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

In summary: If the numerator is larger, then the function will approach -1 as X-> pi/2. If the denominator is larger, then the function will approach 1 as X-> pi/2.
  • #1
LASmith
21
0

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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  • #2
Hi LASmith! :smile:

l'Hôpital's rule? :wink:
 
  • #3
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.
 
  • #4
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: )
 
  • #5
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.
 
  • #6
Consider: What are both of the functions doing as X-> pi/2?

As you are in precalc and not expected to use L'Hopitals 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?
 

1. What is the limit of log sin equal to 0?

The limit of log sin as it approaches 0 is 0. This can be proven using the squeeze theorem, where we can compare the function to sin x and -sin x, and show that they both approach 0 as x approaches 0.

2. What is the limit of log sin equal to 1?

The limit of log sin as it approaches 1 is 0. This can be shown by substituting 1 into the function, which results in log 1, which equals 0.

3. What is the limit of log sin equal to infinity?

The limit of log sin as it approaches infinity does not exist. This is because the function oscillates between -infinity and infinity as x approaches infinity, so it does not approach a single value.

4. How do you determine the limit of log sin?

The limit of log sin can be determined by using the properties of limits, such as the squeeze theorem and substitution, to evaluate the function as x approaches the given value (0, 1, or infinity).

5. Why is the limit of log sin important in mathematics?

The limit of log sin is important in mathematics because it helps us understand the behavior of the logarithmic function and the trigonometric function sin x. It also plays a role in solving various mathematical problems and applications, especially in calculus and differential equations.

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