I need to calculate a limit using Euler number [closed]
$begingroup$
How would one evaluate this limit without using L'Hôpital's rule? What I know for sure is that this limit equals to zero, but I don’t know how to solve it.
$$
lim_{xrightarrow infty}left(frac{2x-5}{2x-2}right)^{4x^{2}} = 0
$$
Edit: This is Wolframalpha solution: https://www.wolframcloud.com/objects/f6d89370-5d6b-44de-9f4e-0377dea18aad , but it uses the L'Hôpital's rule and I think it is a quite complicated solution.
limits limits-without-lhopital eulers-constant
asked Jan 24 at 21:24
makulo123makulo123
33
$endgroup$
closed as off-topic by Rahul, caverac, Stefan4024, Abcd, Shailesh Jan 25 at 0:40
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." – Rahul, caverac, Stefan4024, Abcd, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
How would one evaluate this limit without using L'Hôpital's rule? What I know for sure is that this limit equals to zero, but I don’t know how to solve it.
$$
lim_{xrightarrow infty}left(frac{2x-5}{2x-2}right)^{4x^{2}} = 0
$$
Edit: This is Wolframalpha solution: https://www.wolframcloud.com/objects/f6d89370-5d6b-44de-9f4e-0377dea18aad , but it uses the L'Hôpital's rule and I think it is a quite complicated solution.
limits limits-without-lhopital eulers-constant
asked Jan 24 at 21:24
makulo123makulo123
33
$endgroup$
closed as off-topic by Rahul, caverac, Stefan4024, Abcd, Shailesh Jan 25 at 0:40
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." – Rahul, caverac, Stefan4024, Abcd, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
Telling people you need something ASAP will not earn you any love around here.
$endgroup$
– Doug M
Jan 24 at 21:28
$begingroup$
"I was not present at the time they were doing that" Then read the relevant chapter of the textbook, or ask your fellow students for some notes.
$endgroup$
– Rahul
Jan 24 at 21:30
$begingroup$
@Rahul That is exactly what I have done, but the problem is, they did not encounter such problem during class and the teacher gave us these problems randomly. He did not solve them himself, so I am not sure if it even can be solved without the L'Hospital's rule.
$endgroup$
– makulo123
Jan 24 at 21:32
$begingroup$
@makulo123 hey There! You know your question has been put on hold. Next time if you asked a question, just write your own solution (doesn't matter if it's incomplete, wrong, or unrelated), it gives us the insight to look the question and understand your level and give you the best to your level ;) :) :D
$endgroup$
– Abhas Kumar Sinha
Jan 25 at 15:32
add a comment |
$begingroup$
How would one evaluate this limit without using L'Hôpital's rule? What I know for sure is that this limit equals to zero, but I don’t know how to solve it.
$$
lim_{xrightarrow infty}left(frac{2x-5}{2x-2}right)^{4x^{2}} = 0
$$
Edit: This is Wolframalpha solution: https://www.wolframcloud.com/objects/f6d89370-5d6b-44de-9f4e-0377dea18aad , but it uses the L'Hôpital's rule and I think it is a quite complicated solution.
limits limits-without-lhopital eulers-constant
asked Jan 24 at 21:24
makulo123makulo123
33
$endgroup$
How would one evaluate this limit without using L'Hôpital's rule? What I know for sure is that this limit equals to zero, but I don’t know how to solve it.
$$
lim_{xrightarrow infty}left(frac{2x-5}{2x-2}right)^{4x^{2}} = 0
$$
Edit: This is Wolframalpha solution: https://www.wolframcloud.com/objects/f6d89370-5d6b-44de-9f4e-0377dea18aad , but it uses the L'Hôpital's rule and I think it is a quite complicated solution.
limits limits-without-lhopital eulers-constant
limits limits-without-lhopital eulers-constant
asked Jan 24 at 21:24
makulo123makulo123
33
asked Jan 24 at 21:24
makulo123makulo123
33
edited Jan 24 at 22:02
makulo123
asked Jan 24 at 21:24
makulo123makulo123
33
asked Jan 24 at 21:24
makulo123makulo123
33
asked Jan 24 at 21:24
makulo123makulo123
33
33
closed as off-topic by Rahul, caverac, Stefan4024, Abcd, Shailesh Jan 25 at 0:40
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." – Rahul, caverac, Stefan4024, Abcd, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Rahul, caverac, Stefan4024, Abcd, Shailesh Jan 25 at 0:40
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." – Rahul, caverac, Stefan4024, Abcd, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
Telling people you need something ASAP will not earn you any love around here.
$endgroup$
– Doug M
Jan 24 at 21:28
$begingroup$
"I was not present at the time they were doing that" Then read the relevant chapter of the textbook, or ask your fellow students for some notes.
$endgroup$
– Rahul
Jan 24 at 21:30
$begingroup$
@Rahul That is exactly what I have done, but the problem is, they did not encounter such problem during class and the teacher gave us these problems randomly. He did not solve them himself, so I am not sure if it even can be solved without the L'Hospital's rule.
$endgroup$
– makulo123
Jan 24 at 21:32
$begingroup$
@makulo123 hey There! You know your question has been put on hold. Next time if you asked a question, just write your own solution (doesn't matter if it's incomplete, wrong, or unrelated), it gives us the insight to look the question and understand your level and give you the best to your level ;) :) :D
$endgroup$
– Abhas Kumar Sinha
Jan 25 at 15:32
add a comment |
2
$begingroup$
Telling people you need something ASAP will not earn you any love around here.
$endgroup$
– Doug M
Jan 24 at 21:28
$begingroup$
"I was not present at the time they were doing that" Then read the relevant chapter of the textbook, or ask your fellow students for some notes.
$endgroup$
– Rahul
Jan 24 at 21:30
$begingroup$
@Rahul That is exactly what I have done, but the problem is, they did not encounter such problem during class and the teacher gave us these problems randomly. He did not solve them himself, so I am not sure if it even can be solved without the L'Hospital's rule.
$endgroup$
– makulo123
Jan 24 at 21:32
$begingroup$
@makulo123 hey There! You know your question has been put on hold. Next time if you asked a question, just write your own solution (doesn't matter if it's incomplete, wrong, or unrelated), it gives us the insight to look the question and understand your level and give you the best to your level ;) :) :D
$endgroup$
– Abhas Kumar Sinha
Jan 25 at 15:32
2
2
$begingroup$
Telling people you need something ASAP will not earn you any love around here.
$endgroup$
– Doug M
Jan 24 at 21:28
$begingroup$
Telling people you need something ASAP will not earn you any love around here.
$endgroup$
– Doug M
Jan 24 at 21:28
$begingroup$
"I was not present at the time they were doing that" Then read the relevant chapter of the textbook, or ask your fellow students for some notes.
$endgroup$
– Rahul
Jan 24 at 21:30
$begingroup$
"I was not present at the time they were doing that" Then read the relevant chapter of the textbook, or ask your fellow students for some notes.
$endgroup$
– Rahul
Jan 24 at 21:30
$begingroup$
@Rahul That is exactly what I have done, but the problem is, they did not encounter such problem during class and the teacher gave us these problems randomly. He did not solve them himself, so I am not sure if it even can be solved without the L'Hospital's rule.
$endgroup$
– makulo123
Jan 24 at 21:32
$begingroup$
@Rahul That is exactly what I have done, but the problem is, they did not encounter such problem during class and the teacher gave us these problems randomly. He did not solve them himself, so I am not sure if it even can be solved without the L'Hospital's rule.
$endgroup$
– makulo123
Jan 24 at 21:32
$begingroup$
@makulo123 hey There! You know your question has been put on hold. Next time if you asked a question, just write your own solution (doesn't matter if it's incomplete, wrong, or unrelated), it gives us the insight to look the question and understand your level and give you the best to your level ;) :) :D
$endgroup$
– Abhas Kumar Sinha
Jan 25 at 15:32
$begingroup$
@makulo123 hey There! You know your question has been put on hold. Next time if you asked a question, just write your own solution (doesn't matter if it's incomplete, wrong, or unrelated), it gives us the insight to look the question and understand your level and give you the best to your level ;) :) :D
$endgroup$
– Abhas Kumar Sinha
Jan 25 at 15:32
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$lim_limits {nto infty} (1+frac {x}{n})^n = e^x$
$lim_limits {xto infty} (frac {2x-5}{2x-2})^{4x^2}\
lim_limits {xto infty} (1 +frac{-3}{2x-2})^{4x^2}$
Now $lim_limits {xto infty} (1 +frac{-3}{2x-2})^{2x-2} = e^{-3}$
So, what do you have to do to get a factor like this out of the line above?
$lim_limits {xto infty} ((1 +frac{-3}{2x-2})^{2x-2})^{2x+2}(1 +frac{-3}{2x-2})^4\
lim_limits {xto infty} (e^{-3})^{2x+2}(1 +frac{-3}{2x-2})^4$
$0<e^{-3} < 1$
What I have done here is not rigorous.
But you can use the work above to say:
$0le lim_limits {xto infty}(frac {2x-5}{2x-2})^{4x^2} le lim_limits {xto infty} e^{-6x}$
$lim_limits {xto infty}(frac {2x-5}{2x-2})^{4x^2} = 0$ by the squeeze theorem.
answered Jan 24 at 21:44
Doug MDoug M
45.3k31954
$endgroup$
$begingroup$
You can't take the limit of part of an expression, the take the limit of what remains (with exponents, &c.).
$endgroup$
– Bernard
Jan 24 at 21:48
$begingroup$
True, I have played a bit fast-and-loose. But, It points us in a direction that is rigorous.
$endgroup$
– Doug M
Jan 24 at 22:12
$begingroup$
@doug-m Ok, thanks. But I don't quite understand this line: $0<e^{-3} < 1$ How did we get this statement from the previous line?
$endgroup$
– makulo123
Jan 24 at 22:20
$begingroup$
$0<e^{-3}<1$ because $e>1$
$endgroup$
– Doug M
Jan 24 at 23:29
$begingroup$
Yes, thanks, I used a bit different aproach based on your solution: $lim_limits {xto infty}(e^{-3})^{2x+2}=lim_limits {xto infty}(e^{-6x-6})=lim_limits {xto infty}(e^{-6x})(e^{-6})=0$
$endgroup$
– makulo123
Jan 25 at 0:00
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$lim_limits {nto infty} (1+frac {x}{n})^n = e^x$
$lim_limits {xto infty} (frac {2x-5}{2x-2})^{4x^2}\
lim_limits {xto infty} (1 +frac{-3}{2x-2})^{4x^2}$
Now $lim_limits {xto infty} (1 +frac{-3}{2x-2})^{2x-2} = e^{-3}$
So, what do you have to do to get a factor like this out of the line above?
$lim_limits {xto infty} ((1 +frac{-3}{2x-2})^{2x-2})^{2x+2}(1 +frac{-3}{2x-2})^4\
lim_limits {xto infty} (e^{-3})^{2x+2}(1 +frac{-3}{2x-2})^4$
$0<e^{-3} < 1$
What I have done here is not rigorous.
But you can use the work above to say:
$0le lim_limits {xto infty}(frac {2x-5}{2x-2})^{4x^2} le lim_limits {xto infty} e^{-6x}$
$lim_limits {xto infty}(frac {2x-5}{2x-2})^{4x^2} = 0$ by the squeeze theorem.
answered Jan 24 at 21:44
Doug MDoug M
45.3k31954
$endgroup$
$begingroup$
You can't take the limit of part of an expression, the take the limit of what remains (with exponents, &c.).
$endgroup$
– Bernard
Jan 24 at 21:48
$begingroup$
True, I have played a bit fast-and-loose. But, It points us in a direction that is rigorous.
$endgroup$
– Doug M
Jan 24 at 22:12
$begingroup$
@doug-m Ok, thanks. But I don't quite understand this line: $0<e^{-3} < 1$ How did we get this statement from the previous line?
$endgroup$
– makulo123
Jan 24 at 22:20
$begingroup$
$0<e^{-3}<1$ because $e>1$
$endgroup$
– Doug M
Jan 24 at 23:29
$begingroup$
Yes, thanks, I used a bit different aproach based on your solution: $lim_limits {xto infty}(e^{-3})^{2x+2}=lim_limits {xto infty}(e^{-6x-6})=lim_limits {xto infty}(e^{-6x})(e^{-6})=0$
$endgroup$
– makulo123
Jan 25 at 0:00
add a comment |
$begingroup$
$lim_limits {nto infty} (1+frac {x}{n})^n = e^x$
$lim_limits {xto infty} (frac {2x-5}{2x-2})^{4x^2}\
lim_limits {xto infty} (1 +frac{-3}{2x-2})^{4x^2}$
Now $lim_limits {xto infty} (1 +frac{-3}{2x-2})^{2x-2} = e^{-3}$
So, what do you have to do to get a factor like this out of the line above?
$lim_limits {xto infty} ((1 +frac{-3}{2x-2})^{2x-2})^{2x+2}(1 +frac{-3}{2x-2})^4\
lim_limits {xto infty} (e^{-3})^{2x+2}(1 +frac{-3}{2x-2})^4$
$0<e^{-3} < 1$
What I have done here is not rigorous.
But you can use the work above to say:
$0le lim_limits {xto infty}(frac {2x-5}{2x-2})^{4x^2} le lim_limits {xto infty} e^{-6x}$
$lim_limits {xto infty}(frac {2x-5}{2x-2})^{4x^2} = 0$ by the squeeze theorem.
answered Jan 24 at 21:44
Doug MDoug M
45.3k31954
$endgroup$
$begingroup$
You can't take the limit of part of an expression, the take the limit of what remains (with exponents, &c.).
$endgroup$
– Bernard
Jan 24 at 21:48
$begingroup$
True, I have played a bit fast-and-loose. But, It points us in a direction that is rigorous.
$endgroup$
– Doug M
Jan 24 at 22:12
$begingroup$
@doug-m Ok, thanks. But I don't quite understand this line: $0<e^{-3} < 1$ How did we get this statement from the previous line?
$endgroup$
– makulo123
Jan 24 at 22:20
$begingroup$
$0<e^{-3}<1$ because $e>1$
$endgroup$
– Doug M
Jan 24 at 23:29
$begingroup$
Yes, thanks, I used a bit different aproach based on your solution: $lim_limits {xto infty}(e^{-3})^{2x+2}=lim_limits {xto infty}(e^{-6x-6})=lim_limits {xto infty}(e^{-6x})(e^{-6})=0$
$endgroup$
– makulo123
Jan 25 at 0:00
add a comment |
$begingroup$
$lim_limits {nto infty} (1+frac {x}{n})^n = e^x$
$lim_limits {xto infty} (frac {2x-5}{2x-2})^{4x^2}\
lim_limits {xto infty} (1 +frac{-3}{2x-2})^{4x^2}$
Now $lim_limits {xto infty} (1 +frac{-3}{2x-2})^{2x-2} = e^{-3}$
So, what do you have to do to get a factor like this out of the line above?
$lim_limits {xto infty} ((1 +frac{-3}{2x-2})^{2x-2})^{2x+2}(1 +frac{-3}{2x-2})^4\
lim_limits {xto infty} (e^{-3})^{2x+2}(1 +frac{-3}{2x-2})^4$
$0<e^{-3} < 1$
What I have done here is not rigorous.
But you can use the work above to say:
$0le lim_limits {xto infty}(frac {2x-5}{2x-2})^{4x^2} le lim_limits {xto infty} e^{-6x}$
$lim_limits {xto infty}(frac {2x-5}{2x-2})^{4x^2} = 0$ by the squeeze theorem.
answered Jan 24 at 21:44
Doug MDoug M
45.3k31954
$endgroup$
$lim_limits {nto infty} (1+frac {x}{n})^n = e^x$
$lim_limits {xto infty} (frac {2x-5}{2x-2})^{4x^2}\
lim_limits {xto infty} (1 +frac{-3}{2x-2})^{4x^2}$
Now $lim_limits {xto infty} (1 +frac{-3}{2x-2})^{2x-2} = e^{-3}$
So, what do you have to do to get a factor like this out of the line above?
$lim_limits {xto infty} ((1 +frac{-3}{2x-2})^{2x-2})^{2x+2}(1 +frac{-3}{2x-2})^4\
lim_limits {xto infty} (e^{-3})^{2x+2}(1 +frac{-3}{2x-2})^4$
$0<e^{-3} < 1$
What I have done here is not rigorous.
But you can use the work above to say:
$0le lim_limits {xto infty}(frac {2x-5}{2x-2})^{4x^2} le lim_limits {xto infty} e^{-6x}$
$lim_limits {xto infty}(frac {2x-5}{2x-2})^{4x^2} = 0$ by the squeeze theorem.
answered Jan 24 at 21:44
Doug MDoug M
45.3k31954
edited Jan 24 at 22:11
answered Jan 24 at 21:44
Doug MDoug M
45.3k31954
answered Jan 24 at 21:44
Doug MDoug M
45.3k31954
answered Jan 24 at 21:44
Doug MDoug M
45.3k31954
45.3k31954
$begingroup$
You can't take the limit of part of an expression, the take the limit of what remains (with exponents, &c.).
$endgroup$
– Bernard
Jan 24 at 21:48
$begingroup$
True, I have played a bit fast-and-loose. But, It points us in a direction that is rigorous.
$endgroup$
– Doug M
Jan 24 at 22:12
$begingroup$
@doug-m Ok, thanks. But I don't quite understand this line: $0<e^{-3} < 1$ How did we get this statement from the previous line?
$endgroup$
– makulo123
Jan 24 at 22:20
$begingroup$
$0<e^{-3}<1$ because $e>1$
$endgroup$
– Doug M
Jan 24 at 23:29
$begingroup$
Yes, thanks, I used a bit different aproach based on your solution: $lim_limits {xto infty}(e^{-3})^{2x+2}=lim_limits {xto infty}(e^{-6x-6})=lim_limits {xto infty}(e^{-6x})(e^{-6})=0$
$endgroup$
– makulo123
Jan 25 at 0:00
add a comment |
$begingroup$
You can't take the limit of part of an expression, the take the limit of what remains (with exponents, &c.).
$endgroup$
– Bernard
Jan 24 at 21:48
$begingroup$
True, I have played a bit fast-and-loose. But, It points us in a direction that is rigorous.
$endgroup$
– Doug M
Jan 24 at 22:12
$begingroup$
@doug-m Ok, thanks. But I don't quite understand this line: $0<e^{-3} < 1$ How did we get this statement from the previous line?
$endgroup$
– makulo123
Jan 24 at 22:20
$begingroup$
$0<e^{-3}<1$ because $e>1$
$endgroup$
– Doug M
Jan 24 at 23:29
$begingroup$
Yes, thanks, I used a bit different aproach based on your solution: $lim_limits {xto infty}(e^{-3})^{2x+2}=lim_limits {xto infty}(e^{-6x-6})=lim_limits {xto infty}(e^{-6x})(e^{-6})=0$
$endgroup$
– makulo123
Jan 25 at 0:00
$begingroup$
You can't take the limit of part of an expression, the take the limit of what remains (with exponents, &c.).
$endgroup$
– Bernard
Jan 24 at 21:48
$begingroup$
You can't take the limit of part of an expression, the take the limit of what remains (with exponents, &c.).
$endgroup$
– Bernard
Jan 24 at 21:48
$begingroup$
True, I have played a bit fast-and-loose. But, It points us in a direction that is rigorous.
$endgroup$
– Doug M
Jan 24 at 22:12
$begingroup$
True, I have played a bit fast-and-loose. But, It points us in a direction that is rigorous.
$endgroup$
– Doug M
Jan 24 at 22:12
$begingroup$
@doug-m Ok, thanks. But I don't quite understand this line: $0<e^{-3} < 1$ How did we get this statement from the previous line?
$endgroup$
– makulo123
Jan 24 at 22:20
$begingroup$
@doug-m Ok, thanks. But I don't quite understand this line: $0<e^{-3} < 1$ How did we get this statement from the previous line?
$endgroup$
– makulo123
Jan 24 at 22:20
$begingroup$
$0<e^{-3}<1$ because $e>1$
$endgroup$
– Doug M
Jan 24 at 23:29
$begingroup$
$0<e^{-3}<1$ because $e>1$
$endgroup$
– Doug M
Jan 24 at 23:29
$begingroup$
Yes, thanks, I used a bit different aproach based on your solution: $lim_limits {xto infty}(e^{-3})^{2x+2}=lim_limits {xto infty}(e^{-6x-6})=lim_limits {xto infty}(e^{-6x})(e^{-6})=0$
$endgroup$
– makulo123
Jan 25 at 0:00
$begingroup$
Yes, thanks, I used a bit different aproach based on your solution: $lim_limits {xto infty}(e^{-3})^{2x+2}=lim_limits {xto infty}(e^{-6x-6})=lim_limits {xto infty}(e^{-6x})(e^{-6})=0$
$endgroup$
– makulo123
Jan 25 at 0:00
add a comment |
2
$begingroup$
Telling people you need something ASAP will not earn you any love around here.
$endgroup$
– Doug M
Jan 24 at 21:28
$begingroup$
"I was not present at the time they were doing that" Then read the relevant chapter of the textbook, or ask your fellow students for some notes.
$endgroup$
– Rahul
Jan 24 at 21:30
$begingroup$
@Rahul That is exactly what I have done, but the problem is, they did not encounter such problem during class and the teacher gave us these problems randomly. He did not solve them himself, so I am not sure if it even can be solved without the L'Hospital's rule.
$endgroup$
– makulo123
Jan 24 at 21:32
$begingroup$
@makulo123 hey There! You know your question has been put on hold. Next time if you asked a question, just write your own solution (doesn't matter if it's incomplete, wrong, or unrelated), it gives us the insight to look the question and understand your level and give you the best to your level ;) :) :D
$endgroup$
– Abhas Kumar Sinha
Jan 25 at 15:32