Limit of x/sqrt(1-cosx) Approaching 0 from Negative Side

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SUMMARY

The limit of x/sqrt(1-cos(x)) as x approaches 0 from the negative side can be evaluated using L'Hôpital's Rule and trigonometric identities. Initially, attempts to apply L'Hôpital's Rule directly resulted in an indeterminate form of 0/0. However, by rationalizing the expression and utilizing the identity 1 - cos(x) = 2sin²(x/2), the limit can be simplified to x/sqrt(2)sin(x/2), allowing for a successful application of L'Hôpital's Rule without encountering the indeterminate form.

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aquitaine
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Ok, the problem is find the limit of x/sqrt(1-cosx) as x approaches 0 from the negative side.

First I tried simply applying l'hopital's rule to see what would happen, and it didn't work.

Next I tried rationalizing it by multiplying the numerator and denominator by sqrt(1+cosx), then using a trig identity (1-(cosx)^2=(sinx)^2 to get sqrt((sinx)^2) or simply sinx. Then I applied l'hopital's rule and ended up with a big mess that still ended up with 0/0.

Did I miss something or do something incorrectly?
 
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1 - cos(x) = 2sin^2(x/2)

x/sqrt(1-cos(x)) = x/sqrt(2)sin(x/2)

Using l'hospital's on this does not lead to 0/0.
 
So I was missing something, thanks.
 

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