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fishingspree2
May19-08, 06:03 PM
1. The problem statement, all variables and given/known data
\mathop {\lim }\limits_{x \to 0} (1 - \cos x)

2. The attempt at a solution

well as x goes to 0, cos x goes to 1... ln(1-1) is undefined
now if I forget about plugging in x=0 and think a little bit, the ln argument gets very very small, and the logarithm of a decimal number is a negative number... so I would say the limit is minus infinity.

however, is it possible to get that result analytically.... by transforming/simplyfing/etc. the function?

sorry, I have just finished precalculus and I am beginning calculus, so my calculus skills are crap =\

thank you

rocomath
May19-08, 06:14 PM
hmm ... problem is: \lim_{x\rightarrow0}(1-\cos x)?

Can't you just plug it in?

Or is it:

\lim_{x\rightarrow0}\ln{(1-\cos x)}

Ok so have the graph infront of you. You know as you "approach" coming from the right, it's negative infinity ...

\lim_{x\rightarrow0^{+}}\ln{(1-\cos x)}=-\infty

What about from the left? Because in order for the Limit to exist, the L & R Limits must agree. But for the natural logarithm, how is it uniquely defined?

kt9aq
May20-08, 05:13 PM
Once you have learned L^Hopitals rule, you will be able to apply it to this equation and get the same answer analytically.