Limits of log sin equal to 0/1/infinity?

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Homework Help Overview

The discussion revolves around evaluating the limit of the expression (ln sin x)/((π-2x)²) as x approaches π/2. Participants are exploring the nature of the limit, which initially presents as an indeterminate form 0/0, and are questioning whether the limit could be 1, 0, or infinity.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Some participants suggest using l'Hôpital's rule to resolve the indeterminate form. Others propose evaluating the function at values close to π/2 from both sides to observe behavior. There is also a consideration of the comparative behavior of the numerator and denominator as x approaches π/2.

Discussion Status

The discussion is active, with participants offering various approaches to analyze the limit. There is an acknowledgment that the limit does not have to be confined to the initially suggested values of 1, 0, or infinity. Participants are exploring different methods to understand the limit's behavior without reaching a consensus.

Contextual Notes

Participants note that the problem is situated within a precalculus context, which may limit the tools available for evaluation, such as l'Hôpital's rule. There is also mention of the challenges in calculating values extremely close to π/2 due to practical limitations of calculators.

LASmith
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Homework Statement



limx->[itex]\pi[/itex]/2(ln sin x)/(([itex]\pi[/itex]-2x)2)

Homework Equations



The Attempt at a Solution



Putting [itex]\pi/2[/itex] into this equation gives 0/0, however, the limit of this could be 1,0 or infinity, so how would you distinguish between the three?
 
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Hi LASmith! :smile:

l'Hôpital's rule? :wink:
 
LASmith said:
Putting [itex]\pi/2[/itex] into this equation gives 0/0, however, the limit of this could be 1,0 or infinity, so how would you distinguish between the three?

LASmith, note that it doesn't have to be one of those three options.
 
LASmith said:

Homework Statement



limx->[itex]\pi[/itex]/2(ln sin x)/(([itex]\pi[/itex]-2x)2)

This being precalculus, I think it would be sufficient to evaluate the function for a few values very close to Pi/2, on one side then the other. For example, Pi/2 - 0.001, Pi/2 - 0.002, Pi/2 - 0.005 and also try Pi/2 + 0.001 etc. (Then graph the points, at least in your mind, so you can see what the function is doing in the vicinity of Pi/2.)

Ideally, you might think it would be useful to evaluate Pi/2 - 0.00000000001 etc. so as to be really close to the point of interest, Pi/2, but this many sig figs will exceed the capabilities of your calculator. There is software which will implement a calculator that can work accurately to hundreds of decimal places, but we really don't need such extravagance here. (Though it is fun to experiment with to confirm what happens really really close to difficult points. :wink: )
 
LASmith said:

Homework Statement



limx->[itex]\pi[/itex]/2(ln sin x)/(([itex]\pi[/itex]-2x)2)


Homework Equations






The Attempt at a Solution



Putting [itex]\pi/2[/itex] into this equation gives 0/0, however, the limit of this could be 1,0 or infinity, so how would you distinguish between the three?


That is why we have limx->[itex]\pi[/itex]/2

We consider what happens when x gets very close to [itex]\pi[/itex]/2 [from above and below] but never actually equaling [itex]\pi[/itex]/2.
 
Consider: What are both of the functions doing as X-> pi/2?

As you are in precalc and not expected to use l'hospital's Rule, look at the graphs of these functions and consider which has larger values as X-> pi/2

If you have a very small value in the numerator and a larger value in the denominator, what happens?
 

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