# Thread:

1. ## Finding a limit. Finding Maclaurin series.

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.

2. Originally Posted by skydiver1921
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.
$\lim_{x \to 0}\left(\frac{1}{x} - \frac{1}{\sin{x}}\right) = \lim_{x \to 0}\left(\frac{\sin{x} - x}{x\sin{x}}\right)$.

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

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

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

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

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

$= \frac{0}{-2}$

$= 0$.

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

3. Originally Posted by skydiver1921
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.

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

Assume that $y$ can be written as a polynomial. Then

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

By letting $x = 0$, we can see that $c_0 = 0$.

Differentiate both sides:

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

By letting $x = 0$, we can see that $c_1 = 1$.

Differentiate both sides:

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

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

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

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

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

Differentiate both sides:

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

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

Differentiate both sides:

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

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

So up to the $x^4$ term, we have

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