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I am going through some lecture notes and I came across this limit:

$$\lim_{x\to 0}\frac{\sinh x^4-x^4}{(x-\sin x)^4} $$

In the notes, it says (after introducing L'Hopital's Rule) that this would be difficult to evaluate using L'Hopital's Rule but can be done on sight using Taylor's Theorem. After reading the section on Taylor's Theorem, I don't understand how this can be done in sight.

Would one need to calculate its Taylor expansion? If so, how would one go about doing that as its derivatives aren't defined at 0? I have used Wolfram to see the Taylor expansion is $216+O(x^2)$ which means the limit is equal to 216, but how does one calculate this Taylor expansion?

haqnatural
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jonty1245
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6 Answers6

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Hint Expanding we have that $$\sinh x^4 = x^4 + \frac{(x^4)^3}{3!} + O(x^{13}) ,$$ so the numerator is $$\sinh x^4 - x^4 = \frac{x^{12}}{3!} + O(x^{13}).$$ On the other hand, $$\sin x = x - \frac{x^3}{3!} + O(x^4) ,$$ so the denominator is $$(x - \sin x)^4 = \left(\frac{x^3}{3!} + O(x^4)\right)^4 = \cdots$$

$$ \cdots = \left(\frac{x^3}{3!}\right)^4 + O(x^{13}) = \frac{x^{12}}{(3!)^4} + O(x^{13}).$$ The leading term of the quotient is the quotient of the leading terms, namely, $$\frac{(3!)^4}{3!} = 3!^3 = 216 .$$ The fact that the leading terms are both comparable to $x^{12}$ tells us that we would need to apply l'Hopital's Rule 12 times before being able to evaluate---needless to say, this a much faster method.

Travis Willse
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The Taylor series for sinh is: $$\sinh x = x + \frac {x^3} {3!} + \frac {x^5} {5!} + \frac {x^7} {7!} +\cdots = \sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!}$$ and $$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$$ Therefore $$\lim_{x\rightarrow0}\frac{\sinh x^4-x^4}{(x-\sin x)^4} $$ $$=\lim_{x\rightarrow0}\frac{\frac {x^{12}} {3!} + \frac {x^{20}} {5!} + \frac {x^{28}} {7!} +\cdots}{(\frac{x^3}{3!} - \frac{x^5}{5!} + \frac{x^7}{7!} - \cdots)^4} $$ $$=\lim_{x\rightarrow0}\frac{x^{12}/3!}{x^{12}/(3!)^4}$$ $$=216$$

Paramanand Singh
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Akababa
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    You really want to use Taylor-Young here (Taylor expansion with Peano remainder, not the infinite series), otherwise your argument is not actually a rigorous one. – Clement C. Jun 02 '17 at 20:21
  • The result mentioned by @ClementC. is not as popular as general Taylor's theorem. You may look at this question https://math.stackexchange.com/q/1809293/72031 – Paramanand Singh Jun 03 '17 at 02:10
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From the Taylor expansion, the leading order term of the top and bottom are \begin{eqnarray} \sinh(x^4) - x^4 &\sim& \frac{x^{12}}{6}\\ (x - \sin x)^4 &\sim& \left(\frac{x^3}{6}\right)^4 = \frac{x^{12}}{1296} \end{eqnarray} The ratio of these is 216.

eyeballfrog
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We need only two results which can be easily proved using either Taylor's theorem or L'Hospital's Rule $$\lim_{x\to 0}\frac{x - \sin x} {x^{3}}=\lim_{x\to 0}\frac{\sinh x - x}{x^{3}}=\frac{1}{6}$$ and hence the question can also be solved via L'Hospital's Rule with equal ease (contrary to what your notes mention).

We have $$\lim_{x\to 0}\frac{\sinh x^{4}-x^{4}}{(x-\sin x)^{4}}=\lim_{x\to 0}\dfrac{\dfrac{\sinh x^{4}-x^{4}}{x^{12}}}{\left(\dfrac{x-\sin x}{x^{3}}\right)^{4}}=\frac{1/6}{1/6^{4}}=6^{3}=216$$ The use of any advanced tools like L'Hospital's Rule or Taylor's theorem should always be combined with algebraic manipulation and the related algebra of limits in order to simplify the problem considerably.

Paramanand Singh
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We have that $\sin(x)=x-\frac16x^3+\frac1{120}x^5-\dots\,$ so that

$$x-\sin(x)=-\frac16x^3-\frac1{120}x^5+\dots$$

In particular, the term in ${(x-\sin(x))}^4$ with least degree is $\frac1{6^4}x^{12}$.

Now, $\sinh(x)=x+\frac16x^3+\frac1{120}x^5+\dots\,$ so that

$$\sinh(x^4)-x^4=\frac16x^{12}+\frac1{120}x^{20}+\dots$$

If you divide both expressions by $x^{12}$ you'll get for the numerator

$$\frac16+\frac1{120}x^{8}+\dots$$

and for the denominator $\frac1{6^4}$ plus some terms of positive degree on $x$. It follows that as $x\to 0$, the limit is

$$\frac{\frac16}{\frac1{6^4}}=6^3=216$$

Fimpellizzeri
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  • You really want to use Taylor-Young here (Taylor expansion with Peano remainder, not the infinite series), otherwise your argument is not actually a rigorous one. – Clement C. Jun 02 '17 at 20:21
  • In what sense? Both series converge everywhere. Regardless, the question clearly aims at a more 'How can one apply this technique?', heuristic approach than a technical one. – Fimpellizzeri Jun 02 '17 at 20:25
  • Yes, it's good for a heuristic, but not as a proof. Because, exactly, you have to argue about the convergence, the fact that the sum of the infinitely many low-order terms is negligible in front of the leading term, etc. Exactly what the Peano form states, basically. – Clement C. Jun 02 '17 at 20:26
  • What exactly is your point? – Fimpellizzeri Jun 02 '17 at 20:28
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    That your approach is good as heuristic but not as proof, while using the Peano form is equally good as heuristic, as fast, and works as an actual proof. Since the OP is asking for guidance in their studies, might as well give them directly the right tool. – Clement C. Jun 02 '17 at 20:28
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    Your preference for the Peano remainder is clear, but you have failed to show how exactly the answer is inadequate here, even if the OP had asked for a proof (which he didn't). I believe introducing unnecessary technical language for a question that is clearly not aimed at that is poor judgment, honestly. Would you have been happier if it had been noted that convergence is uniform in some compact ball containing the origin? – Fimpellizzeri Jun 02 '17 at 20:32
  • For a start: your idea won't generalize to functions that are only $C^k$ instead of expressible as power series. Also, I made my point above clear: someone asks for intuition and tools, to handle limits. You give a tool that is very close to what they'd need, but falls short (in case at some point, they actually want to prove something), instead of giving the one that is as simple, as usable, and perfectly adapted. So I'm returning the question to you: why? – Clement C. Jun 02 '17 at 20:34
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    You have probably been studying higher level math for far too long if you think they are 'as simple' for the average calculus student (which, more often than not, aren't even math students). I don't blame you. In any case, the functions in question are analytic near the origin (more is true, they're entire), so the point is moot. – Fimpellizzeri Jun 02 '17 at 20:37
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Expanding to degree 2, $$ \frac{\sinh x^4-x^4}{(x-\sin x)^4}=\frac {x^4+x^{12}/6-x^4}{(x-x+x^3/6)^4}=\frac {26^4x^{12}}{6x^{12}}=6^3=216. $$ Formally, you need to include the third term in each expansion to account for the error.

Martin Argerami
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