Question

Mean Girls Limit Problem (for fun)

Answer 1

Show that the above limit does not exist.

Answer 2

@sillybilly123 @Nnesha

Answer 3

kiddin' right, voca ?!?!

it is this:

https://www.youtube.com/watch?v=oDAKKQuBtDo

Answer 4

\(\lim\limits_{x \to 0} \dfrac{\ln(1-x) - \sin x}{1- \cos^2 x} \)




\(= \lim\limits_{x \to 0} \dfrac{\ln(1-x) - \sin x}{\sin^2 x} = \star\)


Maclaurin Series:


\(\sin x \approx x \)

\( \ln(1-x) \approx - x \)


\(\star = \lim\limits_{x \to 0} \dfrac{- x - x }{x^2 } \)

\(= \lim\limits_{x \to 0} \dfrac{- 2 }{x } \)

That's DNE on both counts

Answer 5

expansion for \(\ln (1-x)\) is a cheat, but easy too :)

Answer 6

Lookin' up stuff is not the best way

\(\lim\limits_{x \to 0} \dfrac{\ln(1-x) - \sin x}{1- \cos^2 x} = \triangle\)


\(= \lim\limits_{x \to 0} \dfrac{\ln(1-x) - \sin x}{\sin^2 x} = \star\)


Maclaurin Series:


\({\sin x =\sum\limits_{n=0}^{\infty }\frac {(-1)^n }{(2n+1)!}}x^{2n+1} = x + \mathcal O (x^3) \)

\({\log(1-x) =-\sum\limits_{n=1}^{\infty }\frac {1}{n}} x^n = - x + \mathcal O (x^2) \)


\(\star = \lim\limits_{x \to 0} \dfrac{- x + \mathcal O (x^2) - (x + \mathcal O (x^3))}{(x + \mathcal O (x^3))^2 } \)

\(= \lim\limits_{x \to 0} \dfrac{- 2x + \mathcal O (x^2) }{x^2 + \mathcal O (x^4) } \)

\(= \lim\limits_{x \to 0} \dfrac{- \frac{2}{x} +\mathcal O(1) }{1 + \mathcal O (x^2) } \)

Break it down as quotient of limits:

\(= \dfrac{ \lim\limits_{x \to 0}( - \frac{2}{x} + \mathcal O(1) ) }{\lim\limits_{x \to 0} (1 + \mathcal O (x^2)) } \)

\(= \dfrac{ \lim\limits_{x \to 0}( -\frac{2}{x} ) }{1 } = - \lim\limits_{x \to 0} \dfrac{2}{x} \)