Does ln of a Fraction Approach Infinity as R Increases?

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SUMMARY

The limit of the expression ln(1 + 1/√(1 + 1/R)) / (1 - 1/√(1 + 1/R)) as R approaches infinity does not approach infinity. As R increases, the term 1/√(1 + R) approaches 0, leading the argument of the natural logarithm to converge to ln(1/1) = ln(1) = 0. Therefore, the limit of the logarithmic expression is 0, not infinity.

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Gamma
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Hi,



What is the limit of
[tex]ln(\frac{1+1/ \sqrt(1+1/R)}{1-1/ \sqrt(1+1/R})[/tex]

as R reaches infinity.


(latex did not show it very well. But the numerator is same as denominator except there is a + sign instead of a - sign.

As R reaches infinity the argument of the ln reaches infinity. Is the limit also infinity?





Thank You.

Gamma.
 
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Gamma said:
Hi,



What is the limit of [itex]ln(\frac{1+1/ \sqrt(1+1/R)}{1-1/ \sqrt(1+1/R})[/itex]

as R reaches infinity.


(latex did not show it very well. But the numerator is same as denominator except there is a + sign instead of a - sign.

As R reaches infinity the argument of the ln reaches infinity. Is the limit also infinity?





Thank You.

Gamma.
If your LaTeX image is big, don't use [ itex ], use [ tex ] instead.
So do you mean:
[tex]\lim_{R \rightarrow \infty} \ln \left( \frac{1 + \frac{1}{\sqrt{1 + R}}}{1 - \frac{1}{\sqrt{1 + R}}} \right)[/tex]? Or what?
If you mean that, then if [tex]R \rightarrow \infty[/tex], then [tex]\sqrt{1 + R} \rightarrow \ ?[/tex], [tex]\frac{1}{\sqrt{1 + R}} \rightarrow \ ?[/tex], [tex]1 \pm \frac{1}{\sqrt{1 + R}} \rightarrow \ ?[/tex].
Can you go from here? :)
 
Gamma, I've taken the liberty of editing your Latex by changing "itex" to "tex". "itex" doesn't work well with complex fractions.

As VietDao29 told you- the argument inside the ln does NOT go to infinity. It should be sufficient to see what happens to 1/R as R goes to infinity.
 

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