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

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In summary, the conversation discussed finding the limit of x/sqrt(1-cosx) as x approaches 0 from the negative side. The first attempted method was using l'hopital's rule, which did not work. The next approach involved rationalizing the expression and using a trig identity to simplify it, but it still resulted in 0/0. The conversation concluded by realizing that a step was missed and the correct approach involved using a different trig identity.
  • #1
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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  • #2
1 - cos(x) = 2sin^2(x/2)

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

Using L'Hopitals on this does not lead to 0/0.
 
  • #3
So I was missing something, thanks.
 

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

What is the limit of x/sqrt(1-cosx) as x approaches 0 from the negative side?

The limit of x/sqrt(1-cosx) as x approaches 0 from the negative side is equal to 0. This can be solved using the L'Hopital's rule or by recognizing that the numerator approaches 0 while the denominator approaches infinity.

How do you prove that the limit of x/sqrt(1-cosx) as x approaches 0 from the negative side is 0?

To prove that the limit is 0, we can use the Squeeze Theorem. We know that -1 ≤ cosx ≤ 1 for all real numbers x. Therefore, we can write the inequality -x ≤ x(1-cosx) ≤ x and take the square root of each term. This gives us -√x ≤ √(x(1-cosx)) ≤ √x. As x approaches 0 from the negative side, the left and right sides of the inequality approach 0, thus proving that the limit of x/sqrt(1-cosx) is 0.

Can we use the L'Hopital's rule to find the limit of x/sqrt(1-cosx) as x approaches 0 from the negative side?

Yes, we can use the L'Hopital's rule to solve this limit. By taking the derivative of the numerator and denominator separately, we get the limit of 1/sqrt(1-cosx) as x approaches 0 from the negative side. This can be further simplified using trigonometric identities to get the final answer of 0.

Is it possible for the limit of x/sqrt(1-cosx) as x approaches 0 from the negative side to be undefined?

No, the limit of x/sqrt(1-cosx) as x approaches 0 from the negative side is always defined and equal to 0. This can be proven using the Squeeze Theorem or L'Hopital's rule.

How does the behavior of the function x/sqrt(1-cosx) differ between approaching 0 from the negative and positive side?

The behavior of the function x/sqrt(1-cosx) is the same when approaching 0 from both the negative and positive side. The limit of the function is 0 in both cases, and the graph of the function approaches the x-axis as x approaches 0 from either side. The only difference is the direction in which the graph approaches the x-axis, with a negative slope for approaching from the negative side and a positive slope for approaching from the positive side.

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