Question

The phrase '"the limit does not exist " can be heard in the movie "Mean Girls." In the final scene at a math contest the moderator ask them to find the limit of this to which Lohan's character says, the limit does not exist In real life, how would you know that. The original limit is indeterminant but a quick application of l'hopital, and it is still indeterminant. That does not mean the limit DNE. Can this be solved for real?

Answer 1

Problem appears on attached file photo.

Answer 2

-infinity. that's one funky looking graph, too!

Answer 3

I don't think the limit is negative infinity. But, I need to look at that graph.

Answer 4

Can't you simplify it a bit?

Answer 5

\[\begin{split}
\lim_{x\to 0}\frac{\ln(x-1)-\sin(x)}{1-\cos^2(x)}
&=\lim_{x\to 0}\frac{\ln(x-1)-\sin(x)}{\sin^2(x)} \\
&=\lim_{x\to 0}\frac{\ln(x-1)}{\sin^2(x)} -\lim_{x\to 0}\frac{\sin(x)}{\sin^2(x)} \\
&=\lim_{x\to 0}\frac{\ln(x-1)}{\sin^2(x)} -\lim_{x\to 0}\frac{1}{\sin(x)} \\

\end{split}
\]And we know \[
\lim_{x\to 0}\frac{1}{\sin(x)}
\]doesn't exist..

Answer 6

@wio
that is not enough to say the limit does not exist

Answer 7

one way to know is when the left hand limit is not the same as the right hand limit
\[\lim_{x \rightarrow 0^+}=-\infty\]
\[\lim_{x \rightarrow 0^-}f(x)=+\infty\]
so \[\lim_{x \rightarrow 0}f(x)=UND\]

Answer 8

from the graph of ln(1-x)
this is ln(x) refleted y -axis
shifted one to the left
we know the graph tends to + infty from the left and - infty from the right

Answer 9

Suggested Answer
----------------
Do a quick approximation by considering the linear approximations. Both ln(1-x) and sin(x) look like y = x at the origin, so you have something like (x-x)/x^2.

Since the numerator approaches zero linearly while the denominator approaches quadratically, the denominator wins. The limit does not exist.