# Solving a Limit Problem

• 22990atinesh
In this case, the restriction is that a*b must be a positive number.So, in summary, the limit as n approaches ##\infty## is not true, and you cannot take the log of both sides of an equation to get a log of a quotient, nor can you take the log of a sum and get the sum of the logs.f

#### 22990atinesh

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##

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.

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?

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##

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.