Evaluation of Trigonometric Limit having 5 terms 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
 A: $$
\frac{f(x+4h)-4f(x+3h)+6f(x+2h)-4f(x+h)+f(x)}{h^4}=\frac{[\Delta^4f](x)}{(\Delta x)^4}
$$
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 $h\to 0$.
A: 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):
$$\sin\left(\frac\pi3+4h\right)-4\sin\left(\frac\pi3+3h\right)+6\sin\left(\frac\pi3+2h\right)-4\sin\left(\frac\pi3+h\right)+\sin\frac\pi3$$$$=\frac{\sqrt3}2h^4+o(h^4).$$
A: 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
$$\lim_{h\to 0} \frac{16\left(\frac{h}{2}\right)^4\left(\frac{\sqrt{3}}{2} + h\right)}{h⁴} = \frac{\sqrt{3}}{2}.$$
