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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: $lim_{xtoinfty} frac{sqrt{x}}{sqrt{x+sqrt{x+sqrt{x}}}}.$

-4

$begingroup$

$$lim_{nto infty} sqrt[n]{frac{2^n+3^n}{3^n+5^n}}$$

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

share|cite|improve this question

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 | 

-4

$begingroup$

$$lim_{nto infty} sqrt[n]{frac{2^n+3^n}{3^n+5^n}}$$

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

share|cite|improve this question

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

share|cite|improve this question

edited Mar 14 at 21:17

abiessu

6,74221541

asked Mar 14 at 21:10

Norbert RychlińskiNorbert Rychliński

1

$endgroup$

share|cite|improve this question

edited Mar 14 at 21:17

abiessu

6,74221541

asked Mar 14 at 21:10

Norbert RychlińskiNorbert Rychliński

1

share|cite|improve this question

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

asked Mar 14 at 21:10

Norbert RychlińskiNorbert Rychliński

1

asked Mar 14 at 21:10

Norbert RychlińskiNorbert Rychliński

1

3 Answers
3

active

oldest

votes

4

$begingroup$

Hint: $displaystylesqrt[n]{frac{2^n+3^n}{3^n+5^n}}=frac35sqrt[n]{frac{left(frac23right)^n+1}{left(frac35right)^n+1}}$.

share|cite|improve this answer

answered Mar 14 at 21:16

José Carlos SantosJosé Carlos Santos

170k23132238

$endgroup$

add a comment | 

2

$begingroup$

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 $frac{3}{5}$.

share|cite|improve this answer

answered Mar 14 at 21:15

Jonathan HebertJonathan Hebert

3,50211031

$endgroup$

add a comment | 

0

$begingroup$

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:

$$expbigg(lim_{ntoinfty} frac{logbigg(frac{2^n+3^n}{3^n+5^n}bigg)}{n}bigg) = exp(log(3)-log(5)) $$

share|cite|improve this answer

edited Mar 15 at 23:33

answered Mar 14 at 21:26

ChocolateRainChocolateRain

1364

$endgroup$

add a comment | 

4

$begingroup$

Hint: $displaystylesqrt[n]{frac{2^n+3^n}{3^n+5^n}}=frac35sqrt[n]{frac{left(frac23right)^n+1}{left(frac35right)^n+1}}$.

share|cite|improve this answer

answered Mar 14 at 21:16

José Carlos SantosJosé Carlos Santos

170k23132238

$endgroup$

add a comment | 

4

$begingroup$

Hint: $displaystylesqrt[n]{frac{2^n+3^n}{3^n+5^n}}=frac35sqrt[n]{frac{left(frac23right)^n+1}{left(frac35right)^n+1}}$.

share|cite|improve this answer

answered Mar 14 at 21:16

José Carlos SantosJosé Carlos Santos

170k23132238

$endgroup$

add a comment | 

$begingroup$

Hint: $displaystylesqrt[n]{frac{2^n+3^n}{3^n+5^n}}=frac35sqrt[n]{frac{left(frac23right)^n+1}{left(frac35right)^n+1}}$.

share|cite|improve this answer

answered Mar 14 at 21:16

José Carlos SantosJosé Carlos Santos

170k23132238

$endgroup$

share|cite|improve this answer

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

2

$begingroup$

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 $frac{3}{5}$.

share|cite|improve this answer

answered Mar 14 at 21:15

Jonathan HebertJonathan Hebert

3,50211031

$endgroup$

add a comment | 

2

$begingroup$

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 $frac{3}{5}$.

share|cite|improve this answer

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 $frac{3}{5}$.

share|cite|improve this answer

answered Mar 14 at 21:15

Jonathan HebertJonathan Hebert

3,50211031

$endgroup$

share|cite|improve this answer

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

0

$begingroup$

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:

$$expbigg(lim_{ntoinfty} frac{logbigg(frac{2^n+3^n}{3^n+5^n}bigg)}{n}bigg) = exp(log(3)-log(5)) $$

share|cite|improve this answer

edited Mar 15 at 23:33

answered Mar 14 at 21:26

ChocolateRainChocolateRain

1364

$endgroup$

add a comment | 

0

$begingroup$

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:

$$expbigg(lim_{ntoinfty} frac{logbigg(frac{2^n+3^n}{3^n+5^n}bigg)}{n}bigg) = exp(log(3)-log(5)) $$

share|cite|improve this answer

edited Mar 15 at 23:33

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:

$$expbigg(lim_{ntoinfty} frac{logbigg(frac{2^n+3^n}{3^n+5^n}bigg)}{n}bigg) = exp(log(3)-log(5)) $$

share|cite|improve this answer

edited Mar 15 at 23:33

answered Mar 14 at 21:26

ChocolateRainChocolateRain

1364

$endgroup$

share|cite|improve this answer

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