Limit of logarithm of square roots function

In summary, the conversation is about proving a limit involving logarithms and square roots. The limit is simplified by pulling out a factor of \sqrt{s} and using a Taylor expansion with x=\frac{4m^2}{s}\rightarrow 0. The conversation ends with the person thanking the other for the help.
  • #1
salparadise
23
0

Homework Statement


I need to prove that the following limit holds when

[tex]\sqrt{s}>>m[/tex]

[tex]\log\left(\frac{\sqrt{s}+\sqrt{s-4m^2}}{\sqrt{s}-\sqrt{s-4m^2}}\right) \rightarrow \log\left(\frac{s}{m^2}\right)[/tex]

The Attempt at a Solution


I've tried several manipulations using logarithms properties but had no success.

Thanks
 
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  • #2
You don't need to bother with the logarithm, just focus on the stuff inside. I'll give you a starting hint; you should provide the rest of your workings after that if you can't get it and want further help. To start, I would pull the factor of [itex]\sqrt{s}[/itex] out from the top and bottom of the fraction inside the logarithm. Find the limiting form of the fraction in your desired limit.
 
  • #3
Hello Mute, thanks for the help. I've managed to do it, after Taylor expansion of [tex]\sqrt{1-x}[/tex], where in my limit:

[tex]x=\frac{4m^2}{s}\rightarrow 0[/tex].

As I'm under some time pressure to finish this, very bigg, home exam I'm already not thinking so clearly..

Thanks again
 

FAQ: Limit of logarithm of square roots function

What is the limit of the logarithm of the square roots function as x approaches infinity?

The limit of the logarithm of the square roots function as x approaches infinity is infinity. This means that the function grows without bound as x gets larger and larger.

Is the limit of the logarithm of the square roots function the same as the limit of the natural logarithm of the square roots function?

Yes, the limit of the logarithm of the square roots function and the limit of the natural logarithm of the square roots function are the same. This is because the natural logarithm is just a special case of the logarithm with base e, and the square roots function remains the same regardless of the base of the logarithm.

How do you calculate the limit of the logarithm of the square roots function?

To calculate the limit of the logarithm of the square roots function, you can use the rules of logarithms and exponentials to rewrite the function and then evaluate the limit. Alternatively, you can use l'Hopital's rule, which states that the limit of the quotient of two functions is equal to the quotient of their derivatives if the limit of the derivatives exists.

Does the limit of the logarithm of the square roots function exist at x=0?

No, the limit of the logarithm of the square roots function does not exist at x=0. This is because the function is undefined at x=0, as the square root of 0 is undefined. Therefore, the limit also does not exist at this point.

How does the limit of the logarithm of the square roots function change if the base of the logarithm is varied?

The limit of the logarithm of the square roots function will not change regardless of the base of the logarithm. This is because the function remains the same regardless of the base, and the limit is determined by the behavior of the function as x approaches a certain value, not the base of the logarithm.

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