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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?

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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.

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  • $\begingroup$ You have a $3$ that should be a $13$. $\endgroup$ – Ian Jun 2 '17 at 20:19
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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 $$\lim_{x\rightarrow0}\frac{\sinh x^4-x^4}{(x-\sin x)^4} $$ $$=\lim_{x\rightarrow0}\frac{\frac {x^{12}} {3!} + \frac {x^{20}} {5!} + \frac {x^{28}} {7!} +\cdots}{(\frac{x^3}{3!} - \frac{x^5}{5!} + \frac{x^7}{7!} - \cdots)^4} $$ $$=\lim_{x\rightarrow0}\frac{x^{12}/3!}{x^{12}/(3!)^4}$$ $$=216$$

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    $\begingroup$ 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. $\endgroup$ – Clement C. Jun 2 '17 at 20:21
  • $\begingroup$ The result mentioned by @ClementC. is not as popular as general Taylor's theorem. You may look at this question math.stackexchange.com/q/1809293/72031 $\endgroup$ – Paramanand Singh Jun 3 '17 at 2:10
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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.

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  • $\begingroup$ It's not similar $(\sim)$, it's approx. $(\approx)$ $\endgroup$ – Jaideep Khare Jun 2 '17 at 20:22
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    $\begingroup$ It is $\sim$. (Equivalent). This is a well-defined symbol, cf. Landau notations. $\endgroup$ – Clement C. Jun 2 '17 at 20:23
  • $\begingroup$ You need to say "as $x \to 0$" when using the $\sim$ notation to make it well-defined. $\endgroup$ – Chappers Jun 3 '17 at 12:09
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    $\begingroup$ @Chappers I believe that is clear from context. $\endgroup$ – eyeballfrog Jun 3 '17 at 16:52
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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.

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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.

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    $\begingroup$ You are missing all $o()$'s. Your equalities are not equalities as written. $\endgroup$ – Clement C. Jun 2 '17 at 20:22
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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

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

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  • $\begingroup$ 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. $\endgroup$ – Clement C. Jun 2 '17 at 20:21
  • $\begingroup$ 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. $\endgroup$ – Fimpellizieri Jun 2 '17 at 20:25
  • $\begingroup$ 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. $\endgroup$ – Clement C. Jun 2 '17 at 20:26
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    $\begingroup$ 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. $\endgroup$ – Clement C. Jun 2 '17 at 20:28
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    $\begingroup$ 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? $\endgroup$ – Fimpellizieri Jun 2 '17 at 20:32

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