Using Taylor Expansion to evaluate limits

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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

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

Answer 5

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 $$\lim_{x\rightarrow0}\frac{\sinh x^4-x^4}{(x-\sin x)^4} $$ $$=\lim_{x\rightarrow0}\frac{\frac {x^{12}} {3!} + \frac {x^{20}} {5!} + \frac {x^{28}} {7!} +\cdots}{(\frac{x^3}{3!} - \frac{x^5}{5!} + \frac{x^7}{7!} - \cdots)^4} $$ $$=\lim_{x\rightarrow0}\frac{x^{12}/3!}{x^{12}/(3!)^4}$$ $$=216$$

Answer 6

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.