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It's from this answer. https://math.stackexchange.com/a/368574/481435

I cannot evaluate the last limit. I arranged that limit to slightly clearer form.

$$\lim\limits_{t\to 0}\left(\frac{2+\cos2t}{3\sin^4t}-\frac{1}{t^4}\right)=\frac{1}{45}$$

I have no idea how to confirm that value. I know in theory I can use L'hospital's theorem but it would be endless derivatives for this problem. I tried Taylor expansion but I failed.

Septacle
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2 Answers2

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After a little algebra/trigonometry the numerator is seen to be $$t^4(2+\cos 2t)-(3/8)(3-4\cos 2t+\cos 4t)$$ and using Taylor series for $\cos $ you can see that the above expression is $t^8/15+o(t^8)$ and denominator is clearly $3t^8+o(t^8)$ so the desired limit is $1/45$.

Paramanand Singh
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  • Best way of course! – user Feb 01 '18 at 19:19
  • Thanks it was helpful. By the way, I found out that you were also one of those replied in the linked question. And by your comment in that post I looked the one of the answers (https://math.stackexchange.com/a/41601/481435) again (at first I thought it was too hard) and I found that's more elegant way. (though it requires me to memorize where to start) Thank you. – Septacle Feb 02 '18 at 03:11
  • @Septacle: I am glad you liked that answer. It is elementary and the credit probably should go to Apostol who gave the hints in the exercises. – Paramanand Singh Feb 02 '18 at 04:06
  • @gimusi: I tried to use trig identities to put the numerator in terms of trig function of multiple angles instead of powers of trig functions. This avoids multiplication of Taylor series. But still showing that the coefficients of $t^4,t^6$ are $0$ and that of $t^8$ is $1/15$ requires some calculation with fractions. – Paramanand Singh Feb 02 '18 at 04:09
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Note that

$$\frac{2+\cos2t}{3\sin^4t}-\frac{1}{t^4}=\frac{2t^4+t^4\cdot\cos2t-3\sin^4t}{3t^4\cdot\sin^4t}$$

thus we need to expand in such way to have $8^{th}$ order terms, then

  • $\cos2t=1-\frac{4t^2}{2}+\frac{16t^4}{24}+o(t^4)=1-2t^2+\frac{2t^4}{3}+o(t^4)$
  • $\sin t=t-\frac{t^3}{6}+o(t^3)$
  • $\sin^4 t=\left(t-\frac{t^3}{6}+o(t^3)\right)^4=t^4-\frac{2t^6}{3}+\frac{t^8}{5}+o(t^8)$

then

$$\frac{2t^4+t^4\cdot\cos2t-3\sin^4t}{3t^4\cdot\sin^4t}=\frac{2t^4+t^4-2t^6+\frac{2t^8}{3}+o(t^8)-3t^4+2t^6-\frac{3t^8}{5}+o(t^8)}{3t^8+o(t^8)}=\frac{\frac{t^8}{15}+o(t^8)}{3t^8}=\frac1{45}+o(1)\to \frac1{45}$$

user
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