# Thread: Maclaurin series

1. ## Maclaurin series

Hi!

I've been trying to solve this problem for two days now without any luck.
$\displaystyle \lim_{x\rightarrow 0}\left( \frac{1}{\sin(x)} -\frac{1}{\ln(x+1)}\right)$

then i get this and now im completely stuck

$\displaystyle \lim_{x\rightarrow 0} \frac{\ln(x+1)-\sin(x)}{\sin(x)\ln(x+1)}$

when i check my notes from class i cant see that our professor has shown us how its done when x->0 or infinity or w/e...

Thanks in advance!

2. Applying L'Hopital's Rule twice will give you the answer.

3. Yeah thanks for the tip, but some in my class tried that but our professor told em he want them to use the Maclaurin.

4. Originally Posted by ogward
Hi!

I've been trying to solve this problem for two days now without any luck.
$\displaystyle \lim_{x\rightarrow 0}\left( \frac{1}{\sin(x)} -\frac{1}{\ln(x+1)}\right)$

then i get this and now im completely stuck

$\displaystyle \lim_{x\rightarrow 0} \frac{\ln(x+1)-\sin(x)}{\sin(x)\ln(x+1)}$

when i check my notes from class i cant see that our professor has shown us how its done when x->0 or infinity or w/e...

Thanks in advance!
What DrSteve says, but given the thread title you can expand $\sin(x)$ and $\ln(x+1)$ as Maclaurin series simplify etc... which will work.

CB

5. this is how far i've got...
is it correct? and how do i continue?

$sinx=\frac{1}{1!}x-\frac{1}{3!}x^{3}+\frac{1}{5!}x^{5}-\frac{1}{7!}x^{7}+..........
\\ =\sum ^{\infty }_{n=1}(-1)^{n+1}\frac{x^{2n-1}}{(2n-1)!}$

$ln(x+1)=\frac{x}{1!}-\frac{x^{2}}{3!}+\frac{2x^{3}}{3!}-\frac{6x^{4}}{4!}+..........
\\ \sum ^{\infty }_{n=0}unsure\frac{x^{n}}{n!}
$

im unsure of this last part.

6. Originally Posted by ogward
this is how far i've got...
is it correct? and how do i continue?

$sinx=\frac{1}{1!}x-\frac{1}{3!}x^{3}+\frac{1}{5!}x^{5}-\frac{1}{7!}x^{7}+..........
\\ =\sum ^{\infty }_{n=1}(-1)^{n+1}\frac{x^{2n-1}}{(2n-1)!}$

$ln(x+1)=\frac{x}{1!}-\frac{x^{2}}{3!}+\frac{2x^{3}}{3!}-\frac{6x^{4}}{4!}+..........
\\ \sum ^{\infty }_{n=0}unsure\frac{x^{n}}{n!}
$

im unsure of this last part.
Does not matter you only need the first two non-zeros terms so put:

$\sin(x)=x-\dfrac{x^3}{6}+O(x^5)$

and:

$\ln(x+1)=x-\dfrac{x^2}{2}+O(x^3)$

(here the big-O notation represents the remainder in the truncated series)

Then procceed.

CB

7. the problem is i have no idea what should be done next...

8. I think you can finish up by "substituting in" what CaptainBlack has given you, dividing numerator and denominator $x^2$, then taking the limit (plug in a 0 for x).

9. this is what I get
$\lim_{x\rightarrow 0}=\frac{\frac{x^3}{6}-\frac{x^2}{2}}{\frac{x^5}{12}-\frac{x^4}{6}-\frac{x^3}{2}+x^2
$

click on get above.

should I put x=0? then I get 0/0, is that good?

10. Originally Posted by ogward
this is what I get
$\lim_{x\rightarrow 0}=\frac{\frac{x^3}{6}-\frac{x^2}{2}}{\frac{x^5}{12}-\frac{x^4}{6}-\frac{x^3}{2}+x^2
$

click on get above.

should I put x=0? then I get 0/0, is that good?
Divide top and bottom of what your input to WA gives by $x^2$, and then let $x$ go to zero

CB

11. does this look good?

$\lim_{x\rightarrow 0}= \frac{2(x-3)}{(x-2)(x^2-6)}=\frac{-6}{12}$

12. Originally Posted by ogward
does this look good?

$\lim_{x\rightarrow 0}= \frac{2(x-3)}{(x-2)(x^2-6)}=\frac{-6}{12}$
Well the number on the right is correct.

CB

13. Originally Posted by ogward
this is what I get
$\lim_{x\rightarrow 0}=\frac{\frac{x^3}{6}-\frac{x^2}{2}}{\frac{x^5}{12}-\frac{x^4}{6}-\frac{x^3}{2}+x^2}
$

click on get above.

should I put x=0? then I get 0/0, is that good?
Divide the numerator & denominator by x² . Then set x=0.