Sequence limit with form $infty cdot 0$ [closed]
Compute $lim_{ntoinfty} 2^n cdot left(frac{2n^2+4n+1}{4n^2+12n+8}right)^{n+1}$.
I tried to reduce this to $e$ raised to some power, but I didn't succeed.
calculus algebra-precalculus limits
edited Nov 26 at 16:56
Servaes
22.3k33793
asked Nov 26 at 16:50
user69503
626
closed as off-topic by amWhy, Shailesh, Cesareo, Jyrki Lahtonen, KReiser Nov 27 at 4:14
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Compute $lim_{ntoinfty} 2^n cdot left(frac{2n^2+4n+1}{4n^2+12n+8}right)^{n+1}$.
I tried to reduce this to $e$ raised to some power, but I didn't succeed.
calculus algebra-precalculus limits
edited Nov 26 at 16:56
Servaes
22.3k33793
asked Nov 26 at 16:50
user69503
626
closed as off-topic by amWhy, Shailesh, Cesareo, Jyrki Lahtonen, KReiser Nov 27 at 4:14
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Shailesh, Cesareo, Jyrki Lahtonen, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
Compute $lim_{ntoinfty} 2^n cdot left(frac{2n^2+4n+1}{4n^2+12n+8}right)^{n+1}$.
I tried to reduce this to $e$ raised to some power, but I didn't succeed.
calculus algebra-precalculus limits
edited Nov 26 at 16:56
Servaes
22.3k33793
asked Nov 26 at 16:50
user69503
626
Compute $lim_{ntoinfty} 2^n cdot left(frac{2n^2+4n+1}{4n^2+12n+8}right)^{n+1}$.
I tried to reduce this to $e$ raised to some power, but I didn't succeed.
calculus algebra-precalculus limits
calculus algebra-precalculus limits
edited Nov 26 at 16:56
Servaes
22.3k33793
asked Nov 26 at 16:50
user69503
626
edited Nov 26 at 16:56
Servaes
22.3k33793
asked Nov 26 at 16:50
user69503
626
edited Nov 26 at 16:56
Servaes
22.3k33793
edited Nov 26 at 16:56
Servaes
22.3k33793
edited Nov 26 at 16:56
Servaes
22.3k33793
22.3k33793
asked Nov 26 at 16:50
user69503
626
asked Nov 26 at 16:50
user69503
626
asked Nov 26 at 16:50
user69503
626
626
closed as off-topic by amWhy, Shailesh, Cesareo, Jyrki Lahtonen, KReiser Nov 27 at 4:14
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Shailesh, Cesareo, Jyrki Lahtonen, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, Shailesh, Cesareo, Jyrki Lahtonen, KReiser Nov 27 at 4:14
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Shailesh, Cesareo, Jyrki Lahtonen, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
2 Answers
2
active
oldest
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HINT: Note that for all $ngeq0$ we have
$$2^n cdotleft(frac{2n^2+4n+1}{4n^2+12n+8}right)^{n+1}=frac{1}{2}left(frac{4n^2+8n+2}{4n^2+12n+8}right)^{n+1}.$$
answered Nov 26 at 16:52
Servaes
22.3k33793
Very good hint at the key point!
– gimusi
Nov 26 at 17:19
add a comment |
HINT
We have that
$$2^n cdot left(frac{2n^2+4n+1}{4n^2+12n+8}right)^{n+1}=$$$$=frac12 cdot left(frac{4n^2+8n+4}{4n^2+12n+8}right)^{n+1}=frac12 cdot left(frac{4n^2+12n+8-4n-4}{4n^2+12n+8}right)^{n+1}=$$
$$=frac12 cdot left[left(1-frac{4n+4}{4n^2+12n+8}right)^{frac{4n^2+12n+8}{4n+4}}right]^{frac{(4n+4)(n+1)}{4n^2+12n+8}}$$
answered Nov 26 at 17:10
gimusi
1
The first linebreak is in a rather unfortunate place. The last expression terrifies me.
– Servaes
Nov 26 at 17:17
@Servaes Yes you are right! I fix that
– gimusi
Nov 26 at 17:18
@Servaes That's only apparently terrifiant :)
– gimusi
Nov 26 at 17:24
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
HINT: Note that for all $ngeq0$ we have
$$2^n cdotleft(frac{2n^2+4n+1}{4n^2+12n+8}right)^{n+1}=frac{1}{2}left(frac{4n^2+8n+2}{4n^2+12n+8}right)^{n+1}.$$
answered Nov 26 at 16:52
Servaes
22.3k33793
Very good hint at the key point!
– gimusi
Nov 26 at 17:19
add a comment |
HINT: Note that for all $ngeq0$ we have
$$2^n cdotleft(frac{2n^2+4n+1}{4n^2+12n+8}right)^{n+1}=frac{1}{2}left(frac{4n^2+8n+2}{4n^2+12n+8}right)^{n+1}.$$
answered Nov 26 at 16:52
Servaes
22.3k33793
Very good hint at the key point!
– gimusi
Nov 26 at 17:19
add a comment |
HINT: Note that for all $ngeq0$ we have
$$2^n cdotleft(frac{2n^2+4n+1}{4n^2+12n+8}right)^{n+1}=frac{1}{2}left(frac{4n^2+8n+2}{4n^2+12n+8}right)^{n+1}.$$
answered Nov 26 at 16:52
Servaes
22.3k33793
HINT: Note that for all $ngeq0$ we have
$$2^n cdotleft(frac{2n^2+4n+1}{4n^2+12n+8}right)^{n+1}=frac{1}{2}left(frac{4n^2+8n+2}{4n^2+12n+8}right)^{n+1}.$$
answered Nov 26 at 16:52
Servaes
22.3k33793
answered Nov 26 at 16:52
Servaes
22.3k33793
answered Nov 26 at 16:52
Servaes
22.3k33793
answered Nov 26 at 16:52
Servaes
22.3k33793
22.3k33793
Very good hint at the key point!
– gimusi
Nov 26 at 17:19
add a comment |
Very good hint at the key point!
– gimusi
Nov 26 at 17:19
Very good hint at the key point!
– gimusi
Nov 26 at 17:19
Very good hint at the key point!
– gimusi
Nov 26 at 17:19
add a comment |
HINT
We have that
$$2^n cdot left(frac{2n^2+4n+1}{4n^2+12n+8}right)^{n+1}=$$$$=frac12 cdot left(frac{4n^2+8n+4}{4n^2+12n+8}right)^{n+1}=frac12 cdot left(frac{4n^2+12n+8-4n-4}{4n^2+12n+8}right)^{n+1}=$$
$$=frac12 cdot left[left(1-frac{4n+4}{4n^2+12n+8}right)^{frac{4n^2+12n+8}{4n+4}}right]^{frac{(4n+4)(n+1)}{4n^2+12n+8}}$$
answered Nov 26 at 17:10
gimusi
1
The first linebreak is in a rather unfortunate place. The last expression terrifies me.
– Servaes
Nov 26 at 17:17
@Servaes Yes you are right! I fix that
– gimusi
Nov 26 at 17:18
@Servaes That's only apparently terrifiant :)
– gimusi
Nov 26 at 17:24
add a comment |
HINT
We have that
$$2^n cdot left(frac{2n^2+4n+1}{4n^2+12n+8}right)^{n+1}=$$$$=frac12 cdot left(frac{4n^2+8n+4}{4n^2+12n+8}right)^{n+1}=frac12 cdot left(frac{4n^2+12n+8-4n-4}{4n^2+12n+8}right)^{n+1}=$$
$$=frac12 cdot left[left(1-frac{4n+4}{4n^2+12n+8}right)^{frac{4n^2+12n+8}{4n+4}}right]^{frac{(4n+4)(n+1)}{4n^2+12n+8}}$$
answered Nov 26 at 17:10
gimusi
1
The first linebreak is in a rather unfortunate place. The last expression terrifies me.
– Servaes
Nov 26 at 17:17
@Servaes Yes you are right! I fix that
– gimusi
Nov 26 at 17:18
@Servaes That's only apparently terrifiant :)
– gimusi
Nov 26 at 17:24
add a comment |
HINT
We have that
$$2^n cdot left(frac{2n^2+4n+1}{4n^2+12n+8}right)^{n+1}=$$$$=frac12 cdot left(frac{4n^2+8n+4}{4n^2+12n+8}right)^{n+1}=frac12 cdot left(frac{4n^2+12n+8-4n-4}{4n^2+12n+8}right)^{n+1}=$$
$$=frac12 cdot left[left(1-frac{4n+4}{4n^2+12n+8}right)^{frac{4n^2+12n+8}{4n+4}}right]^{frac{(4n+4)(n+1)}{4n^2+12n+8}}$$
answered Nov 26 at 17:10
gimusi
1
HINT
We have that
$$2^n cdot left(frac{2n^2+4n+1}{4n^2+12n+8}right)^{n+1}=$$$$=frac12 cdot left(frac{4n^2+8n+4}{4n^2+12n+8}right)^{n+1}=frac12 cdot left(frac{4n^2+12n+8-4n-4}{4n^2+12n+8}right)^{n+1}=$$
$$=frac12 cdot left[left(1-frac{4n+4}{4n^2+12n+8}right)^{frac{4n^2+12n+8}{4n+4}}right]^{frac{(4n+4)(n+1)}{4n^2+12n+8}}$$
answered Nov 26 at 17:10
gimusi
1
edited Nov 26 at 17:18
answered Nov 26 at 17:10
gimusi
1
answered Nov 26 at 17:10
gimusi
1
answered Nov 26 at 17:10
gimusi
1
1
The first linebreak is in a rather unfortunate place. The last expression terrifies me.
– Servaes
Nov 26 at 17:17
@Servaes Yes you are right! I fix that
– gimusi
Nov 26 at 17:18
@Servaes That's only apparently terrifiant :)
– gimusi
Nov 26 at 17:24
add a comment |
The first linebreak is in a rather unfortunate place. The last expression terrifies me.
– Servaes
Nov 26 at 17:17
@Servaes Yes you are right! I fix that
– gimusi
Nov 26 at 17:18
@Servaes That's only apparently terrifiant :)
– gimusi
Nov 26 at 17:24
The first linebreak is in a rather unfortunate place. The last expression terrifies me.
– Servaes
Nov 26 at 17:17
The first linebreak is in a rather unfortunate place. The last expression terrifies me.
– Servaes
Nov 26 at 17:17
@Servaes Yes you are right! I fix that
– gimusi
Nov 26 at 17:18
@Servaes Yes you are right! I fix that
– gimusi
Nov 26 at 17:18
@Servaes That's only apparently terrifiant :)
– gimusi
Nov 26 at 17:24
@Servaes That's only apparently terrifiant :)
– gimusi
Nov 26 at 17:24
add a comment |
