How to calculate $\lim_{x \to 0} \frac{x^{6000} - (\sin x)^{6000}}{x^2(\sin x)^{6000}}$? [closed]

Evaluate this limit $$\lim_{x \to 0} \frac{x^{6000} - (\sin x)^{6000}}{x^2(\sin x)^{6000}}$$

What's the method?

The answer is $1000$.

closed as off-topic by heropup, Claude Leibovici, user26857, Watson, ClarinetistMay 16 '16 at 11:52

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• Did you try applying L'Hopital's rule (i.e. taking the limit of the ratio of the derivatives of the numerator and denominator)? Note: you may need to use the rule more than once if possible. – Dave May 15 '16 at 13:16
• Taylor series and binomial approximations look easier. – Aritra Das May 15 '16 at 13:19
• Are you (rather sadistically) forced to use L'H? – Did May 15 '16 at 20:59

Ok, one way to intimidate students is to use large numbers like 6000 in questions. A simple tool to beat this strategy of examiner is to replace the big number by a generic symbol say $n$. We thus calculate the limit $$f(n) = \lim_{x \to 0}\frac{x^{n} - \sin^{n}x}{x^{2}\sin^{n}x}$$ where $n$ is a positive integer. The answer for the question is $f(6000)$.

We have \begin{align} f(n) &= \lim_{x \to 0}\frac{x^{n} - \sin^{n}x}{x^{2}\sin^{n}x}\notag\\ &= \lim_{x \to 0}\dfrac{x - \sin x}{x^{3}\cdot\dfrac{\sin x}{x}}\cdot\dfrac{{\displaystyle \sum_{i = 0}^{n - 1}x^{i}\sin^{n - 1 - i}x}}{\sin^{n - 1}x}\notag\\ &= \lim_{x \to 0}\frac{x - \sin x}{x^{3}}\sum_{i = 0}^{n - 1}\left(\frac{x}{\sin x}\right)^{i}\notag\\ &= \sum_{i = 0}^{n - 1} 1\cdot \lim_{x \to 0}\frac{1 - \cos x}{3x^{2}}\text{ (via L'Hospital's Rule)}\notag\\ &= \frac{n}{6}\notag \end{align} and hence the desired answer is $f(6000) = 1000$.

• I could not understand it from the step where you used the summation sign. How did you do that? Also can you please tell me to solve it using L Hopital rule. – Harry Karwasra May 16 '16 at 7:53
• @HarryKarwasra: The first step where I introduce the summation sign in easy. I split $x^{n} - \sin^{n}x$ using the formula $$a^{n} - b^{n} = (a - b)(a^{n - 1} + a^{n - 2}b + \cdots + ab^{n - 2} + b^{n - 1}$$ which is written compactly as $$a^{n} - b^{n} = (a - b) \sum_{i = 0}^{n - 1}a^{i}b^{n - 1 - i}$$ This is how numerator is handled. For denominator I write $$x^{2}\sin^{n}x = x^{3}\cdot\frac{\sin x}{x}\cdot \sin^{n - 1}x$$ – Paramanand Singh May 16 '16 at 8:11
• @HarryKarwasra: Further the $\sin^{n - 1}x$ in denominator is put with the summation and then it can be taken as denominator of each term in summation and this leads to $$\sum_{i = 0}^{n - 1}\frac{x^{i}\sin^{n - 1 - i}x}{\sin^{n - 1}x} = \sum_{n = 0}^{n - 1}\left(\frac{x}{\sin x}\right)^{i}$$ Now each term in summation tends to limit $1$ and hence this factor I take out separately from the limit operation. This leads to $n$ (summing $1$ for $n$ times). Next factor is $(x - \sin x)/x^{3}$ and this on L'Hospital gives $(1 - \cos x)/(3x^{2})$. – Paramanand Singh May 16 '16 at 8:15
• @HarryKarwasra: The limit of $(1 - \cos x)/x^{2}$ is almost standard and can be obtained as $$\frac{1 - \cos x}{x^{2}}= \frac{(1 - \cos x)(1 + \cos x)}{x^{2}(1 + \cos x)} = \frac{1 - \cos^{2}x}{x^{2}(1 + \cos x)}$$ which simplifies to $$\frac{\sin^{2}x}{x^{2}}\cdot\frac{1}{1 + \cos x}$$ and it tends to $$1\cdot\frac{1}{1 + 1} = \frac{1}{2}$$ In total we get the desired limit as $$n \cdot\frac{1}{3}\cdot\frac{1}{2} = \frac{n}{6}$$ – Paramanand Singh May 16 '16 at 8:18
• @HarryKarwasra: When you are told to solve a limit via L'Hospital Rule, it does not mean that you start applying L'Hospital Rule immediately on whatever expression is there, but it means that you first transform your expression into a form where it is possible, simple and useful to apply L'Hospital Rule and then apply it. Applying L'Hospital blindly is so so bad an idea to evaluate limits. – Paramanand Singh May 16 '16 at 8:22

You have $$\sin x=x-x^3/6+o(x^3)$$ so $$(\sin x)^{6000}=x^{6000}-6000\frac{x^{6002}}{6}+o(x^{6002})$$ Hence your limit is $$\lim_{x\to0}\frac{1000x^{6002}+o(x^{6002})}{x^2(\sin x)^{6000}}$$

• Can you please elaborate this? I can't understand it. – Harry Karwasra May 15 '16 at 15:03
• @HarryKarwasra It's a standard application of Taylor expansion. – egreg May 15 '16 at 15:28

Since $\sin(n+1)x=2\sin nx\cos x-\sin(n-1)x$, we have \begin{align}\sin3x&=2\sin2x\cos x-\sin x=2(2\sin x\cos x)\cos x-\sin x\\ &=4\sin x(1-\sin^2 x)-\sin x=3\sin x-4\sin^3x\end{align} Then $\sin x=\sin(3(x/3))=3\sin(x/3)-4\sin^3(x/3)$. Then \begin{align}\frac{x^n-\sin^nx}{x^2\sin^nx}&=\frac{(3(x/3))^n-(3(\sin(x/3)-(4/3)\sin^3(x/3)))^n}{x^2\sin^nx}\\ &=\frac{(3(x/3))^n-3^n\sin^n(x/3)+n\cdot3^n(4/3)\sin^{n+2}(x/3)+O(\sin^{n+4}(x/3))}{x^2\sin^nx}\\ &=\frac{(3(x/3))^n-3^n\sin^n(x/3)}{x^2\sin^nx}+\frac{n\cdot3^n(4/3)\sin^{n+2}(x/3)}{x^2\sin^nx}+\frac{O(\sin^{n+4}(x/3)}{x^2\sin^nx}\\ &=\frac{(3(x/3))^n-3^n\sin^n(x/3)}{(3(x/3))^2(3^n\sin^n(x/3))+O(\sin^{n+2}(x/3)))}\\ &+\frac{n\cdot3^n(4/3)\sin^{n+2}(x/3)}{(3(x/3))^2(3^n\sin^n(x/3))+O(\sin^{n+2}(x/3)))}+\frac{O(\sin^{n+4}(x/3)}{x^2\sin^nx}\end{align} So \begin{align}\lim_{x\rightarrow0}\frac{x^n-\sin^nx}{x^2\sin^nx}&=\lim_{x\rightarrow0}\frac{3^n}{3^{n+2}}\frac{(x/3)^n-\sin^n(x/3)}{(x/3)^2\sin^n(x/3)}\frac{1}{\left(1+\frac{O(\sin^{n+2}(x/3))}{3^n\sin^n(x/3)}\right)}\\&+\lim_{x\rightarrow0}\frac{4n\cdot3^{n-1}}{3^{n+2}}\frac{\sin^{n+2}(x/3)}{(x/3)^2\sin^n(x/3)}\frac{1}{\left(1+\frac{O(\sin^{n+2}(x/3))}{3^n\sin^n(x/3)}\right)}+\lim_{x\rightarrow0}\frac{O(\sin^{n+4}(x/3)}{x^2\sin^nx}\\ &=\frac19\lim_{x\rightarrow0}\frac{x^n-\sin^nx}{x^2\sin^nx}\frac1{1+0}+\frac{4n}{27}\frac1{1+0}+0=\frac98\frac{4n}{27}=\frac n6=\frac{6000}6=1000\end{align} Sort of like the Taylor series approach but using trigonometric identities instead. That $O(\sin^{n+4}(x/3))$ term represents further terms of the binomial expansion which have at least the factor $\sin^{n+4}(x/3)$, so their limits when divided merely by $\sin^{n+2}(x/3)$ are all zero. We are, of course, using $$\lim_{x\rightarrow0}\frac{\sin x}x=\lim_{x\rightarrow0}\frac{\sin(x3)}{(x/3)}=1$$

HINT Divide denominator and numerator by $x^{6000}$ then use $\lim_{x \to 0} \frac {\sin(x)} x =1$

You get: $$\lim_{x \to 0} \frac{1 - (\frac {\sin x} x)^{6000}}{x^2}$$ Now apply L'Hospital: $$\lim_{x \to 0} \frac{- 6000 (\frac {\sin x} x)^{5999} (\frac {\sin x} x )'} {2x} = \lim_{x \to 0} \frac{- 6000 ( (\frac {\sin x} x )'} {2x}$$ then again L'Hospital after calculating $(\frac {\sin x} x )'$ etc.

• That is not enough to handle the difference of terms in the numerator, is it? – Clement C. May 15 '16 at 14:49
• @ClementC. No, L'Hospital must be repeatedly applied. – user261263 May 15 '16 at 15:36

This answer is pretty useless for the results per se, much shorter proofs have been given. Honestly, it took me some time to write it, and I could non just abandon it in the forest.

[EDIT] Its intent is to play with standard inequalities, that are useful to keep in mind, and keep them as long as possible, without using de l'Hôpital rule, which I revere for the insight. However it requires some care, and probably should mention Johan Bernoulli in its name. Keeping inequalities may provide you with convergence rate estimates.

The function (call it $f(x)$) is even. Borrowing from @Paramanand Singh notation, it can be factorized as:

$$f(x) = \frac{1 - \left(\frac{\sin x }{x}\right)^n}{x^2 \left(\frac{\sin x }{x}\right)^n}\,.$$

We use standard inequalities close enough to $0^+$: $$1-ny \le (1-y)^n \le 1-ny + \frac{n(n+1)}{2} y^2$$ for $n>0$ and $$x - \frac{x^3}{6} \le \sin x \le x - \frac{x^3}{6} + \frac{x^5}{5!}$$ hence $$1 - \frac{x^2}{6} \le \frac{\sin x }{x} \le 1 - \frac{x^2}{6} + \frac{x^4}{5!}\,.$$

From the last one, we see that $\left(\frac{\sin x }{x}\right)^n \to 1$, so we forget about it and study $g(x) = f(x)\left(\frac{\sin x }{x}\right)^n$.

By replacing above $y$ by $\frac{x^2}{6}$ or $\frac{x^2}{6} - \frac{x^4}{5!}$ , we have: $$n x^2 \left(\frac{1}{6} - \frac{x^2}{5!}\right) - \frac{n(n+1)x^4}{2}\left(\frac{1}{6} - \frac{x^2}{5!}\right) ^2\le 1 - \left(\frac{\sin x }{x}\right)^n \le n\left(\frac{x^2}{6} \right)$$ hence $$\frac{n}{6}+a_nx^2+b_n x^4+c_nx^6\le g(x) \le \frac{n}{6}$$ with constants $a_n,b_n,c_n$ which you can compute explicitly, especially if you want some rate of convergence, for instance I got $a_n = -\frac{n(5n+8)}{360}$. And for your question of course the limit is $1000 = 6000/6$.

Here is a visual simulation of the bounds for function $g(x)$, only for $n=4$ for numerical reasons: $$\lim_{x \to 0} \frac{x^{6000} - (\sin x)^{6000}}{x^2(\sin x)^{6000}}=\lim_{x \to 0} \frac{\left(\frac{x}{\sin x}\right)^{6000} - 1}{x^2}$$

Lets call $u(x)=\frac{x}{\sin x}$ and apply L'Hospital twice:

$$\lim_{x \to 0} \frac{\left(\frac{x}{\sin x}\right)^{6000} - 1}{x^2}=3000\lim_{x \to 0}\left[ u'^2+u\cdot u''(x)\right]$$

Evaluate the last limit and note that $\lim_{x\to 0}\frac{x}{\sin x}=1$.

$$\lim_{x \to 0} \frac{x^{6000} - (\sin x)^{6000}}{x^2(\sin x)^{6000}} =\lim_{x \to 0} \frac{x-\sin x}{x^3}\cdot\lim_{x \to 0}\dfrac{\sum_{r=0}^{6000-1-1}x^r\sin^{6000-r}}{(\sin x)^{6000}}$$

Now, $$\lim_{x \to 0}\dfrac{\sum_{r=0}^{6000-1}x^r\sin^{6000-1-r}}{(\sin x)^{6000}}=\sum_{r=0}^{6000-1}\left(\dfrac x{\sin x}\right)^r$$

• There's something wrong: you should have $x^3$ in the denominator of the first limit, because $\lim_{x\to0}\frac{x-\sin x}{x^2}=0$. – egreg May 15 '16 at 14:25
• @egreg, Yes, what's wrong with this as the other limit is finite – lab bhattacharjee May 16 '16 at 9:04
• You have an off by one exponent in the sines under the summation: it should be $(\sin x)^{6000-1-r}$. Since the limit is $1000$, there must be an error! See Paramanand Singh's answer, where the computation is carried out correctly. – egreg May 16 '16 at 9:23
• @egreg, Hey, that's a rectified & generalized copy of my answer:) – lab bhattacharjee May 16 '16 at 9:59
• Downvote removed – egreg May 16 '16 at 10:10

A lovely cheat that complex analysis offers:

Recall that the Laurent expansion of $\csc(z)$ is $\frac{1}{z}(1+\frac{z^2}{6}+o(z^4))$.

Apply this mindlessly to the question to obtain

$\frac{z^{5998}}{(\sin(x))^{6000}}-\frac{1}{z^2}=\frac{z^{5998}}{z^{6000}}(1+\frac{6000}{6}z^{2}+o(z^4))-\frac{1}{z^2}$ and the answer pops right out.

• The expansion $\csc x=\frac1x\left(1+\frac16x^2+O(x^4)\right)$ (but with $O(x^4)$, not $o(x^4)$...) does not require complex analysis, only to know the standard expansions $\sin x=x-\frac16x^3+O(x^5)$ and $\frac1{1-x}=1+x+O(x^2)$. – Did May 16 '16 at 20:32
• Yes, but those who have seen Laurent series in action are more sensitive to them. – Yon Teh May 23 '16 at 12:41