Finding a limit. Finding Maclaurin series.

May 2010
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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.
 
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Prove It

MHF Helper
Aug 2008
12,883
4,999
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.
\(\displaystyle \lim_{x \to 0}\left(\frac{1}{x} - \frac{1}{\sin{x}}\right) = \lim_{x \to 0}\left(\frac{\sin{x} - x}{x\sin{x}}\right)\).

Since this \(\displaystyle \to \frac{0}{0}\) you can use L'Hospital's Rule


\(\displaystyle \lim_{x \to 0}\left(\frac{\sin{x} - x}{x\sin{x}}\right) = \lim_{x \to 0}\left(\frac{\cos{x} - 1}{x\cos{x} + \sin{x}}\right)\).

This still \(\displaystyle \to \frac{0}{0}\), so use L'Hospital's Rule again...


\(\displaystyle \lim_{x \to 0}\left(\frac{\cos{x} - 1}{x\cos{x} + \sin{x}}\right) = \lim_{x \to 0}\left(\frac{-\sin{x}}{-x\sin{x} + \cos{x} + \cos{x}}\right)\)

\(\displaystyle = \lim_{x \to 0}\left(\frac{\sin{x}}{x\sin{x} - 2\cos{x}}\right)\)

\(\displaystyle = \frac{0}{-2}\)

\(\displaystyle = 0\).



So \(\displaystyle \lim_{x \to 0}\left(\frac{1}{x} - \frac{1}{\sin{x}}\right) = 0\).
 
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Prove It

MHF Helper
Aug 2008
12,883
4,999
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

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