Show that $lim_{ntoinfty}frac{log_an}{n} = 0$ for $0<a<1$
$begingroup$
Let $0<a<1$, prove that:
$$
lim_{ntoinfty}frac{log_an}{n} = 0
$$
I've started with proving a simpler case for $a>1$. Choose some $varepsilon >0$ such that:
$$
frac{log_an}{n} < varepsilon iff log_an < nvarepsilon iff n<a^{nvarepsilon}
$$
We know that for $a>1$ and $kin mathbb N$:
$$
lim_{ntoinfty}frac{n^k}{a^n}=0
$$
Using that fact we may show that:
$$
exists N in mathbb N:forall n ge N => frac{n}{(a^varepsilon)^n} < 1
$$
Since the above yields a true statement starting from $n ge N$ we may conclude that initial assumption is also true and hence:
$$
lim_{ntoinfty} frac{log_an}{n} = 0
$$
Now consider the case for $0<a<1$:
$$
left|frac{log_an}{n}right| < varepsilon
$$
And this is where I got stuck, since $0 < a < 1$ the value of logarithm is negative and I'm not sure how to proceed. What are the next steps?
Also one of my thoughts was to present $a$ as:
$$
a = frac{1}{1+r} , r in mathbb R
$$
and then try to use Bernoulli's, but that didn't yield anything I could use.
Please note that i'm free to use anything before the definition of a derivative.
Update
Based on the hint by MathLover I think this is how I can proceed. Use the fact that:
$$
log_ax = - log_{1/a}x
$$
Hence:
$$
log_{1/a}n < nvarepsilon
$$
Define $b = {1 over a} > 1$, so:
$$
n < frac{1}{a^{nvarepsilon}} = b^{nvarepsilon}
$$
Now based on the reasoning for case when $a > 1$ we may as well conclude that:
$$
exists Nin mathbb N : forall n ge N implies frac{n}{b^{nvarepsilon}} < 1
$$
Thus:
$$
begin{cases}
lim_{ntoinfty}frac{log_an}{n} = 0 \
ain (0, 1) bigcup (1, +infty)
end{cases}
$$
calculus limits logarithms limits-without-lhopital epsilon-delta
asked Dec 4 '18 at 17:07
romanroman
2,17321224
$endgroup$
add a comment |
$begingroup$
Let $0<a<1$, prove that:
$$
lim_{ntoinfty}frac{log_an}{n} = 0
$$
I've started with proving a simpler case for $a>1$. Choose some $varepsilon >0$ such that:
$$
frac{log_an}{n} < varepsilon iff log_an < nvarepsilon iff n<a^{nvarepsilon}
$$
We know that for $a>1$ and $kin mathbb N$:
$$
lim_{ntoinfty}frac{n^k}{a^n}=0
$$
Using that fact we may show that:
$$
exists N in mathbb N:forall n ge N => frac{n}{(a^varepsilon)^n} < 1
$$
Since the above yields a true statement starting from $n ge N$ we may conclude that initial assumption is also true and hence:
$$
lim_{ntoinfty} frac{log_an}{n} = 0
$$
Now consider the case for $0<a<1$:
$$
left|frac{log_an}{n}right| < varepsilon
$$
And this is where I got stuck, since $0 < a < 1$ the value of logarithm is negative and I'm not sure how to proceed. What are the next steps?
Also one of my thoughts was to present $a$ as:
$$
a = frac{1}{1+r} , r in mathbb R
$$
and then try to use Bernoulli's, but that didn't yield anything I could use.
Please note that i'm free to use anything before the definition of a derivative.
Update
Based on the hint by MathLover I think this is how I can proceed. Use the fact that:
$$
log_ax = - log_{1/a}x
$$
Hence:
$$
log_{1/a}n < nvarepsilon
$$
Define $b = {1 over a} > 1$, so:
$$
n < frac{1}{a^{nvarepsilon}} = b^{nvarepsilon}
$$
Now based on the reasoning for case when $a > 1$ we may as well conclude that:
$$
exists Nin mathbb N : forall n ge N implies frac{n}{b^{nvarepsilon}} < 1
$$
Thus:
$$
begin{cases}
lim_{ntoinfty}frac{log_an}{n} = 0 \
ain (0, 1) bigcup (1, +infty)
end{cases}
$$
calculus limits logarithms limits-without-lhopital epsilon-delta
asked Dec 4 '18 at 17:07
romanroman
2,17321224
$endgroup$
add a comment |
$begingroup$
Let $0<a<1$, prove that:
$$
lim_{ntoinfty}frac{log_an}{n} = 0
$$
I've started with proving a simpler case for $a>1$. Choose some $varepsilon >0$ such that:
$$
frac{log_an}{n} < varepsilon iff log_an < nvarepsilon iff n<a^{nvarepsilon}
$$
We know that for $a>1$ and $kin mathbb N$:
$$
lim_{ntoinfty}frac{n^k}{a^n}=0
$$
Using that fact we may show that:
$$
exists N in mathbb N:forall n ge N => frac{n}{(a^varepsilon)^n} < 1
$$
Since the above yields a true statement starting from $n ge N$ we may conclude that initial assumption is also true and hence:
$$
lim_{ntoinfty} frac{log_an}{n} = 0
$$
Now consider the case for $0<a<1$:
$$
left|frac{log_an}{n}right| < varepsilon
$$
And this is where I got stuck, since $0 < a < 1$ the value of logarithm is negative and I'm not sure how to proceed. What are the next steps?
Also one of my thoughts was to present $a$ as:
$$
a = frac{1}{1+r} , r in mathbb R
$$
and then try to use Bernoulli's, but that didn't yield anything I could use.
Please note that i'm free to use anything before the definition of a derivative.
Update
Based on the hint by MathLover I think this is how I can proceed. Use the fact that:
$$
log_ax = - log_{1/a}x
$$
Hence:
$$
log_{1/a}n < nvarepsilon
$$
Define $b = {1 over a} > 1$, so:
$$
n < frac{1}{a^{nvarepsilon}} = b^{nvarepsilon}
$$
Now based on the reasoning for case when $a > 1$ we may as well conclude that:
$$
exists Nin mathbb N : forall n ge N implies frac{n}{b^{nvarepsilon}} < 1
$$
Thus:
$$
begin{cases}
lim_{ntoinfty}frac{log_an}{n} = 0 \
ain (0, 1) bigcup (1, +infty)
end{cases}
$$
calculus limits logarithms limits-without-lhopital epsilon-delta
asked Dec 4 '18 at 17:07
romanroman
2,17321224
$endgroup$
Let $0<a<1$, prove that:
$$
lim_{ntoinfty}frac{log_an}{n} = 0
$$
I've started with proving a simpler case for $a>1$. Choose some $varepsilon >0$ such that:
$$
frac{log_an}{n} < varepsilon iff log_an < nvarepsilon iff n<a^{nvarepsilon}
$$
We know that for $a>1$ and $kin mathbb N$:
$$
lim_{ntoinfty}frac{n^k}{a^n}=0
$$
Using that fact we may show that:
$$
exists N in mathbb N:forall n ge N => frac{n}{(a^varepsilon)^n} < 1
$$
Since the above yields a true statement starting from $n ge N$ we may conclude that initial assumption is also true and hence:
$$
lim_{ntoinfty} frac{log_an}{n} = 0
$$
Now consider the case for $0<a<1$:
$$
left|frac{log_an}{n}right| < varepsilon
$$
And this is where I got stuck, since $0 < a < 1$ the value of logarithm is negative and I'm not sure how to proceed. What are the next steps?
Also one of my thoughts was to present $a$ as:
$$
a = frac{1}{1+r} , r in mathbb R
$$
and then try to use Bernoulli's, but that didn't yield anything I could use.
Please note that i'm free to use anything before the definition of a derivative.
Update
Based on the hint by MathLover I think this is how I can proceed. Use the fact that:
$$
log_ax = - log_{1/a}x
$$
Hence:
$$
log_{1/a}n < nvarepsilon
$$
Define $b = {1 over a} > 1$, so:
$$
n < frac{1}{a^{nvarepsilon}} = b^{nvarepsilon}
$$
Now based on the reasoning for case when $a > 1$ we may as well conclude that:
$$
exists Nin mathbb N : forall n ge N implies frac{n}{b^{nvarepsilon}} < 1
$$
Thus:
$$
begin{cases}
lim_{ntoinfty}frac{log_an}{n} = 0 \
ain (0, 1) bigcup (1, +infty)
end{cases}
$$
calculus limits logarithms limits-without-lhopital epsilon-delta
calculus limits logarithms limits-without-lhopital epsilon-delta
asked Dec 4 '18 at 17:07
romanroman
2,17321224
asked Dec 4 '18 at 17:07
romanroman
2,17321224
edited Dec 4 '18 at 17:39
roman
asked Dec 4 '18 at 17:07
romanroman
2,17321224
asked Dec 4 '18 at 17:07
romanroman
2,17321224
asked Dec 4 '18 at 17:07
romanroman
2,17321224
2,17321224
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
We know that for arbitrary function $f$, $$lim_{ntoinfty}f(n)=0Longleftrightarrowlim_{ntoinfty}a^{f(n)}=1$$
So, if $f(n) = frac{log_a(n)}n$, then $$lim_{ntoinfty}frac{log_a(n)}n=0Longleftrightarrowlim_{ntoinfty}a^{frac{log_a(n)}n}=1$$But, if the second limit exists, then $$lim_{ntoinfty}a^{frac{log_a(n)}n}=lim_{ntoinfty}bigg(a^{log_a(n)}bigg)^{frac1n}=lim_{ntoinfty}n^{frac1n}=1$$
(If you aren't convinced of the last step, see this)
As such, if the limit exists, it has the value of 0.
answered Dec 4 '18 at 17:16
Don ThousandDon Thousand
4,332734
$endgroup$
add a comment |
$begingroup$
Hint:
$$log_a(n) = -log_{1/a}(n).$$
answered Dec 4 '18 at 17:18
Math LoverMath Lover
14k31436
$endgroup$
add a comment |
$begingroup$
Well, the cases $a<1$ or $a>1$ hardly have any difference as $$log_an=frac{log n}{log a}$$ and we know that $(1/n)log nto 0$.
answered Dec 4 '18 at 17:43
Paramanand SinghParamanand Singh
49.9k556163
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We know that for arbitrary function $f$, $$lim_{ntoinfty}f(n)=0Longleftrightarrowlim_{ntoinfty}a^{f(n)}=1$$
So, if $f(n) = frac{log_a(n)}n$, then $$lim_{ntoinfty}frac{log_a(n)}n=0Longleftrightarrowlim_{ntoinfty}a^{frac{log_a(n)}n}=1$$But, if the second limit exists, then $$lim_{ntoinfty}a^{frac{log_a(n)}n}=lim_{ntoinfty}bigg(a^{log_a(n)}bigg)^{frac1n}=lim_{ntoinfty}n^{frac1n}=1$$
(If you aren't convinced of the last step, see this)
As such, if the limit exists, it has the value of 0.
answered Dec 4 '18 at 17:16
Don ThousandDon Thousand
4,332734
$endgroup$
add a comment |
$begingroup$
We know that for arbitrary function $f$, $$lim_{ntoinfty}f(n)=0Longleftrightarrowlim_{ntoinfty}a^{f(n)}=1$$
So, if $f(n) = frac{log_a(n)}n$, then $$lim_{ntoinfty}frac{log_a(n)}n=0Longleftrightarrowlim_{ntoinfty}a^{frac{log_a(n)}n}=1$$But, if the second limit exists, then $$lim_{ntoinfty}a^{frac{log_a(n)}n}=lim_{ntoinfty}bigg(a^{log_a(n)}bigg)^{frac1n}=lim_{ntoinfty}n^{frac1n}=1$$
(If you aren't convinced of the last step, see this)
As such, if the limit exists, it has the value of 0.
answered Dec 4 '18 at 17:16
Don ThousandDon Thousand
4,332734
$endgroup$
add a comment |
$begingroup$
We know that for arbitrary function $f$, $$lim_{ntoinfty}f(n)=0Longleftrightarrowlim_{ntoinfty}a^{f(n)}=1$$
So, if $f(n) = frac{log_a(n)}n$, then $$lim_{ntoinfty}frac{log_a(n)}n=0Longleftrightarrowlim_{ntoinfty}a^{frac{log_a(n)}n}=1$$But, if the second limit exists, then $$lim_{ntoinfty}a^{frac{log_a(n)}n}=lim_{ntoinfty}bigg(a^{log_a(n)}bigg)^{frac1n}=lim_{ntoinfty}n^{frac1n}=1$$
(If you aren't convinced of the last step, see this)
As such, if the limit exists, it has the value of 0.
answered Dec 4 '18 at 17:16
Don ThousandDon Thousand
4,332734
$endgroup$
We know that for arbitrary function $f$, $$lim_{ntoinfty}f(n)=0Longleftrightarrowlim_{ntoinfty}a^{f(n)}=1$$
So, if $f(n) = frac{log_a(n)}n$, then $$lim_{ntoinfty}frac{log_a(n)}n=0Longleftrightarrowlim_{ntoinfty}a^{frac{log_a(n)}n}=1$$But, if the second limit exists, then $$lim_{ntoinfty}a^{frac{log_a(n)}n}=lim_{ntoinfty}bigg(a^{log_a(n)}bigg)^{frac1n}=lim_{ntoinfty}n^{frac1n}=1$$
(If you aren't convinced of the last step, see this)
As such, if the limit exists, it has the value of 0.
answered Dec 4 '18 at 17:16
Don ThousandDon Thousand
4,332734
answered Dec 4 '18 at 17:16
Don ThousandDon Thousand
4,332734
answered Dec 4 '18 at 17:16
Don ThousandDon Thousand
4,332734
answered Dec 4 '18 at 17:16
Don ThousandDon Thousand
4,332734
4,332734
add a comment |
add a comment |
$begingroup$
Hint:
$$log_a(n) = -log_{1/a}(n).$$
answered Dec 4 '18 at 17:18
Math LoverMath Lover
14k31436
$endgroup$
add a comment |
$begingroup$
Hint:
$$log_a(n) = -log_{1/a}(n).$$
answered Dec 4 '18 at 17:18
Math LoverMath Lover
14k31436
$endgroup$
add a comment |
$begingroup$
Hint:
$$log_a(n) = -log_{1/a}(n).$$
answered Dec 4 '18 at 17:18
Math LoverMath Lover
14k31436
$endgroup$
Hint:
$$log_a(n) = -log_{1/a}(n).$$
answered Dec 4 '18 at 17:18
Math LoverMath Lover
14k31436
answered Dec 4 '18 at 17:18
Math LoverMath Lover
14k31436
answered Dec 4 '18 at 17:18
Math LoverMath Lover
14k31436
answered Dec 4 '18 at 17:18
Math LoverMath Lover
14k31436
14k31436
add a comment |
add a comment |
$begingroup$
Well, the cases $a<1$ or $a>1$ hardly have any difference as $$log_an=frac{log n}{log a}$$ and we know that $(1/n)log nto 0$.
answered Dec 4 '18 at 17:43
Paramanand SinghParamanand Singh
49.9k556163
$endgroup$
add a comment |
$begingroup$
Well, the cases $a<1$ or $a>1$ hardly have any difference as $$log_an=frac{log n}{log a}$$ and we know that $(1/n)log nto 0$.
answered Dec 4 '18 at 17:43
Paramanand SinghParamanand Singh
49.9k556163
$endgroup$
add a comment |
$begingroup$
Well, the cases $a<1$ or $a>1$ hardly have any difference as $$log_an=frac{log n}{log a}$$ and we know that $(1/n)log nto 0$.
answered Dec 4 '18 at 17:43
Paramanand SinghParamanand Singh
49.9k556163
$endgroup$
Well, the cases $a<1$ or $a>1$ hardly have any difference as $$log_an=frac{log n}{log a}$$ and we know that $(1/n)log nto 0$.
answered Dec 4 '18 at 17:43
Paramanand SinghParamanand Singh
49.9k556163
answered Dec 4 '18 at 17:43
Paramanand SinghParamanand Singh
49.9k556163
answered Dec 4 '18 at 17:43
Paramanand SinghParamanand Singh
49.9k556163
answered Dec 4 '18 at 17:43
Paramanand SinghParamanand Singh
49.9k556163
49.9k556163
add a comment |
add a comment |
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