Mixing
How to prove that $lim_{ntoinfty} sqrt[n]{a^n+b^n}=operatorname{max}(a,b)$? [duplicate]
$begingroup$
This question already has an answer here:
Convergence of $sqrt[n]{x^n+y^n}$ (for $x, y > 0$)
2 answers
How to compute the limit $lim_{n to infty} sqrt[n]{a_1^n + cdots + a_m^n}$?
2 answers
How do I show that $$lim_{ntoinfty} sqrt[n]{a^n+b^n}=operatorname{max}(a,b)$$
with $a,bge0$.
I tried to do this by dividing it in two cases, when $a=b$ and $agt b$.
In the case $agt b$ I factored $a^n$ like this: $$lim_{ntoinfty} sqrt[n]{a^n left(1+{b^nover a^n} right)} = lim_{ntoinfty} asqrt[n]{1+{b^nover a^n}} = alim_{ntoinfty} sqrt[n]{1+{b^nover a^n}}$$
Then I expressed it in exponential way. $$alim_{ntoinfty} left({1+{b^nover a^n}}right)^{1/n}$$
Now I need to prove that $lim_{ntoinfty} left({1+{b^nover a^n}}right)^{1/n}=1$ so the whole limit is equal to a. The problem is that I don't know how to take the limit of $lim_{ntoinfty} left({1+{b^nover a^n}}right)^{1/n}$ I tried to use the natural logarithm, but it ended up like this:
$$lim_{ntoinfty} logleft({1+{b^nover a^n}}right)^{1/n}=lim_{ntoinfty} {1over n} logleft({1+{b^nover a^n}}right)=lim_{ntoinfty} {1over n} lim_{ntoinfty}logleft({1+{b^nover a^n}}right)=0$$
What did I do wrong and how can I do it right?
calculus sequences-and-series limits
edited Jan 23 at 5:42
user549397
1,5061418
asked Jan 23 at 3:38
davidllerenavdavidllerenav
1888
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marked as duplicate by StubbornAtom, Martin R, Cesareo, egreg calculus
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Jan 23 at 10:12
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Convergence of $sqrt[n]{x^n+y^n}$ (for $x, y > 0$)
2 answers
How to compute the limit $lim_{n to infty} sqrt[n]{a_1^n + cdots + a_m^n}$?
2 answers
How do I show that $$lim_{ntoinfty} sqrt[n]{a^n+b^n}=operatorname{max}(a,b)$$
with $a,bge0$.
I tried to do this by dividing it in two cases, when $a=b$ and $agt b$.
In the case $agt b$ I factored $a^n$ like this: $$lim_{ntoinfty} sqrt[n]{a^n left(1+{b^nover a^n} right)} = lim_{ntoinfty} asqrt[n]{1+{b^nover a^n}} = alim_{ntoinfty} sqrt[n]{1+{b^nover a^n}}$$
Then I expressed it in exponential way. $$alim_{ntoinfty} left({1+{b^nover a^n}}right)^{1/n}$$
Now I need to prove that $lim_{ntoinfty} left({1+{b^nover a^n}}right)^{1/n}=1$ so the whole limit is equal to a. The problem is that I don't know how to take the limit of $lim_{ntoinfty} left({1+{b^nover a^n}}right)^{1/n}$ I tried to use the natural logarithm, but it ended up like this:
$$lim_{ntoinfty} logleft({1+{b^nover a^n}}right)^{1/n}=lim_{ntoinfty} {1over n} logleft({1+{b^nover a^n}}right)=lim_{ntoinfty} {1over n} lim_{ntoinfty}logleft({1+{b^nover a^n}}right)=0$$
What did I do wrong and how can I do it right?
calculus sequences-and-series limits
edited Jan 23 at 5:42
user549397
1,5061418
asked Jan 23 at 3:38
davidllerenavdavidllerenav
1888
$endgroup$
marked as duplicate by StubbornAtom, Martin R, Cesareo, egreg calculus
Users with the calculus badge can single-handedly close calculus questions as duplicates and reopen them as needed.
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Jan 23 at 10:12
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
$begingroup$
You have shown that the logarithm of the limit is $0$, so the limit is $1$, which I believe is what you were trying to prove.
$endgroup$
– greelious
Jan 23 at 3:50
$begingroup$
I don't understand how the fact that logarithm of the limit is $0$ implies that the limit is $1$, can you explain please?
$endgroup$
– davidllerenav
Jan 23 at 3:55
1
$begingroup$
$x=1$ is the solution to the equation $log(x)=0$. Just replace $x$ with your limit.
$endgroup$
– greelious
Jan 23 at 3:57
$begingroup$
Oh, so I was correct all this time, thanks!
$endgroup$
– davidllerenav
Jan 23 at 4:05
add a comment |
$begingroup$
This question already has an answer here:
Convergence of $sqrt[n]{x^n+y^n}$ (for $x, y > 0$)
2 answers
How to compute the limit $lim_{n to infty} sqrt[n]{a_1^n + cdots + a_m^n}$?
2 answers
How do I show that $$lim_{ntoinfty} sqrt[n]{a^n+b^n}=operatorname{max}(a,b)$$
with $a,bge0$.
I tried to do this by dividing it in two cases, when $a=b$ and $agt b$.
In the case $agt b$ I factored $a^n$ like this: $$lim_{ntoinfty} sqrt[n]{a^n left(1+{b^nover a^n} right)} = lim_{ntoinfty} asqrt[n]{1+{b^nover a^n}} = alim_{ntoinfty} sqrt[n]{1+{b^nover a^n}}$$
Then I expressed it in exponential way. $$alim_{ntoinfty} left({1+{b^nover a^n}}right)^{1/n}$$
Now I need to prove that $lim_{ntoinfty} left({1+{b^nover a^n}}right)^{1/n}=1$ so the whole limit is equal to a. The problem is that I don't know how to take the limit of $lim_{ntoinfty} left({1+{b^nover a^n}}right)^{1/n}$ I tried to use the natural logarithm, but it ended up like this:
$$lim_{ntoinfty} logleft({1+{b^nover a^n}}right)^{1/n}=lim_{ntoinfty} {1over n} logleft({1+{b^nover a^n}}right)=lim_{ntoinfty} {1over n} lim_{ntoinfty}logleft({1+{b^nover a^n}}right)=0$$
What did I do wrong and how can I do it right?
calculus sequences-and-series limits
edited Jan 23 at 5:42
user549397
1,5061418
asked Jan 23 at 3:38
davidllerenavdavidllerenav
1888
$endgroup$
This question already has an answer here:
Convergence of $sqrt[n]{x^n+y^n}$ (for $x, y > 0$)
2 answers
How to compute the limit $lim_{n to infty} sqrt[n]{a_1^n + cdots + a_m^n}$?
2 answers
How do I show that $$lim_{ntoinfty} sqrt[n]{a^n+b^n}=operatorname{max}(a,b)$$
with $a,bge0$.
I tried to do this by dividing it in two cases, when $a=b$ and $agt b$.
In the case $agt b$ I factored $a^n$ like this: $$lim_{ntoinfty} sqrt[n]{a^n left(1+{b^nover a^n} right)} = lim_{ntoinfty} asqrt[n]{1+{b^nover a^n}} = alim_{ntoinfty} sqrt[n]{1+{b^nover a^n}}$$
Then I expressed it in exponential way. $$alim_{ntoinfty} left({1+{b^nover a^n}}right)^{1/n}$$
Now I need to prove that $lim_{ntoinfty} left({1+{b^nover a^n}}right)^{1/n}=1$ so the whole limit is equal to a. The problem is that I don't know how to take the limit of $lim_{ntoinfty} left({1+{b^nover a^n}}right)^{1/n}$ I tried to use the natural logarithm, but it ended up like this:
$$lim_{ntoinfty} logleft({1+{b^nover a^n}}right)^{1/n}=lim_{ntoinfty} {1over n} logleft({1+{b^nover a^n}}right)=lim_{ntoinfty} {1over n} lim_{ntoinfty}logleft({1+{b^nover a^n}}right)=0$$
What did I do wrong and how can I do it right?
This question already has an answer here:
Convergence of $sqrt[n]{x^n+y^n}$ (for $x, y > 0$)
2 answers
How to compute the limit $lim_{n to infty} sqrt[n]{a_1^n + cdots + a_m^n}$?
2 answers
calculus sequences-and-series limits
calculus sequences-and-series limits
edited Jan 23 at 5:42
user549397
1,5061418
asked Jan 23 at 3:38
davidllerenavdavidllerenav
1888
edited Jan 23 at 5:42
user549397
1,5061418
asked Jan 23 at 3:38
davidllerenavdavidllerenav
1888
edited Jan 23 at 5:42
user549397
1,5061418
edited Jan 23 at 5:42
user549397
1,5061418
edited Jan 23 at 5:42
user549397
1,5061418
1,5061418
asked Jan 23 at 3:38
davidllerenavdavidllerenav
1888
asked Jan 23 at 3:38
davidllerenavdavidllerenav
1888
asked Jan 23 at 3:38
davidllerenavdavidllerenav
1888
1888
marked as duplicate by StubbornAtom, Martin R, Cesareo, egreg calculus
Users with the calculus badge can single-handedly close calculus questions as duplicates and reopen them as needed.
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Jan 23 at 10:12
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by StubbornAtom, Martin R, Cesareo, egreg calculus
Users with the calculus badge can single-handedly close calculus questions as duplicates and reopen them as needed.
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Jan 23 at 10:12
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
$begingroup$
You have shown that the logarithm of the limit is $0$, so the limit is $1$, which I believe is what you were trying to prove.
$endgroup$
– greelious
Jan 23 at 3:50
$begingroup$
I don't understand how the fact that logarithm of the limit is $0$ implies that the limit is $1$, can you explain please?
$endgroup$
– davidllerenav
Jan 23 at 3:55
1
$begingroup$
$x=1$ is the solution to the equation $log(x)=0$. Just replace $x$ with your limit.
$endgroup$
– greelious
Jan 23 at 3:57
$begingroup$
Oh, so I was correct all this time, thanks!
$endgroup$
– davidllerenav
Jan 23 at 4:05
add a comment |
3
$begingroup$
You have shown that the logarithm of the limit is $0$, so the limit is $1$, which I believe is what you were trying to prove.
$endgroup$
– greelious
Jan 23 at 3:50
$begingroup$
I don't understand how the fact that logarithm of the limit is $0$ implies that the limit is $1$, can you explain please?
$endgroup$
– davidllerenav
Jan 23 at 3:55
1
$begingroup$
$x=1$ is the solution to the equation $log(x)=0$. Just replace $x$ with your limit.
$endgroup$
– greelious
Jan 23 at 3:57
$begingroup$
Oh, so I was correct all this time, thanks!
$endgroup$
– davidllerenav
Jan 23 at 4:05
3
3
$begingroup$
You have shown that the logarithm of the limit is $0$, so the limit is $1$, which I believe is what you were trying to prove.
$endgroup$
– greelious
Jan 23 at 3:50
$begingroup$
You have shown that the logarithm of the limit is $0$, so the limit is $1$, which I believe is what you were trying to prove.
$endgroup$
– greelious
Jan 23 at 3:50
$begingroup$
I don't understand how the fact that logarithm of the limit is $0$ implies that the limit is $1$, can you explain please?
$endgroup$
– davidllerenav
Jan 23 at 3:55
$begingroup$
I don't understand how the fact that logarithm of the limit is $0$ implies that the limit is $1$, can you explain please?
$endgroup$
– davidllerenav
Jan 23 at 3:55
1
1
$begingroup$
$x=1$ is the solution to the equation $log(x)=0$. Just replace $x$ with your limit.
$endgroup$
– greelious
Jan 23 at 3:57
$begingroup$
$x=1$ is the solution to the equation $log(x)=0$. Just replace $x$ with your limit.
$endgroup$
– greelious
Jan 23 at 3:57
$begingroup$
Oh, so I was correct all this time, thanks!
$endgroup$
– davidllerenav
Jan 23 at 4:05
$begingroup$
Oh, so I was correct all this time, thanks!
$endgroup$
– davidllerenav
Jan 23 at 4:05
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Assume that $a geq b$. Else, relabel the numbers $a leftrightarrow b$. This being understood, we have $a = max(a,b)$. Then $$a =sqrt[n]{a^n} leq sqrt{a^n+b^n} leq sqrt{a^n + a^n} = sqrt[n]{2} a. $$Now apply $lim_{n to +infty}$ on everything, noting that $sqrt[n]{2} to 1$. It follows from the squeeze theorem that $$lim_{nto +infty}sqrt[n]{a^n+b^n} = a,$$ as wanted.
answered Jan 23 at 5:13
Ivo TerekIvo Terek
46.4k954142
$endgroup$
add a comment |
$begingroup$
For your specific question you may proceed as follows using
$lim_{xto 0}left(1+x right)^{frac{1}{x}} = e$ and- Let $a>b > 0$ (For $b= 0$ there is nothing to show.) $Rightarrow 0< q:= frac{b}{a} < 1 Rightarrow q^n stackrel{n to infty}{longrightarrow} 0$
begin{eqnarray*} left({1+{b^nover a^n}}right)^{1/n}
& = & (1+q^n)^{frac{1}{n}}\
& = & left( underbrace{(1+q^n)^{frac{1}{q^n}}}_{stackrel{n to infty}{longrightarrow}e} right)^{underbrace{frac{q^n}{n}}_{stackrel{n to infty}{longrightarrow}0}}\
& stackrel{n to infty}{longrightarrow} & e^0 = 1
end{eqnarray*}
answered Jan 23 at 7:22
trancelocationtrancelocation
12.6k1826
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Assume that $a geq b$. Else, relabel the numbers $a leftrightarrow b$. This being understood, we have $a = max(a,b)$. Then $$a =sqrt[n]{a^n} leq sqrt{a^n+b^n} leq sqrt{a^n + a^n} = sqrt[n]{2} a. $$Now apply $lim_{n to +infty}$ on everything, noting that $sqrt[n]{2} to 1$. It follows from the squeeze theorem that $$lim_{nto +infty}sqrt[n]{a^n+b^n} = a,$$ as wanted.
answered Jan 23 at 5:13
Ivo TerekIvo Terek
46.4k954142
$endgroup$
add a comment |
$begingroup$
Assume that $a geq b$. Else, relabel the numbers $a leftrightarrow b$. This being understood, we have $a = max(a,b)$. Then $$a =sqrt[n]{a^n} leq sqrt{a^n+b^n} leq sqrt{a^n + a^n} = sqrt[n]{2} a. $$Now apply $lim_{n to +infty}$ on everything, noting that $sqrt[n]{2} to 1$. It follows from the squeeze theorem that $$lim_{nto +infty}sqrt[n]{a^n+b^n} = a,$$ as wanted.
answered Jan 23 at 5:13
Ivo TerekIvo Terek
46.4k954142
$endgroup$
add a comment |
$begingroup$
Assume that $a geq b$. Else, relabel the numbers $a leftrightarrow b$. This being understood, we have $a = max(a,b)$. Then $$a =sqrt[n]{a^n} leq sqrt{a^n+b^n} leq sqrt{a^n + a^n} = sqrt[n]{2} a. $$Now apply $lim_{n to +infty}$ on everything, noting that $sqrt[n]{2} to 1$. It follows from the squeeze theorem that $$lim_{nto +infty}sqrt[n]{a^n+b^n} = a,$$ as wanted.
answered Jan 23 at 5:13
Ivo TerekIvo Terek
46.4k954142
$endgroup$
Assume that $a geq b$. Else, relabel the numbers $a leftrightarrow b$. This being understood, we have $a = max(a,b)$. Then $$a =sqrt[n]{a^n} leq sqrt{a^n+b^n} leq sqrt{a^n + a^n} = sqrt[n]{2} a. $$Now apply $lim_{n to +infty}$ on everything, noting that $sqrt[n]{2} to 1$. It follows from the squeeze theorem that $$lim_{nto +infty}sqrt[n]{a^n+b^n} = a,$$ as wanted.
answered Jan 23 at 5:13
Ivo TerekIvo Terek
46.4k954142
answered Jan 23 at 5:13
Ivo TerekIvo Terek
46.4k954142
answered Jan 23 at 5:13
Ivo TerekIvo Terek
46.4k954142
answered Jan 23 at 5:13
Ivo TerekIvo Terek
46.4k954142
46.4k954142
add a comment |
add a comment |
$begingroup$
For your specific question you may proceed as follows using
$lim_{xto 0}left(1+x right)^{frac{1}{x}} = e$ and- Let $a>b > 0$ (For $b= 0$ there is nothing to show.) $Rightarrow 0< q:= frac{b}{a} < 1 Rightarrow q^n stackrel{n to infty}{longrightarrow} 0$
begin{eqnarray*} left({1+{b^nover a^n}}right)^{1/n}
& = & (1+q^n)^{frac{1}{n}}\
& = & left( underbrace{(1+q^n)^{frac{1}{q^n}}}_{stackrel{n to infty}{longrightarrow}e} right)^{underbrace{frac{q^n}{n}}_{stackrel{n to infty}{longrightarrow}0}}\
& stackrel{n to infty}{longrightarrow} & e^0 = 1
end{eqnarray*}
answered Jan 23 at 7:22
trancelocationtrancelocation
12.6k1826
$endgroup$
add a comment |
$begingroup$
For your specific question you may proceed as follows using
$lim_{xto 0}left(1+x right)^{frac{1}{x}} = e$ and- Let $a>b > 0$ (For $b= 0$ there is nothing to show.) $Rightarrow 0< q:= frac{b}{a} < 1 Rightarrow q^n stackrel{n to infty}{longrightarrow} 0$
begin{eqnarray*} left({1+{b^nover a^n}}right)^{1/n}
& = & (1+q^n)^{frac{1}{n}}\
& = & left( underbrace{(1+q^n)^{frac{1}{q^n}}}_{stackrel{n to infty}{longrightarrow}e} right)^{underbrace{frac{q^n}{n}}_{stackrel{n to infty}{longrightarrow}0}}\
& stackrel{n to infty}{longrightarrow} & e^0 = 1
end{eqnarray*}
answered Jan 23 at 7:22
trancelocationtrancelocation
12.6k1826
$endgroup$
add a comment |
$begingroup$
For your specific question you may proceed as follows using
$lim_{xto 0}left(1+x right)^{frac{1}{x}} = e$ and- Let $a>b > 0$ (For $b= 0$ there is nothing to show.) $Rightarrow 0< q:= frac{b}{a} < 1 Rightarrow q^n stackrel{n to infty}{longrightarrow} 0$
begin{eqnarray*} left({1+{b^nover a^n}}right)^{1/n}
& = & (1+q^n)^{frac{1}{n}}\
& = & left( underbrace{(1+q^n)^{frac{1}{q^n}}}_{stackrel{n to infty}{longrightarrow}e} right)^{underbrace{frac{q^n}{n}}_{stackrel{n to infty}{longrightarrow}0}}\
& stackrel{n to infty}{longrightarrow} & e^0 = 1
end{eqnarray*}
answered Jan 23 at 7:22
trancelocationtrancelocation
12.6k1826
$endgroup$
For your specific question you may proceed as follows using
$lim_{xto 0}left(1+x right)^{frac{1}{x}} = e$ and- Let $a>b > 0$ (For $b= 0$ there is nothing to show.) $Rightarrow 0< q:= frac{b}{a} < 1 Rightarrow q^n stackrel{n to infty}{longrightarrow} 0$
begin{eqnarray*} left({1+{b^nover a^n}}right)^{1/n}
& = & (1+q^n)^{frac{1}{n}}\
& = & left( underbrace{(1+q^n)^{frac{1}{q^n}}}_{stackrel{n to infty}{longrightarrow}e} right)^{underbrace{frac{q^n}{n}}_{stackrel{n to infty}{longrightarrow}0}}\
& stackrel{n to infty}{longrightarrow} & e^0 = 1
end{eqnarray*}
answered Jan 23 at 7:22
trancelocationtrancelocation
12.6k1826
edited Jan 23 at 9:52
answered Jan 23 at 7:22
trancelocationtrancelocation
12.6k1826
answered Jan 23 at 7:22
trancelocationtrancelocation
12.6k1826
answered Jan 23 at 7:22
trancelocationtrancelocation
12.6k1826
12.6k1826
add a comment |
add a comment |
3
$begingroup$
You have shown that the logarithm of the limit is $0$, so the limit is $1$, which I believe is what you were trying to prove.
$endgroup$
– greelious
Jan 23 at 3:50
$begingroup$
I don't understand how the fact that logarithm of the limit is $0$ implies that the limit is $1$, can you explain please?
$endgroup$
– davidllerenav
Jan 23 at 3:55
1
$begingroup$
$x=1$ is the solution to the equation $log(x)=0$. Just replace $x$ with your limit.
$endgroup$
– greelious
Jan 23 at 3:57
$begingroup$
Oh, so I was correct all this time, thanks!
$endgroup$
– davidllerenav
Jan 23 at 4:05