How to Solve a Limit Problem Involving Logarithms?

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The discussion centers on evaluating the limit of the expression lim_{n → ∞} (n^{1.74}) / (n × (log n)^9). Participants debate the validity of taking logarithms of both the numerator and denominator, with some asserting it alters the original problem. It is clarified that while logarithmic properties allow for log(a * b) = log(a) + log(b) and log(a/b) = log(a) - log(b), the transformation used in the limit evaluation is incorrect. The conversation also touches on the complexity level of the problem, questioning whether it is suitable for high school or college students. Ultimately, the correct approach to comparing growth rates of functions is emphasized.
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Suppose there is a limit
##\lim_{n \to \infty} \frac{n^{1.74}}{n \times (\log n)^9}##
Taking logs both on numerator and denominator
##=\lim_{n \to \infty} \frac{1.74 \times \log n}{\log n + 9 \log \log n}##
What can we say about the limit as n approaches ##\infty##
 
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How can you take log on both numerator and denominator. You are changing the question.
1/2= 0.5 is not the same as
log(1)/log(2) , which is 0.
 
Raghav Gupta said:
How can you take log on both numerator and denominator. You are changing the question.
1/2= 0.5 is not the same as
log(1)/log(2) , which is 0.
In order to compare growth rate of two functions we can do that.
 
Can you give an example?
Algebraically , in question you have stated how one can do that on numbers?
Is that a university( college ) level problem
Or a high school one?
 
22990atinesh said:
Suppose there is a limit
##\lim_{n \to \infty} \frac{n^{1.74}}{n \times (\log n)^9}##
Taking logs both on numerator and denominator
##=\lim_{n \to \infty} \frac{1.74 \times \log n}{\log n + 9 \log \log n}##
What can we say about the limit as n approaches ##\infty##

22990atinesh said:
In order to compare growth rate of two functions we can do that.
In your first post (quoted above) you are claiming that ##\lim_{n \to \infty} \frac{n^{1.74}}{n \times (\log n)^9} = \lim_{n \to \infty} \frac{1.74 \times \log n}{\log n + 9 \log \log n}##. Most emphatically, this is NOT TRUE! You can take the log of both sides of an equation, but you cannot take the log of a quotient to get a quotient of logs, nor can you take the log of a sum and get the sum of the logs.

What IS true is that log(a * b) = log(a) + log(b), and that log(a/b) = log(a) - log(b), assuming suitable restrictions on the values of a and b.
 

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