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Limit problem from "Mean girls" TV program?

Posted 7 days ago
5 Replies
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I watched mean girls recently and i got interested in the math problem presented. evaluate the limit:

(lim)(x→0)  ( ln⁡(1-x)-sin⁡x)/(1-cos^2⁡x )

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One girl got the answer -1, which was wrong. The main character reached the right answer in 13 seconds, the limit does not exist. What intrigues me is what different ways could this be solved? How can you reach the answer in 13 seconds, and how did the other girl get -1?

5 Replies

Here are three methods:

(1) Plug it into Wolfram|Alpha.

(2) Plug it into Mathematica.

Or the off-topic response, (3) use Taylor series and don't mess up the signs like I did at first.

If the minus in the numerator became a plus, and the square in the denom term was lost, that would give -1. Offhand i do not see a way to get that result with only one typography change.

Interesting. Applying LHopital's rule twice gives -1/2:

expr = (Log[1 - x] - Sin[x])/(1 - Cos[x]^2);

d2num = D[Numerator[expr], {x, 2}]

(* -(1/(1-x)^2)+Sin[x] *)

d2denom = D[Denominator[expr], {x, 2}]

(* 2 Cos[x]^2-2 Sin[x]^2 *)

d2num/d2denom /. x -> 0

(* -(1/2) *)

But this is not correct, so LHopital's rule must arrive at an indeterminate form. But why is this indeterminate?

Recalling an admonition to mind the signs, the first derivative of the numerator does not vanish at 0. So you don't (sob) get to apply a second time.

Yes! I had forgotten that! Thanks, Daniel

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