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

#### arsenaler

##### New member
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?

Thanks.

#### Country Boy

##### Well-known member
MHB Math Helper
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
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
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?

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
\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?