x$ as the denominator. This forces things in such a way that it becomes very difficult (or perhaps impossible) to find the third method you want. The problem might grow in difficulty (only for L'Hôpital's rule) but your two ways of finding the solution would still be efficient.

So we can enunciate: Find the limit when $x\to0$ of $$\frac{\sin(x)-x+\frac{x^3}{6}-\frac{x^5}{5!}+\frac{x^7}{7!}}{x^9}$$ which clearly is equal to $\dfrac{1}{9!}$ before calculation.

Ataulfo
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Using methods similar to those indicated here, assuming that limit $L$ exists

$$L=\lim\limits_{x\to 0} \frac{\sin x-x+\frac{x^3}{6}}{x^5}\implies L=\lim\limits_{x\to 0} \frac{\sin(2x)-2x+\frac{8x^3}{6}}{2^5x^5}$$

$$\implies 2^4L=\lim\limits_{x\to 0} \frac{\frac12\sin(2x)-x+\frac{2x^3}{3}}{x^5}\implies (2^4-1)L=\lim\limits_{x\to 0} \frac{\frac12\sin(2x)-\sin(x)+\frac{x^3}{2}}{x^5}$$

$$\implies (2^4-1)L=\lim\limits_{x\to 0} \frac{\sin x}{x}\frac{\cos x-1+\frac x{\sin x}\frac{x^2}{2}}{x^4}=\frac18\implies L=\frac1{120}$$

indeed

$$\frac{\cos x-1+\frac x{\sin x}\frac{x^2}{2}}{x^4}=\frac{\cos x-1+\frac{x^2}{2}}{x^4}+\frac12\frac{\left(\frac x{\sin x}-1\right)}{x^2}\to \frac1{24}+\frac1{12}=\frac18 $$

where the second one comes from the well known $\frac{x-\sin x}{x^3}\to \frac16$ and for the first one, assuming that limit $M$ exists

$$M=\lim\limits_{x\to 0}\frac{\cos x-1+\frac{x^2}{2}}{x^4}\implies 2^4M=\lim\limits_{x\to 0}\frac{\cos(2x)-1+2x^2}{x^4}$$

$$\implies 2^3M=\lim\limits_{x\to 0}\frac{x^2-\sin^2x}{x^4} =\frac13 \implies M=\frac1{24}$$

indeed

$$\frac{x^2-\sin^2x}{x^4}=\frac{x+\sin x}{x}\frac{x-\sin x}{x^2}\to 2 \cdot \frac16=\frac13$$

user
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You can check that $$\sin(x) - x + \frac{x^3}{6} = \frac{1}{24} \int_0^x (x-t)^4 \cos t \, dt.$$

Now, if $0 < x < \frac{\pi}{2}$, then $\cos x \le \cos t \le 1$ for $t \in [0, x]$. Therefore, $$\frac{1}{24} \int_0^x (x-t)^4 \cos x \, dt \le \sin(x) - x + \frac{x^3}{6} \le \frac{1}{24} \int_0^x (x-t)^4 \, dt,$$ or in other words, $$\frac{1}{120} x^5 \cos x \le \sin(x) - x + \frac{x^3}{6} \le \frac{1}{120} x^5.$$ Applying the squeeze theorem, we thus get $\lim_{x\to 0^+} \frac{\sin(x) - x + \frac{x^3}{6}}{x^5} = \frac{1}{120}$. Since the function $\frac{\sin(x) - x + \frac{x^3}{6}}{x^5}$ is even on $\mathbb{R} \setminus \{ 0 \}$, that implies that in fact $\lim_{x\to 0} \frac{\sin(x) - x + \frac{x^3}{6}}{x^5} = \frac{1}{120}$.

(Of course, you could view this as cheating since it's just extracting an important part of a certain proof of Taylor's theorem, in the form of an integral expression for the remainder term.)

Daniel Schepler
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