Limit as x approaches infinity.

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SUMMARY

The limit as x approaches infinity of the expression (logx)(log(logx)) / x is confirmed to be zero. Attempts to solve this using L'Hospital's rule were ineffective, leading to the suggestion of using substitutions such as rewriting (logx)log(log(x)) as 10(log(logx))^2 and x as 10logx. The discussion emphasizes the importance of understanding logarithmic notation, specifically the distinction between log(x) as the natural logarithm (ln(x)) and its base-10 counterpart.

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Homework Statement


Compute Limit as x--> infinity of (logx)(log(logx)) / x

The Attempt at a Solution



Graphically, I see that the answer is perhaps zero, but I am not sure how to approach this algebraically. I worked at this for a couple hours, trying L'Hospital's rule but that did not really lead me anywhere. Is there some tricky substitution I am missing?

Can I do this?

Rewrite (logx)log(log(x))
as

10(log(logx))2

and then x as

10logx

and evaluate the limit of the exponents? (as in as a ratio)?
 
Last edited:
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I would try rescaling the independent variable as ##x=e^{\xi}##.
 
SPhy said:

Homework Statement


Compute Limit as x--> infinity of (logx)(log(logx)) / x

The Attempt at a Solution



Graphically, I see that the answer is perhaps zero, but I am not sure how to approach this algebraically. I worked at this for a couple hours, trying L'Hospital's rule but that did not really lead me anywhere. Is there some tricky substitution I am missing?

Can I do this?

Rewrite (logx)log(log(x))
as

10(log(logx))2

and then x as

10logx

and evaluate the limit of the exponents? (as in as a ratio)?

Isn't ##\log(x)## the natural logarithm ##\ln (x)##? Nowadays that is usually the case, with ##\log## and ##\ln## just being two alternative notations for the same function. Then, if alternative bases are needed they are indicated explicitly, such as ##\log[10](x)## or ##\log_{10}(x).##

However, if you are looking at a 50-100 year-old book that might not be the case, and even today there may be some exceptions: "usual" does not mean 100%.
 
Last edited:

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