Question

evaluate the following:

Answer 1

ok, where's the problem?

Answer 2

\[\lim_{x \rightarrow 0}\frac{ \ln(1-x)-\sin(x) }{ 1-\cos ^{2}(x) }\]

Answer 3

I smell L hôpital

Answer 4

do you remember l hopital?

Answer 5

yes L'Hopitals

Answer 6

derivate above and below each separated, until you can evaluate. Sometimes you might have to apply other properties. I don't think this is the case

Answer 7

This is the limit from the movie "Mean Girls" I'd like to see if there are any other approaches to evaluate the limit. Cady evaluates the answer in her head.

Answer 8

lol, I only know two methods bro. Grahically and with L hôpital

Answer 9

graphically*

Answer 10

I found out the limit is zero. I gotta check if it's true.

Answer 11

The limit should not exist

Answer 12

http://www.wolframalpha.com/input/?i=limit+as+x+approaches+0+%28ln%281-x%29-sin%28x%29%29%2F%281-cos%5E2%28x%29%29

Answer 13

:sss I will cry then :C

Answer 14

How comfortable are you with Taylor series?

Answer 15

@SithsAndGiggles I want to learn how to apply Taylor here, please

Answer 16

that would be interesting.

Answer 17

Actually the method might not work. The series I had in mind are centered around \(x=0\) :/

Answer 18

For \(x\) near 0, we have \(\sin x\approx x\) and \(\ln(1-x)\approx -x\).
\[\lim_{x\to0}\frac{\ln(1-x)-\sin x}{1-\cos^2x}=\lim_{x\to0}\frac{\ln(1-x)-\sin x}{\sin^2x}=\lim_{x\to0}\frac{-x-x}{x^2}=\text{DNE}\]

Answer 19

Not the most rigorous, but I think you can actually reach this conclusion using Taylor series, I must have done something wrong in the series manipulation.

Answer 20

interesting solution. I'm going to have to understand how you can make the assumptions for sin and natural log

Answer 21

I think I see that you took the first two terms in the Mclauren series for sin(x) and ln(1-x)

Answer 22

That's essentially what it amounts to, yeah

Answer 23

Thank you very much. @SithsAndGiggles