The value of $\lim_{h\to 0}\frac{\sin(a+3h)-3\sin(a+2h)+3\sin(a+h)-\sin a}{h^3}$ is?

Question

Here was my approach: I grouped the terms $\frac{\sin(a+3h)-\sin a}{h^3} $ and $\frac{3\sin(a+h)-3\sin(a+2h)}{h^3}$. By using first principle and got $\frac{3\cos a-3\cos(a+h)}{h^2}$. However, now when I expand $\cos(a+h)$ using the identity, my final answer isn't quite right. The answer is supposed to be $-\cos a$. I cannot understand where I went wrong so please help me with this.

Answer 1

$\lim_\limits{h\to 0}\frac{\sin(a+3h)-3\sin(a+2h)+3\sin(a+h)-\sin a}{h^3}\\ \lim_\limits{h\to 0}\frac{\sin(a+3h)-\sin(a+2h)-2\sin(a+2h)+2\sin(a+h) + \sin(a+h)-\sin a}{h^3}$

From the definition of the derivative
$\lim_\limits{h\to 0} \frac {\sin(a+h) - \sin a}{h} = \frac {d}{dx} \sin a = \cos a$

Similarly

$\lim_\limits{h\to 0} \frac {\sin(a+2h) - \sin (a+h)}{h} = \frac {d}{dx} \sin (a+h) = \cos (a+h)$
and
$\lim_\limits{h\to 0} \frac {\sin(a+3h) - \sin (a+2h)}{h} = \frac {d}{dx} \sin (a+2h) = \cos (a+2h)$

$\lim_\limits{h\to 0}\frac{\cos(a+2h)-2\cos(a+h)+\cos a}{h^2}$

Doing a similar trick...

$\lim_\limits{h\to 0}\frac{\cos(a+2h)-\cos(a+h) -\cos(a+h) +\cos a}{h^2}\\ \lim_\limits{h\to 0}\frac{-\sin(a+h)+\sin(a)}{h} = -\cos a$

Answer 2

I unsuccessfully attempted to brute force the limit by expanding all of the terms using the angle addition identity for sine. Instead, using L'Hôpital's Rule three times produces the same result,

$$ \begin{align} L &=\tag{1}\lim_{h \to 0} \left[\frac{\sin(a + 3h) - 3\sin(a + 2h) + 3\sin(a + h) - \sin a}{h^3} \right] \\ &=\tag{2} \lim_{h \to 0} \left[\frac{\frac{\mathrm{d}}{\mathrm{d}h}[\sin(a + 3h) - 3\sin(a + 2h) + 3\sin(a + h) - \sin a]}{\frac{\mathrm{d}}{\mathrm{d}h}[h^3]} \right] \\ &=\tag{3}\lim_{h \to 0} \left[\frac{3\cos(a + 3h) - 6\cos(a + 2h) + 3\cos(a + h)}{3h^2} \right] \\ &=\tag{4}\lim_{h \to 0} \left[\frac{\cos(a + 3h) - 2\cos(a + 2h) + \cos(a + h)}{h^2} \right] \\ &=\tag{5}\lim_{h \to 0} \left[\frac{\frac{\mathrm{d}}{\mathrm{d}h}[\cos(a + 3h) - 2\cos(a + 2h) + \cos(a + h)]}{\frac{\mathrm{d}}{\mathrm{d}h}[h^2]} \right] \\ &=\tag{6}\lim_{h \to 0} \left[\frac{-3\sin(a + 3h) - 4\sin(a + 2h) - \sin(a + h)}{2h} \right] \\ &=\tag{7}\lim_{h \to 0} \left[\frac{\frac{\mathrm{d}}{\mathrm{d}h}[-3\sin(a + 3h) - 4\sin(a + 2h) - \sin(a + h)]}{\frac{\mathrm{d}}{\mathrm{d}h}[2h]} \right] \\ &=\tag{8}\lim_{h \to 0} \left[\frac{-9\cos(a + 3h) + 8\cos(a + 2h) - \cos(a + h)}{2} \right] \\ &=\tag{9} \frac{1}{2} \lim_{h \to 0} \left[-9\cos(a + 3h) + 8\cos(a + 2h) - \cos(a + h) \right] \\ &=\tag{10} \frac{1}{2} (-9 \cos a + 8 \cos a - \cos a) \\ &=\tag{11} \frac{1}{2} (-2 \cos a) \\ &=\tag{12} \boxed{-\cos a} \end{align}$$

L'Hôpital's was used in steps $(2)$-$(3)$, $(5)$-$(6)$, and $(7)$-$(8)$.

Answer 3

You could use the sum-to-product formulas .

$\sin(a\!+\!3h)\!-\!3\sin(a\!+\!2h)\!+\!3\sin(a\!+\!h)\!-\!\sin a=$

$=\sin(a\!+\!3h)\!-\!\sin(a\!+\!2h)\!-\!2\big[\sin(a\!+\!2h)\!-\!\sin(a\!+\!h)\big]\!+\\\quad+\sin(a\!+\!h)\!-\!\sin a=$

$=2\cos\left(\!a\!+\!\dfrac{5h}2\!\right)\!\sin\left(\!\dfrac h2\!\right)\!-\!4\cos\left(\!a\!+\!\dfrac{3h}2\!\right)\!\sin\left(\!\dfrac h2\!\right)\!+\\\quad+2\cos\left(\!a\!+\!\dfrac h2\!\right)\!\sin\left(\!\dfrac h2\!\right)=$

$\!\!=2\sin\left(\!\dfrac h2\!\right)\!\!\left[\cos\left(\!a\!+\!\dfrac{5h}2\!\right)\!\!-\!2\cos\left(\!a\!+\!\dfrac{3h}2\!\right)\!\!+\!\cos\left(\!a\!+\!\dfrac h2\!\right)\right]\!=$

$=2\sin\left(\!\dfrac h2\!\right)\!\!\left[\cos\left(\!a\!+\!\dfrac{5h}2\!\right)\!-\!\cos\left(\!a\!+\!\dfrac{3h}2\!\right)\!-\!\cos\left(\!a\!+\!\dfrac{3h}2\!\right)\!+\\\quad+\cos\left(\!a\!+\!\dfrac h2\!\right)\right]\!=$

$\!\!\!\!\!=2\sin\left(\!\dfrac h2\!\right)\!\!\left[-2\sin\left(a\!+\!2h\right)\sin\left(\!\dfrac h2\!\right)\!\!+\!2\sin\left(a\!+\!h\right)\sin\left(\!\dfrac h2\!\right)\right]\!=$

$=-4\sin^2\!\!\left(\!\dfrac h2\!\right)\!\!\bigg[\sin\left(a\!+\!2h\right)\!-\!\sin\left(a\!+\!h\right)\bigg]\!=$

$=-8\sin^2\!\!\left(\!\dfrac h2\!\right)\!\cos\left(\!a\!+\!\dfrac{3h}2\!\right)\!\sin\left(\!\dfrac h2\!\right)=$

$=-8\sin^3\!\!\left(\!\dfrac h2\!\right)\!\cos\left(\!a\!+\!\dfrac{3h}2\!\right)\,.$

Consequently ,

$\lim\limits_{h\to 0}\dfrac{\sin(a+3h)-3\sin(a+2h)+3\sin(a+h)-\sin a}{h^3}=$

$=\lim\limits_{h\to 0}\dfrac{-8\sin^3\!\!\left(\frac h2\right)\cos\left(a\!+\!\frac{3h}2\right)}{h^3}=$

$=-\lim\limits_{h\to 0}\dfrac{\sin^3\!\!\left(\frac h2\right)\cos\left(a\!+\!\frac{3h}2\right)}{\left(\frac h2\right)^3}=$

$=-\lim\limits_{h\to 0}\left[\dfrac{\sin\!\left(\frac h2\right)} {\frac h2}\right]^3\!\!\!\cos\left(a\!+\!\frac{3h}2\right)=-1^3\!\cdot\cos a=$

$=-\cos a\;.$