Can I get some help in finding the limit of the following trigonometric function?

Ltsin(a+3h)-3sin(a+h)+3sin(a+h)-sin a

h-->0h^3

the h in the denominator is cube.

the denominator is sin(a+3h)-3sin(a+2h)+3sin(a+h)-sin a

Printable View

- Sep 13th 2008, 10:36 AMtariq_h_tauheedLimit of trigonometric function
Can I get some help in finding the limit of the following trigonometric function?

*Lt*__sin(a+3h)-3sin(a+h)+3sin(a+h)-sin a__

*h-->0*h^3

the h in the denominator is cube.

the denominator is sin(a+3h)-3sin(a+2h)+3sin(a+h)-sin a - Sep 13th 2008, 12:33 PMShyam
Is your question like this:

$\displaystyle \mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {a + 3h} \right) - 3\sin \left( {a + h} \right) + 3\sin \left( {a + h} \right) - \sin a}}

{{h^3 }}

\$

or is your question like this:

$\displaystyle \mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {a + 3h} \right) - \sin a}}

{{h^3 }}

\$ - Sep 13th 2008, 12:55 PMSoroban
Hello, tariq_h_tauheed!

Where did this problem come from?

It looks like we walked in on the*middle*of someone's work . . .

Quote:

$\displaystyle \lim_{h\to0}\frac{\sin(a+3h) - \overbrace{3\sin(a+h) + 3\sin(a+h)}^{\text{Don't these cancel?}} - \sin(h)}{h^3} $

Could it be: . $\displaystyle \frac{\sin(a+3h) - \sin({\color{red}a})}{h^3}$

If so, we have a chance of solving it . . . otherwise, forget it!