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

In summary, the proof of the limit theorem in $\epsilon-delta$ language requires the converse implication, which is not stated in the proof.
  • #1
arsenaler
1
0
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.
 
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  • #2
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.
 
  • #3
Since n and $log_a(n)$, for n> 1, are positive we can write that as $\frac{log_a(n)}{n}< \epsilon$ and then $log_a(n)< n\epsilon$.
 
  • #4
arsenaler said:
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.
 
  • #5
Prove It said:
$\displaystyle \begin{align*} 1 < \left| \ln{ \left( n \right) } \right| &< \epsilon \left| \ln{ \left( a \right) } \right| \left| n \right| \\
1 &< \epsilon \left| \ln{ \left( a \right) } \right| \left| n \right|\end{align*}$
And what is the relationship between these two lines?
 
  • #6
Evgeny.Makarov said:
And what is the relationship between these two lines?

What do you mean?
 
  • #7
Well, a proof is not a collection of random statements, is it? There must be a logical relationship between statements, for example, all of them should be equivalent or each of them should imply the following one. I am asking about the logical relationship between the two statements in post 5 and why this relationship is sufficient for proving that $\left|\dfrac{\log_a(n)}{n}\right|<\epsilon$ for all $n>\dfrac{1}{\epsilon|\ln(a)|}$.
 
  • #8
The "logical relationship" between those two lines is
"If a< b< c then a< c".
 
  • #9
Country Boy said:
The "logical relationship" between those two lines is
"If a< b< c then a< c".

OK, but the proof claims that if $n>\dfrac{1}{\epsilon|\ln(a)|}$, then $\left|\dfrac{\log_a(n)}{n}\right|<\epsilon$, and this requires the converse implication. More precisely, $n>\dfrac{1}{\epsilon|\ln(a)|}$ guarantees that $1<\epsilon|\ln(a)|n$, while one needs $\ln(n)<\epsilon|\ln(a)|n$.
 
  • #10
Evgeny.Makarov said:
OK, but the proof claims that if $n>\dfrac{1}{\epsilon|\ln(a)|}$, then $\left|\dfrac{\log_a(n)}{n}\right|<\epsilon$, and this requires the converse implication. More precisely, $n>\dfrac{1}{\epsilon|\ln(a)|}$ guarantees that $1<\epsilon|\ln(a)|n$, while one needs $\ln(n)<\epsilon|\ln(a)|n$.

Every step is reversible.
 
  • #11
Let $a=e$, so $\ln(a)=1$. Are you claiming that if $n>\dfrac{1}{\epsilon}$, then $\dfrac{\ln(n)}{n}<\epsilon$? The assumption only guarantees that $\dfrac{1}{n}<\epsilon$, without $\ln(n)$ in the numerator. For an example, let $n=8>e^2$, then $\ln(n)>2$. Also let $\epsilon=\dfrac{1}{7}$. Then $n>\dfrac{1}{\epsilon}$, but $\dfrac{\ln(n)}{n}>\dfrac{2}{8}=\dfrac{1}{4}>\dfrac{1}{7}=\epsilon$.
 
  • #12
I see what your issue is now. Generally with $\displaystyle \begin{align*} \epsilon\delta \end{align*}$ proofs it's assumed that as long as you can find a relationship from a boundary, that every step should be able to be reversed. I'll see if I can prove it the other way as well.

Note, I'm leaving off the absolute values on the terms that are obviously nonnegative.

It seems to work fine for bases $\displaystyle \begin{align*} a > \mathbf{e} \end{align*}$, because then

$\displaystyle \begin{align*} a &> \mathbf{e} \\
\ln{ \left( a \right) } &> \ln{ \left( \mathbf{e} \right) } \\
\ln{ \left( a \right) } &> 1 \\
n \ln{ \left( a \right) } &> n \end{align*}$

and it's well known that $\displaystyle \begin{align*} n > \ln{ \left( n \right) } \end{align*}$ for all $\displaystyle \begin{align*} n \in \mathbf{R} \end{align*}$, so that gives

$\displaystyle \begin{align*} n \ln{ \left( a \right) } &> n > \ln{ \left( n \right) } \\ n\ln {\left( a \right) } &> \ln{ \left( n \right) } \end{align*}$

It also seems to work for bases $\displaystyle \begin{align*} b = \frac{1}{a} \end{align*}$, where $\displaystyle \begin{align*} a \end{align*}$ is as before defined to be $\displaystyle \begin{align*} a > \mathbf{e} \end{align*}$

$\displaystyle \begin{align*} \left| \ln{ \left( b \right) } \right| &= \left| \ln{ \left( \frac{1}{a} \right) } \right| \\
&= \left| - \ln{ \left( a \right) } \right| \\ &= \ln{ \left( a \right) } \end{align*}$

so the same inequality would apply.

I will have to keep thinking about bases in $\displaystyle \begin{align*} \left( \frac{1}{\mathbf{e}} , 1 \right) \cup \left( 1 , \mathbf{e} \right) \end{align*}$.
 

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

1. What is the definition of a limit in epsilon-delta language?

A limit in epsilon-delta language is a mathematical concept used to describe the behavior of a function as its input approaches a specific value. It is defined by two parameters: epsilon (ε) and delta (δ). Epsilon represents the desired level of accuracy, while delta represents the distance from the input value at which the function must stay within epsilon. In other words, a limit in epsilon-delta language means that the function's output gets arbitrarily close to a specific value as its input gets closer to a given value.

2. How is the limit of a logarithmic function defined in epsilon-delta language?

The limit of a logarithmic function in epsilon-delta language can be defined as follows: for any given epsilon greater than 0, there exists a delta greater than 0 such that if the absolute value of the difference between the input value and the limit point is less than delta, then the absolute value of the difference between the function's output and the limit value is less than epsilon. In other words, the function's output gets arbitrarily close to the limit value as its input gets closer to the limit point.

3. What is the proof that lim loga_n/n = 0 in epsilon-delta language?

The proof that lim loga_n/n = 0 in epsilon-delta language involves using the definition of a limit to show that for any given epsilon greater than 0, there exists a delta greater than 0 such that if the absolute value of the difference between the input value and the limit point is less than delta, then the absolute value of the difference between loga_n/n and 0 is less than epsilon. This can be achieved by manipulating the logarithmic function and using properties of limits.

4. Why is it important to prove that lim loga_n/n = 0 in epsilon-delta language?

Proving that lim loga_n/n = 0 in epsilon-delta language is important because it helps us understand the behavior of logarithmic functions as their input values approach a specific value. This is a fundamental concept in calculus and is used in many applications, such as in the study of derivatives and integrals. Additionally, understanding the proof can help us develop a deeper understanding of the properties of logarithmic functions.

5. Can the concept of a limit in epsilon-delta language be applied to other types of functions?

Yes, the concept of a limit in epsilon-delta language can be applied to various types of functions, including polynomial, exponential, and trigonometric functions. The definition of a limit remains the same, but the specific proof for each type of function may vary. This concept is an essential tool in calculus and is used to study the behavior of functions in various contexts.

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