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Proof that lim loga_n/n = 0 in epsilon delta language

arsenaler

New member
Jun 23, 2021
1
Let $\,a>0\,,\,a\neq1\,$ be a real number. We can prove by using the continuity of $\ln n$ function that $\;\lim\limits_{n\to\infty}\dfrac{\log_an}n=0\;$

However, this problem appears in my problems book quite early right after the definition of $\epsilon$-language definition of limit of a sequence, the reader is supposed not to know anything about continuity.

My question is: Is there any proof for this result in $\epsilon-delta$ language that is more elementary?

Please help me.

Thanks.
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
787
There won't be any "$\delta$" because the limit is being taken as n goes to infinity, not any finite value. What you want instead is to show that, given $\epsilon> 0$ there exist N such that if n>N then $|\frac{ln_a(n)}{n}|< \epsilon$.

Further, the use of "n" rather than "x" implies that this a sequence, not a function of x.
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
787
Since n and $log_a(n)$, for n> 1, are positive we can write that as $\frac{log_a(n)}{n}< \epsilon$ and then $log_a(n)< n\epsilon$.
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,433
Let $\,a>0\,,\,a\neq1\,$ be a real number. We can prove by using the continuity of $\ln n$ function that $\;\lim\limits_{n\to\infty}\dfrac{\log_an}n=0\;$

However, this problem appears in my problems book quite early right after the definition of $\epsilon$-language definition of limit of a sequence, the reader is supposed not to know anything about continuity.

My question is: Is there any proof for this result in $\epsilon-delta$ language that is more elementary?

Please help me.

Thanks.
To prove that $\displaystyle \begin{align*} \lim_{n \to \infty}{ \frac{\log_a{\left( n \right) }}{n} } = 0 \end{align*}$ we would need to prove that:

For any $\displaystyle \begin{align*} \epsilon > 0 \end{align*}$ there exists an $\displaystyle \begin{align*} N > 0 \end{align*}$ such that $\displaystyle \begin{align*} n > N \implies \left| \frac{\log_a{\left( n \right) }}{n} - 0 \right| < \epsilon \end{align*}$.

So to do this:

$\displaystyle \begin{align*} \left| \frac{\log_a{\left( n \right) }}{n} - 0 \right| &< \epsilon \\
\left| \log_a{\left( n \right) } \right| &< \epsilon \left| n \right| \\
\left| \frac{\ln{ \left( n \right) }}{\ln{ \left( a \right) }} \right| &< \epsilon \left| n \right| \\
\left| \ln{ \left( n \right) } \right| &< \epsilon \left| \ln{ \left( a \right) } \right| \left| n \right| \end{align*}$

Now, obviously we are considering what happens for very large $\displaystyle \begin{align*} n \end{align*}$, so it is obvious that there is no reason why we need to consider all $\displaystyle \begin{align*} n > 0 \end{align*}$, so why don't we consider say $\displaystyle \begin{align*} n > \mathrm{e} \implies \ln{ \left( n \right) } > 1 \end{align*}$.

Therefore, if $\displaystyle \begin{align*} n > \mathrm{e} \end{align*}$

$\displaystyle \begin{align*} 1 < \left| \ln{ \left( n \right) } \right| &< \epsilon \left| \ln{ \left( a \right) } \right| \left| n \right| \\
1 &< \epsilon \left| \ln{ \left( a \right) } \right| \left| n \right| \\
\frac{1}{\epsilon \left| \ln{ \left( a \right) } \right| } &< \left| n \right| \\
\left| n \right| &> \frac{1}{\epsilon \left| \ln{ \left( a \right) } \right| } \end{align*}$

So provided that we ensure we start with an $\displaystyle \begin{align*} n > \mathrm{e} \end{align*}$, we can set $\displaystyle \begin{align*} N = \frac{1}{\epsilon \left| \ln{ \left( a \right) } \right| } \end{align*}$ and then you can write the proof.
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,530
$\displaystyle \begin{align*} 1 < \left| \ln{ \left( n \right) } \right| &< \epsilon \left| \ln{ \left( a \right) } \right| \left| n \right| \\
1 &< \epsilon \left| \ln{ \left( a \right) } \right| \left| n \right|\end{align*}$
And what is the relationship between these two lines?