How can these problems be worked out?

1) lim as x goes to 0 of (1/x - 1/sinx)

2) find the 1st 4 derivatives of y= ln(1+sinx) and the Mclaurin series up to the term containing x^4

I know some calculus, but I need help with this. Thank you.

\(\displaystyle y = \ln{(1 + \sin{x})}\).

Assume that \(\displaystyle y\) can be written as a polynomial. Then

\(\displaystyle \ln{(1 + \sin{x})} = c_0 + c_1x + c_2x^2 + c_3x^3 + c_4x^4 + \dots\).

By letting \(\displaystyle x = 0\), we can see that \(\displaystyle c_0 = 0\).

Differentiate both sides:

\(\displaystyle \frac{\cos{x}}{1 + \sin{x}} = c_1 + 2c_2x + 3c_3x^2 + 4c_4x^3 + 5c_5x^4 + \dots\).

By letting \(\displaystyle x = 0\), we can see that \(\displaystyle c_1 = 1\).

Differentiate both sides:

\(\displaystyle \frac{-\sin{x}(1 + \sin{x}) - \cos^2{x}}{(1 + \sin{x})^2} = 2c_2 + 3\cdot 2c_3x + 4\cdot 3c_4x^2 + 5\cdot 4c_5x^3 + 6\cdot 5c_6x^4 + \dots\)

\(\displaystyle \frac{-\sin{x} - \sin^2{x} - \cos^2{x}}{(1 + \sin{x})^2} = 2c_2 + 3\cdot 2c_3x + 4\cdot 3c_4x^2 + 5\cdot 4c_5x^3 + 6\cdot 5c_6x^4 + \dots\)

\(\displaystyle \frac{-(1 + \sin{x})}{(1 + \sin{x})^2} = 2c_2 + 3\cdot 2c_3x + 4\cdot 3c_4x^2 + 5\cdot 4c_5x^3 + 6\cdot 5c_6x^4 + \dots\)

\(\displaystyle -(1 + \sin{x})^{-1} = 2c_2 + 3\cdot 2c_3x + 4\cdot 3c_4x^2 + 5\cdot 4c_5x^3 + 6\cdot 5c_6x^4 + \dots\)

By letting \(\displaystyle x = 0\) we can see that \(\displaystyle c_2 = -\frac{1}{2}\).

Differentiate both sides:

\(\displaystyle \cos{x}(1 + \sin{x})^{-2} = 3\cdot 2c_3 + 4\cdot 3\cdot 2c_4x + 5\cdot 4\cdot 3c_5x^2 + 6\cdot 5\cdot 4c_6x^3 + 7\cdot 6\cdot 5c_7x^4 + \dots\)

By letting \(\displaystyle x = 0\), we can see that \(\displaystyle c_3 = \frac{1}{3\cdot 2} = \frac{1}{6}\).

Differentiate both sides:

\(\displaystyle -2\cos^2{x}(1 + \sin{x})^{-3} - \sin{x}(1 + \sin{x})^{-2} = 4\cdot 3\cdot 2c_4 + 5\cdot 4\cdot 3\cdot 2c_5x + 6\cdot 5\cdot 4\cdot 3c_6x^2 + \dots\)

By letting \(\displaystyle x = 0\), we can see that \(\displaystyle -\frac{1}{4\cdot 3} = -\frac{1}{12}\).

So up to the \(\displaystyle x^4\) term, we have

\(\displaystyle \ln{(1 + \sin{x})} = x - \frac{1}{2}x^2 + \frac{1}{6}x^3 - \frac{1}{12}x^4 + \dots\).