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I was given a task to find $$\lim_{x\to0}\frac{x-\sin{x}}{x^3}$$ at my school today. I thought it was an easy problem and started differentiating denominator and numerator to calculate the limit but the teacher then said we aren't allowed to use L'Hopital's rule, but to "play around" with known limits and limit definition. I got stuck here since I can't really think of a way to do this, and according to my teacher, there are at least 4 ways. A subtle hint would be enough. Thanks.

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  • $\begingroup$ Is the Taylor expansion of $\sin$ known? $\endgroup$ Dec 22, 2014 at 14:07
  • $\begingroup$ @DanielFischer Since Taylor works anywhere where L'Hospital works, I presume not... $\endgroup$
    – 5xum
    Dec 22, 2014 at 14:09
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    $\begingroup$ @Transcendental Just a warning: L'Hospital is basically equivalent to taking the Taylor expansion up to only the first two terms, so if your teacher wants you to calculate without L'Hospital, that probably also includes no Taylor. $\endgroup$
    – 5xum
    Dec 22, 2014 at 14:20
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    $\begingroup$ That is not a duplicate... $\tan x$ is different enough from $\sin x$ that the questions should be separate. Yes, they should be linked, but not duplicate. $\endgroup$
    – apnorton
    Dec 29, 2014 at 3:41
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    $\begingroup$ @anorton: This one then: math.stackexchange.com/questions/856030/… $\endgroup$
    – Aryabhata
    Dec 30, 2014 at 23:06

4 Answers 4

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$$L=\lim_{x\to0}\frac{x-\sin x}{x^3}\\ =\lim_{x\to0}\frac{2x-\sin2x}{8x^3}\\ 4L-L = \lim_{x\to0}\frac{x-\frac12\sin2x-x+\sin x}{x^3}$$
which simplifies to the product of three expressions of the form $\frac{\sin y}{y}$

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  • $\begingroup$ +10 for actually not using the L'Hospital rule (even in its "oh I'm not L'hospital, I'm Taylor" disguise). $\endgroup$
    – 5xum
    Dec 22, 2014 at 14:28
  • $\begingroup$ I find this proof looks very neat, however I don't understand the last step you did. Do you mind explaining it? I don't find it obvious, what 4L is. Ah ok, problem solved. Haha, nice. $\endgroup$
    – Imago
    Dec 22, 2014 at 14:45
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    $\begingroup$ The only thing this proof assume is the existence of $L$. $\endgroup$
    – clark
    Dec 22, 2014 at 15:52
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    $\begingroup$ So for instance $ \lim_{x\rightarrow \infty}\cos (x)=\lim_{x\rightarrow \infty}\cos (2x)= \lim_{x\rightarrow}2\cos^2 x -1$. So $$ L=2L^2 -1$$ $\endgroup$
    – clark
    Dec 22, 2014 at 16:46
  • $\begingroup$ a related post: math.stackexchange.com/questions/508733/… $\endgroup$
    – Math-fun
    Dec 22, 2014 at 20:42
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Note: This method is 'Taylor series inspired,' but I think I can say in good faith that it fits your requirements of not using l'Hospital's Rule or Taylor's Theorem.

We observe first that $$\lim_{x\to 0^-} \frac{x-\sin x}{x^3} = \lim_{x\to0^+} \frac{(-x) - \sin(-x)}{(-x)^3} = \lim_{x\to0^+} \frac{x - \sin x}{x^3}$$

Thus, $$\lim_{x \to 0}\frac{x-\sin x}{x^3} = \lim_{x\to 0^+} \frac{x-\sin x}{x^3}$$ so we can focus on behavior of $\frac{x-\sin x}{x^3}$ for $x>0$ to evaluate the limit.

Set $f(x) = x - \sin x$. We're going to try to bound $f$ with polynomials so we can use the Squeeze Theorem. Since integration preserves inequalities, in the sense that if $f(t) \le g(t)$ for $0 \le t \le x$ then $\int_0^x f(t) \,dt \le \int_0^x g(t) \,dt$, we will find these polynomials by differentiating $f$ some number of times, finding polynomials that bound the derivative, then integrating. Since we want to find the limit of $\frac{f}{x^3}$, we'll differentiate $f$ three times, based on the reasoning that if $P(x) \le f^{(3)} \le Q(x)$ for polynomials $P$ and $Q$, then after integrating three times we will have an inequality involving polynomials with no terms of degree less than $3$. We calculate $$f^{(3)}(x) = \cos x$$ For $x>0$, we have that $$1-x \le \cos x \le 1$$ Let's integrate three times: $$\int_0^x\int_0^y\int_0^z 1-w\, dw\,dz\,dy \le \int_0^x\int_0^y\int_0^z \cos w\, dw\,dz\,dy \le \int_0^x\int_0^y\int_0^z 1\, dw\,dz\,dy$$ Don't worry if you haven't seen triple integrals before: all we're doing is taking three integrals, one after the other. This gives us $$\frac{1}{6}x^3 - \frac{1}{24}x^4 \le x - \sin x \le \frac{1}{6}x^3$$

To finish, just have we divide by $x^3$ and take the limit as $x\to 0^+$.

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$\sin x = x - \dfrac{x^3}{3!} + o(x^5) \to \dfrac{x-\sin x}{x^3} = \dfrac{1}{6} + o(x^2)$. From this the limit is $\dfrac{1}{6}$.

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  • $\begingroup$ Note: the downvote is probably explained by this meta question. $\endgroup$ Dec 22, 2014 at 16:41
  • $\begingroup$ How does one know whether to use big-O or little-o in these kinds of things? $\endgroup$ May 12, 2018 at 2:28
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I suppose you can always use Taylor expansion near $x=0$ (as long as it is known to you), that is: $$ \sin x = x - \frac1{3!}x^3 + o(x^5) $$

Hence we will have: $$ \lim\limits_{x\to 0}\frac{x-\sin x }{x^3}=\frac16 $$

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  • $\begingroup$ Note: the downvote is probably explained by this meta question. $\endgroup$ Dec 22, 2014 at 16:40

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