Why simple limits are so difficult? [closed]Calculate difficult limitsCalculating limits of degree 'n'Why...
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Why simple limits are so difficult? [closed]
Calculate difficult limitsCalculating limits of degree 'n'Why $lim_{nto infty}{frac{1}{1cdot3}+frac{1}{2cdot4}+dots+frac{1}{ncdot(n+2)}}=frac{3}{4}$?Find limits with natural logsLimits and factorialsCalculus (Limits) Doubt: $theta - cfrac{theta^3}{3!} < sin theta < theta$ use to solve limit.Why are the limits different?$ text{Prove:} , {a_n}>0, displaystyle{lim_{n to infty}} { a_n }=L Rightarrow displaystyle{ lim_{n to infty } {a_n }^frac{1}{n}}=1 $Evaluating the $lim_{x to infty}(e^x + x)^{frac{1}{x}}$ as $x$ approaches $infty$Find: $lim_{xtoinfty} frac{sqrt{x}}{sqrt{x+sqrt{x+sqrt{x}}}}.$
$begingroup$
$$lim_{nto infty} sqrt[n]{frac{2^n+3^n}{3^n+5^n}}$$
After quick math, I got $cfrac {2}{5}$, but wolfram shows me $cfrac {3}{5}$ and I can't find any solution at all.
Any suggestions?
calculus limits
edited Mar 14 at 21:17
abiessu
6,74221541
asked Mar 14 at 21:10
Norbert RychlińskiNorbert Rychliński
1
$endgroup$
closed as off-topic by Saad, YiFan, Lord Shark the Unknown, Song, RRL Mar 15 at 6:08
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, YiFan, Lord Shark the Unknown, Song, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
$$lim_{nto infty} sqrt[n]{frac{2^n+3^n}{3^n+5^n}}$$
After quick math, I got $cfrac {2}{5}$, but wolfram shows me $cfrac {3}{5}$ and I can't find any solution at all.
Any suggestions?
calculus limits
edited Mar 14 at 21:17
abiessu
6,74221541
asked Mar 14 at 21:10
Norbert RychlińskiNorbert Rychliński
1
$endgroup$
closed as off-topic by Saad, YiFan, Lord Shark the Unknown, Song, RRL Mar 15 at 6:08
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, YiFan, Lord Shark the Unknown, Song, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
5
$begingroup$
can you write your quick math?
$endgroup$
– user376343
Mar 14 at 21:13
$begingroup$
Why do you call this limit "simple" ?
$endgroup$
– Peter
Mar 14 at 21:14
1
$begingroup$
Of the terms in the numerator, which one did you pick as increasing the fastest?
$endgroup$
– abiessu
Mar 14 at 21:14
$begingroup$
It looks like you made a very simple mistake wherein, although you got the leading-order behaviour right for the denominator, you accidentally focused on the wrong exponential function in the numerator. However, without seeing your method I can't be sure if this.
$endgroup$
– J.G.
Mar 14 at 22:22
add a comment |
$begingroup$
$$lim_{nto infty} sqrt[n]{frac{2^n+3^n}{3^n+5^n}}$$
After quick math, I got $cfrac {2}{5}$, but wolfram shows me $cfrac {3}{5}$ and I can't find any solution at all.
Any suggestions?
calculus limits
edited Mar 14 at 21:17
abiessu
6,74221541
asked Mar 14 at 21:10
Norbert RychlińskiNorbert Rychliński
1
$endgroup$
$$lim_{nto infty} sqrt[n]{frac{2^n+3^n}{3^n+5^n}}$$
After quick math, I got $cfrac {2}{5}$, but wolfram shows me $cfrac {3}{5}$ and I can't find any solution at all.
Any suggestions?
calculus limits
calculus limits
edited Mar 14 at 21:17
abiessu
6,74221541
asked Mar 14 at 21:10
Norbert RychlińskiNorbert Rychliński
1
edited Mar 14 at 21:17
abiessu
6,74221541
asked Mar 14 at 21:10
Norbert RychlińskiNorbert Rychliński
1
edited Mar 14 at 21:17
abiessu
6,74221541
edited Mar 14 at 21:17
abiessu
6,74221541
edited Mar 14 at 21:17
abiessu
6,74221541
6,74221541
asked Mar 14 at 21:10
Norbert RychlińskiNorbert Rychliński
1
asked Mar 14 at 21:10
Norbert RychlińskiNorbert Rychliński
1
asked Mar 14 at 21:10
Norbert RychlińskiNorbert Rychliński
1
1
closed as off-topic by Saad, YiFan, Lord Shark the Unknown, Song, RRL Mar 15 at 6:08
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, YiFan, Lord Shark the Unknown, Song, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, YiFan, Lord Shark the Unknown, Song, RRL Mar 15 at 6:08
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, YiFan, Lord Shark the Unknown, Song, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
5
$begingroup$
can you write your quick math?
$endgroup$
– user376343
Mar 14 at 21:13
$begingroup$
Why do you call this limit "simple" ?
$endgroup$
– Peter
Mar 14 at 21:14
1
$begingroup$
Of the terms in the numerator, which one did you pick as increasing the fastest?
$endgroup$
– abiessu
Mar 14 at 21:14
$begingroup$
It looks like you made a very simple mistake wherein, although you got the leading-order behaviour right for the denominator, you accidentally focused on the wrong exponential function in the numerator. However, without seeing your method I can't be sure if this.
$endgroup$
– J.G.
Mar 14 at 22:22
add a comment |
5
$begingroup$
can you write your quick math?
$endgroup$
– user376343
Mar 14 at 21:13
$begingroup$
Why do you call this limit "simple" ?
$endgroup$
– Peter
Mar 14 at 21:14
1
$begingroup$
Of the terms in the numerator, which one did you pick as increasing the fastest?
$endgroup$
– abiessu
Mar 14 at 21:14
$begingroup$
It looks like you made a very simple mistake wherein, although you got the leading-order behaviour right for the denominator, you accidentally focused on the wrong exponential function in the numerator. However, without seeing your method I can't be sure if this.
$endgroup$
– J.G.
Mar 14 at 22:22
5
5
$begingroup$
can you write your quick math?
$endgroup$
– user376343
Mar 14 at 21:13
$begingroup$
can you write your quick math?
$endgroup$
– user376343
Mar 14 at 21:13
$begingroup$
Why do you call this limit "simple" ?
$endgroup$
– Peter
Mar 14 at 21:14
$begingroup$
Why do you call this limit "simple" ?
$endgroup$
– Peter
Mar 14 at 21:14
1
1
$begingroup$
Of the terms in the numerator, which one did you pick as increasing the fastest?
$endgroup$
– abiessu
Mar 14 at 21:14
$begingroup$
Of the terms in the numerator, which one did you pick as increasing the fastest?
$endgroup$
– abiessu
Mar 14 at 21:14
$begingroup$
It looks like you made a very simple mistake wherein, although you got the leading-order behaviour right for the denominator, you accidentally focused on the wrong exponential function in the numerator. However, without seeing your method I can't be sure if this.
$endgroup$
– J.G.
Mar 14 at 22:22
$begingroup$
It looks like you made a very simple mistake wherein, although you got the leading-order behaviour right for the denominator, you accidentally focused on the wrong exponential function in the numerator. However, without seeing your method I can't be sure if this.
$endgroup$
– J.G.
Mar 14 at 22:22
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint: $displaystylesqrt[n]{frac{2^n+3^n}{3^n+5^n}}=frac35sqrt[n]{frac{left(frac23right)^n+1}{left(frac35right)^n+1}}$.
answered Mar 14 at 21:16
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
add a comment |
$begingroup$
Assuming you meant $n rightarrow infty$, as $n$ grows large, the numerator is dominated by $3^{n}$ and the denominator is dominated by $5^{n}$. You're taking the $n$th root of that ratio, which is just $frac{3}{5}$.
answered Mar 14 at 21:15
Jonathan HebertJonathan Hebert
3,50211031
$endgroup$
add a comment |
$begingroup$
EDIT:
José Carlos Santos his answer is much better and simpler than mine.
Original answer
Try starting with this
$$
lim_{ntoinfty} sqrt[n]{frac{2^n+3^n}{3^n+5^n}} = lim_{ntoinfty} exp(logbigg(sqrt[n]{frac{2^n+3^n}{3^n+5^n}}bigg))
$$
If you need to know what to do next, I added some further hints below. Try solving it yourself first. Hover with your mouse over them, if you want to know what the hints are. Good luck!
Hint 1:
$$ lim_{ntoinfty} exp(logbigg(sqrt[n]{frac{2^n+3^n}{3^n+5^n}}bigg))=lim_{ntoinfty} exp(frac{logbigg(frac{2^n+3^n}{3^n+5^n}bigg)}{n})$$
Hint 2:
$$ lim_{ntoinfty} exp(frac{logbigg(frac{2^n+3^n}{3^n+5^n}bigg)}{n})=expbigg(lim_{ntoinfty} frac{logbigg(frac{2^n+3^n}{3^n+5^n}bigg)}{n}bigg)$$
Hint 3:
l'Hôpital on the limit
Hint 4:
Divide both sides with 15^n
Hint 5:
$lim_{ntoinfty}frac{a}{b} = 0$ for $a<b$
Hint 6:
$$expbigg(lim_{ntoinfty} frac{logbigg(frac{2^n+3^n}{3^n+5^n}bigg)}{n}bigg) = exp(log(3)-log(5)) $$
answered Mar 14 at 21:26
ChocolateRainChocolateRain
1364
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: $displaystylesqrt[n]{frac{2^n+3^n}{3^n+5^n}}=frac35sqrt[n]{frac{left(frac23right)^n+1}{left(frac35right)^n+1}}$.
answered Mar 14 at 21:16
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
add a comment |
$begingroup$
Hint: $displaystylesqrt[n]{frac{2^n+3^n}{3^n+5^n}}=frac35sqrt[n]{frac{left(frac23right)^n+1}{left(frac35right)^n+1}}$.
answered Mar 14 at 21:16
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
add a comment |
$begingroup$
Hint: $displaystylesqrt[n]{frac{2^n+3^n}{3^n+5^n}}=frac35sqrt[n]{frac{left(frac23right)^n+1}{left(frac35right)^n+1}}$.
answered Mar 14 at 21:16
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
Hint: $displaystylesqrt[n]{frac{2^n+3^n}{3^n+5^n}}=frac35sqrt[n]{frac{left(frac23right)^n+1}{left(frac35right)^n+1}}$.
answered Mar 14 at 21:16
José Carlos SantosJosé Carlos Santos
170k23132238
answered Mar 14 at 21:16
José Carlos SantosJosé Carlos Santos
170k23132238
answered Mar 14 at 21:16
José Carlos SantosJosé Carlos Santos
170k23132238
answered Mar 14 at 21:16
José Carlos SantosJosé Carlos Santos
170k23132238
170k23132238
add a comment |
add a comment |
$begingroup$
Assuming you meant $n rightarrow infty$, as $n$ grows large, the numerator is dominated by $3^{n}$ and the denominator is dominated by $5^{n}$. You're taking the $n$th root of that ratio, which is just $frac{3}{5}$.
answered Mar 14 at 21:15
Jonathan HebertJonathan Hebert
3,50211031
$endgroup$
add a comment |
$begingroup$
Assuming you meant $n rightarrow infty$, as $n$ grows large, the numerator is dominated by $3^{n}$ and the denominator is dominated by $5^{n}$. You're taking the $n$th root of that ratio, which is just $frac{3}{5}$.
answered Mar 14 at 21:15
Jonathan HebertJonathan Hebert
3,50211031
$endgroup$
add a comment |
$begingroup$
Assuming you meant $n rightarrow infty$, as $n$ grows large, the numerator is dominated by $3^{n}$ and the denominator is dominated by $5^{n}$. You're taking the $n$th root of that ratio, which is just $frac{3}{5}$.
answered Mar 14 at 21:15
Jonathan HebertJonathan Hebert
3,50211031
$endgroup$
Assuming you meant $n rightarrow infty$, as $n$ grows large, the numerator is dominated by $3^{n}$ and the denominator is dominated by $5^{n}$. You're taking the $n$th root of that ratio, which is just $frac{3}{5}$.
answered Mar 14 at 21:15
Jonathan HebertJonathan Hebert
3,50211031
answered Mar 14 at 21:15
Jonathan HebertJonathan Hebert
3,50211031
answered Mar 14 at 21:15
Jonathan HebertJonathan Hebert
3,50211031
answered Mar 14 at 21:15
Jonathan HebertJonathan Hebert
3,50211031
3,50211031
add a comment |
add a comment |
$begingroup$
EDIT:
José Carlos Santos his answer is much better and simpler than mine.
Original answer
Try starting with this
$$
lim_{ntoinfty} sqrt[n]{frac{2^n+3^n}{3^n+5^n}} = lim_{ntoinfty} exp(logbigg(sqrt[n]{frac{2^n+3^n}{3^n+5^n}}bigg))
$$
If you need to know what to do next, I added some further hints below. Try solving it yourself first. Hover with your mouse over them, if you want to know what the hints are. Good luck!
Hint 1:
$$ lim_{ntoinfty} exp(logbigg(sqrt[n]{frac{2^n+3^n}{3^n+5^n}}bigg))=lim_{ntoinfty} exp(frac{logbigg(frac{2^n+3^n}{3^n+5^n}bigg)}{n})$$
Hint 2:
$$ lim_{ntoinfty} exp(frac{logbigg(frac{2^n+3^n}{3^n+5^n}bigg)}{n})=expbigg(lim_{ntoinfty} frac{logbigg(frac{2^n+3^n}{3^n+5^n}bigg)}{n}bigg)$$
Hint 3:
l'Hôpital on the limit
Hint 4:
Divide both sides with 15^n
Hint 5:
$lim_{ntoinfty}frac{a}{b} = 0$ for $a<b$
Hint 6:
$$expbigg(lim_{ntoinfty} frac{logbigg(frac{2^n+3^n}{3^n+5^n}bigg)}{n}bigg) = exp(log(3)-log(5)) $$
answered Mar 14 at 21:26
ChocolateRainChocolateRain
1364
$endgroup$
add a comment |
$begingroup$
EDIT:
José Carlos Santos his answer is much better and simpler than mine.
Original answer
Try starting with this
$$
lim_{ntoinfty} sqrt[n]{frac{2^n+3^n}{3^n+5^n}} = lim_{ntoinfty} exp(logbigg(sqrt[n]{frac{2^n+3^n}{3^n+5^n}}bigg))
$$
If you need to know what to do next, I added some further hints below. Try solving it yourself first. Hover with your mouse over them, if you want to know what the hints are. Good luck!
Hint 1:
$$ lim_{ntoinfty} exp(logbigg(sqrt[n]{frac{2^n+3^n}{3^n+5^n}}bigg))=lim_{ntoinfty} exp(frac{logbigg(frac{2^n+3^n}{3^n+5^n}bigg)}{n})$$
Hint 2:
$$ lim_{ntoinfty} exp(frac{logbigg(frac{2^n+3^n}{3^n+5^n}bigg)}{n})=expbigg(lim_{ntoinfty} frac{logbigg(frac{2^n+3^n}{3^n+5^n}bigg)}{n}bigg)$$
Hint 3:
l'Hôpital on the limit
Hint 4:
Divide both sides with 15^n
Hint 5:
$lim_{ntoinfty}frac{a}{b} = 0$ for $a<b$
Hint 6:
$$expbigg(lim_{ntoinfty} frac{logbigg(frac{2^n+3^n}{3^n+5^n}bigg)}{n}bigg) = exp(log(3)-log(5)) $$
answered Mar 14 at 21:26
ChocolateRainChocolateRain
1364
$endgroup$
add a comment |
$begingroup$
EDIT:
José Carlos Santos his answer is much better and simpler than mine.
Original answer
Try starting with this
$$
lim_{ntoinfty} sqrt[n]{frac{2^n+3^n}{3^n+5^n}} = lim_{ntoinfty} exp(logbigg(sqrt[n]{frac{2^n+3^n}{3^n+5^n}}bigg))
$$
If you need to know what to do next, I added some further hints below. Try solving it yourself first. Hover with your mouse over them, if you want to know what the hints are. Good luck!
Hint 1:
$$ lim_{ntoinfty} exp(logbigg(sqrt[n]{frac{2^n+3^n}{3^n+5^n}}bigg))=lim_{ntoinfty} exp(frac{logbigg(frac{2^n+3^n}{3^n+5^n}bigg)}{n})$$
Hint 2:
$$ lim_{ntoinfty} exp(frac{logbigg(frac{2^n+3^n}{3^n+5^n}bigg)}{n})=expbigg(lim_{ntoinfty} frac{logbigg(frac{2^n+3^n}{3^n+5^n}bigg)}{n}bigg)$$
Hint 3:
l'Hôpital on the limit
Hint 4:
Divide both sides with 15^n
Hint 5:
$lim_{ntoinfty}frac{a}{b} = 0$ for $a<b$
Hint 6:
$$expbigg(lim_{ntoinfty} frac{logbigg(frac{2^n+3^n}{3^n+5^n}bigg)}{n}bigg) = exp(log(3)-log(5)) $$
answered Mar 14 at 21:26
ChocolateRainChocolateRain
1364
$endgroup$
EDIT:
José Carlos Santos his answer is much better and simpler than mine.
Original answer
Try starting with this
$$
lim_{ntoinfty} sqrt[n]{frac{2^n+3^n}{3^n+5^n}} = lim_{ntoinfty} exp(logbigg(sqrt[n]{frac{2^n+3^n}{3^n+5^n}}bigg))
$$
If you need to know what to do next, I added some further hints below. Try solving it yourself first. Hover with your mouse over them, if you want to know what the hints are. Good luck!
Hint 1:
$$ lim_{ntoinfty} exp(logbigg(sqrt[n]{frac{2^n+3^n}{3^n+5^n}}bigg))=lim_{ntoinfty} exp(frac{logbigg(frac{2^n+3^n}{3^n+5^n}bigg)}{n})$$
Hint 2:
$$ lim_{ntoinfty} exp(frac{logbigg(frac{2^n+3^n}{3^n+5^n}bigg)}{n})=expbigg(lim_{ntoinfty} frac{logbigg(frac{2^n+3^n}{3^n+5^n}bigg)}{n}bigg)$$
Hint 3:
l'Hôpital on the limit
Hint 4:
Divide both sides with 15^n
Hint 5:
$lim_{ntoinfty}frac{a}{b} = 0$ for $a<b$
Hint 6:
$$expbigg(lim_{ntoinfty} frac{logbigg(frac{2^n+3^n}{3^n+5^n}bigg)}{n}bigg) = exp(log(3)-log(5)) $$
answered Mar 14 at 21:26
ChocolateRainChocolateRain
1364
edited Mar 15 at 23:33
answered Mar 14 at 21:26
ChocolateRainChocolateRain
1364
answered Mar 14 at 21:26
ChocolateRainChocolateRain
1364
answered Mar 14 at 21:26
ChocolateRainChocolateRain
1364
1364
add a comment |
add a comment |
5
$begingroup$
can you write your quick math?
$endgroup$
– user376343
Mar 14 at 21:13
$begingroup$
Why do you call this limit "simple" ?
$endgroup$
– Peter
Mar 14 at 21:14
1
$begingroup$
Of the terms in the numerator, which one did you pick as increasing the fastest?
$endgroup$
– abiessu
Mar 14 at 21:14
$begingroup$
It looks like you made a very simple mistake wherein, although you got the leading-order behaviour right for the denominator, you accidentally focused on the wrong exponential function in the numerator. However, without seeing your method I can't be sure if this.
$endgroup$
– J.G.
Mar 14 at 22:22