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Evaluation of

$\displaystyle \lim_{h\rightarrow 0}\bigg[\frac{\sin(60^\circ+4h)-4\sin(60^\circ+3h)+6\sin(60^\circ+2h)-4\sin(60^\circ+h)+\sin(60^\circ)}{h^4}\bigg]$

Here above limit is in $(0/0)$ form

So we have using D, L Hopital rule

$\displaystyle \lim_{h\rightarrow 0}\bigg[\frac{4\cos(60^\circ+4h)-12\cos(60^\circ+3h)+12\cos(60^\circ+2h)-4\cos(60^\circ+h)+0}{4h^3}\bigg]$

Again above limit is in $(0/0)$ form

So using D, L Hopital rule

$\displaystyle \lim_{h\rightarrow 0}\frac{-16\sin(60^\circ+4h)+36\sin(60^\circ+3h)-24\sin(60^\circ+2h)+4\sin(60^\circ+h)}{12h^2}$

Above limit is in $(0/0)$ form

So agian using D, L Rule

$\displaystyle \lim_{h\rightarrow 0}\frac{-64\cos(60^\circ+4h)+108\cos(60^\circ+3h)-48\cos(60^\circ+2h)+4\cos(60+h)}{24h}$

Again using D, L rule , We get

$\displaystyle \lim_{h\rightarrow 0}\frac{256\sin(60^\circ+4h)-324\sin(60^\circ+3h)+96\sin(60^\circ+2h)-4\sin(60+h)}{24}$

$\displaystyle \lim_{h\rightarrow 0}\frac{24\sin(60^\circ)}{24}=\frac{\sqrt{3}}{2}$

Above is very lengthy way

Please explain me some short way

Thanks

jacky
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    First of all, if your angles are in degrees, the derivatives of cos and sin are not what you wrote (you must multiply them by $\pi/180$). – Anne Bauval Dec 28 '22 at 14:17
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    Also, don't be deterred by the "lengthy" calculation. On one hand, it is a bit long (though in the grand scheme of things it is very short, many things that you take for granted at your stage take years to learn and books to explain). On the other hand - what is the complexity of it? Very low - all you did was persistently apply L'Hopital rule. So it is an extremely straightforward solution that you should be proud of. –  Dec 28 '22 at 14:39

3 Answers3

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$$ \frac{f(x+4h)-4f(x+3h)+6f(x+2h)-4f(x+h)+f(x)}{h^4}=\frac{[\Delta^4f](x)}{(\Delta x)^4} $$ is the forward divided-difference approximation of $f^{(4)}(x)$. As the derivatives of the sine have a period of 4, you get back the same sine function in the limit $h\to 0$.

Lutz Lehmann
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  • (+1) Except that the OP's function is not exactly the sine (see comment above). – Anne Bauval Dec 28 '22 at 14:30
  • I do not see your objection. If it were $\sin(60+h)$ or $\sin(60^°+h^°)$ there would be something to object, but $60^°=\frac\pi3$, so everything is correct. – Lutz Lehmann Dec 28 '22 at 14:33
  • OK. I understood "$60°+h$" as a clumsy formulation for "$(60+h)°$". I find "$60°=\frac\pi3$" (instead of $=\frac\pi3$rad) equally awkward. – Anne Bauval Dec 28 '22 at 14:37
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    But angles are just unit-less numbers. 1rad is just 1, while $1^°=\frac{\pi}{180}$. So to explicitly write "rad" is empty, as you say, it just prevents an awkward feeling for some. – Lutz Lehmann Dec 28 '22 at 14:55
  • I don't agree but never mind, I don't want to argue. – Anne Bauval Dec 28 '22 at 15:01
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    What does the OP need to know in order to understand this (otherwise correct) answer? Are difference approximations part of a standard curriculum in either Calculus or Analysis? –  Dec 28 '22 at 15:43
  • @StinkingBishop : That I do not know. It could be mentioned if Newton interpolation is somewhat extensively discussed, in detecting errors via the divided-difference table or in its connection to the Taylor series. But as far as I divine, numerical approximation of higher-order derivatives is not a part of a standard curriculum. – Lutz Lehmann Dec 28 '22 at 17:59
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I prefer radians everywhere and non-ambiguously. Applying

$$\sin\left(\frac\pi3+t\right)=\sin\frac\pi3\cos t+\cos\frac\pi3\sin t=\frac{\sqrt3}2\left(1-\frac{t^2}2+\frac{t^4}{24}\right)+\frac12\left(t-\frac{t^3}6\right)+o(t^4)$$

to $t=4h,$ $3h,$ $2h,$ $h,$ there only remains (after cancellation):

$$\sin\left(\frac\pi3+4h\right)-4\sin\left(\frac\pi3+3h\right)+6\sin\left(\frac\pi3+2h\right)-4\sin\left(\frac\pi3+h\right)+\sin\frac\pi3$$$$=\frac{\sqrt3}2h^4+o(h^4).$$

Anne Bauval
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The numerator can be restated as $$16\sin^4\left(\frac{h}{2}\right) \sin(2h + 60^\circ).$$ Now, applying the first-order expansion and using radians instead of degrees (so $60^\circ$ transforms into $\frac{\sqrt{3}}{2}$), the limit is $$\lim_{h\to 0} \frac{16\left(\frac{h}{2}\right)^4\left(\frac{\sqrt{3}}{2} + h\right)}{h⁴} = \frac{\sqrt{3}}{2}.$$

user326159
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