Calculate the limit $lim_{xto+infty}xe^xsinleft(e^{-x}sinfrac{2}{x}right)$ [closed]
-3
$begingroup$
Please help me to calcute the follow limit:
$$lim_{xto+infty}xe^xsinleft(e^{-x}sinfrac{2}{x}right)$$
calculus
asked Jan 7 at 12:36
brisk tutibrisk tuti
83
$endgroup$
closed as off-topic by Key Flex, Nosrati, KM101, José Carlos Santos, Shailesh Jan 7 at 12:57
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." – Key Flex, Nosrati, KM101, José Carlos Santos, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
-3
$begingroup$
Please help me to calcute the follow limit:
$$lim_{xto+infty}xe^xsinleft(e^{-x}sinfrac{2}{x}right)$$
calculus
asked Jan 7 at 12:36
brisk tutibrisk tuti
83
$endgroup$
closed as off-topic by Key Flex, Nosrati, KM101, José Carlos Santos, Shailesh Jan 7 at 12:57
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." – Key Flex, Nosrati, KM101, José Carlos Santos, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
Please share how you have tried to solve by editing the post
$endgroup$
– Shubham Johri
Jan 7 at 12:39
2
$begingroup$
Please show your work. Do you have any ideas?
$endgroup$
– KM101
Jan 7 at 12:40
$begingroup$
It's unclear what you are allowed to use. How about Taylor expansion?
$endgroup$
– roman
Jan 7 at 12:41
add a comment |
-3
-3
-3
$begingroup$
Please help me to calcute the follow limit:
$$lim_{xto+infty}xe^xsinleft(e^{-x}sinfrac{2}{x}right)$$
calculus
asked Jan 7 at 12:36
brisk tutibrisk tuti
83
$endgroup$
Please help me to calcute the follow limit:
$$lim_{xto+infty}xe^xsinleft(e^{-x}sinfrac{2}{x}right)$$
calculus
calculus
asked Jan 7 at 12:36
brisk tutibrisk tuti
83
asked Jan 7 at 12:36
brisk tutibrisk tuti
83
asked Jan 7 at 12:36
brisk tutibrisk tuti
83
asked Jan 7 at 12:36
brisk tutibrisk tuti
83
asked Jan 7 at 12:36
brisk tutibrisk tuti
83
83
closed as off-topic by Key Flex, Nosrati, KM101, José Carlos Santos, Shailesh Jan 7 at 12:57
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." – Key Flex, Nosrati, KM101, José Carlos Santos, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Key Flex, Nosrati, KM101, José Carlos Santos, Shailesh Jan 7 at 12:57
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." – Key Flex, Nosrati, KM101, José Carlos Santos, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
Please share how you have tried to solve by editing the post
$endgroup$
– Shubham Johri
Jan 7 at 12:39
2
$begingroup$
Please show your work. Do you have any ideas?
$endgroup$
– KM101
Jan 7 at 12:40
$begingroup$
It's unclear what you are allowed to use. How about Taylor expansion?
$endgroup$
– roman
Jan 7 at 12:41
add a comment |
2
$begingroup$
Please share how you have tried to solve by editing the post
$endgroup$
– Shubham Johri
Jan 7 at 12:39
2
$begingroup$
Please show your work. Do you have any ideas?
$endgroup$
– KM101
Jan 7 at 12:40
$begingroup$
It's unclear what you are allowed to use. How about Taylor expansion?
$endgroup$
– roman
Jan 7 at 12:41
2
2
$begingroup$
Please share how you have tried to solve by editing the post
$endgroup$
– Shubham Johri
Jan 7 at 12:39
$begingroup$
Please share how you have tried to solve by editing the post
$endgroup$
– Shubham Johri
Jan 7 at 12:39
2
2
$begingroup$
Please show your work. Do you have any ideas?
$endgroup$
– KM101
Jan 7 at 12:40
$begingroup$
Please show your work. Do you have any ideas?
$endgroup$
– KM101
Jan 7 at 12:40
$begingroup$
It's unclear what you are allowed to use. How about Taylor expansion?
$endgroup$
– roman
Jan 7 at 12:41
$begingroup$
It's unclear what you are allowed to use. How about Taylor expansion?
$endgroup$
– roman
Jan 7 at 12:41
add a comment |
1 Answer
1
active
oldest
votes
0
$begingroup$
Hints:
- $lim_{xto +infty}e^{-x}sinfrac{2}{x} = 0$
- $lim_{xto +infty} frac{sin left(e^{-x}sinfrac{2}{x} right)}{e^{-x}sinfrac{2}{x}} = 1$
answered Jan 7 at 12:51
trancelocationtrancelocation
9,6801722
$endgroup$
$begingroup$
$lim_{xto+infty}xe^xsinleft(e^{-x}sinfrac{2}{x}right)$ is not to equal for the second $cdot$
$endgroup$
– brisk tuti
Jan 7 at 13:06
$begingroup$
Just use the hints and write: $$xe^xsinleft(e^{-x}sinfrac{2}{x}right) = xe^x cdot e^{-x}sinfrac{2}{x} cdot frac{sin left(e^{-x}sinfrac{2}{x} right)}{e^{-x}sinfrac{2}{x}} = 2cdot frac{sinfrac{2}{x}}{frac{2}{x}} cdot frac{sin left(e^{-x}sinfrac{2}{x} right)}{e^{-x}sinfrac{2}{x}}$$
$endgroup$
– trancelocation
Jan 7 at 13:11
$begingroup$
finally the given lim is equal to 2
$endgroup$
– brisk tuti
Jan 7 at 13:22
$begingroup$
Exactly. This is the limit.
$endgroup$
– trancelocation
Jan 7 at 13:26
$begingroup$
thanks sir, I didnt now the solve and another lim, can you help me please $$lim_{xto-infty}xe^{sin x}$$
$endgroup$
– brisk tuti
Jan 7 at 13:35
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
$begingroup$
Hints:
- $lim_{xto +infty}e^{-x}sinfrac{2}{x} = 0$
- $lim_{xto +infty} frac{sin left(e^{-x}sinfrac{2}{x} right)}{e^{-x}sinfrac{2}{x}} = 1$
answered Jan 7 at 12:51
trancelocationtrancelocation
9,6801722
$endgroup$
$begingroup$
$lim_{xto+infty}xe^xsinleft(e^{-x}sinfrac{2}{x}right)$ is not to equal for the second $cdot$
$endgroup$
– brisk tuti
Jan 7 at 13:06
$begingroup$
Just use the hints and write: $$xe^xsinleft(e^{-x}sinfrac{2}{x}right) = xe^x cdot e^{-x}sinfrac{2}{x} cdot frac{sin left(e^{-x}sinfrac{2}{x} right)}{e^{-x}sinfrac{2}{x}} = 2cdot frac{sinfrac{2}{x}}{frac{2}{x}} cdot frac{sin left(e^{-x}sinfrac{2}{x} right)}{e^{-x}sinfrac{2}{x}}$$
$endgroup$
– trancelocation
Jan 7 at 13:11
$begingroup$
finally the given lim is equal to 2
$endgroup$
– brisk tuti
Jan 7 at 13:22
$begingroup$
Exactly. This is the limit.
$endgroup$
– trancelocation
Jan 7 at 13:26
$begingroup$
thanks sir, I didnt now the solve and another lim, can you help me please $$lim_{xto-infty}xe^{sin x}$$
$endgroup$
– brisk tuti
Jan 7 at 13:35
add a comment |
0
$begingroup$
Hints:
- $lim_{xto +infty}e^{-x}sinfrac{2}{x} = 0$
- $lim_{xto +infty} frac{sin left(e^{-x}sinfrac{2}{x} right)}{e^{-x}sinfrac{2}{x}} = 1$
answered Jan 7 at 12:51
trancelocationtrancelocation
9,6801722
$endgroup$
$begingroup$
$lim_{xto+infty}xe^xsinleft(e^{-x}sinfrac{2}{x}right)$ is not to equal for the second $cdot$
$endgroup$
– brisk tuti
Jan 7 at 13:06
$begingroup$
Just use the hints and write: $$xe^xsinleft(e^{-x}sinfrac{2}{x}right) = xe^x cdot e^{-x}sinfrac{2}{x} cdot frac{sin left(e^{-x}sinfrac{2}{x} right)}{e^{-x}sinfrac{2}{x}} = 2cdot frac{sinfrac{2}{x}}{frac{2}{x}} cdot frac{sin left(e^{-x}sinfrac{2}{x} right)}{e^{-x}sinfrac{2}{x}}$$
$endgroup$
– trancelocation
Jan 7 at 13:11
$begingroup$
finally the given lim is equal to 2
$endgroup$
– brisk tuti
Jan 7 at 13:22
$begingroup$
Exactly. This is the limit.
$endgroup$
– trancelocation
Jan 7 at 13:26
$begingroup$
thanks sir, I didnt now the solve and another lim, can you help me please $$lim_{xto-infty}xe^{sin x}$$
$endgroup$
– brisk tuti
Jan 7 at 13:35
add a comment |
0
0
0
$begingroup$
Hints:
- $lim_{xto +infty}e^{-x}sinfrac{2}{x} = 0$
- $lim_{xto +infty} frac{sin left(e^{-x}sinfrac{2}{x} right)}{e^{-x}sinfrac{2}{x}} = 1$
answered Jan 7 at 12:51
trancelocationtrancelocation
9,6801722
$endgroup$
Hints:
- $lim_{xto +infty}e^{-x}sinfrac{2}{x} = 0$
- $lim_{xto +infty} frac{sin left(e^{-x}sinfrac{2}{x} right)}{e^{-x}sinfrac{2}{x}} = 1$
answered Jan 7 at 12:51
trancelocationtrancelocation
9,6801722
answered Jan 7 at 12:51
trancelocationtrancelocation
9,6801722
answered Jan 7 at 12:51
trancelocationtrancelocation
9,6801722
answered Jan 7 at 12:51
trancelocationtrancelocation
9,6801722
9,6801722
$begingroup$
$lim_{xto+infty}xe^xsinleft(e^{-x}sinfrac{2}{x}right)$ is not to equal for the second $cdot$
$endgroup$
– brisk tuti
Jan 7 at 13:06
$begingroup$
Just use the hints and write: $$xe^xsinleft(e^{-x}sinfrac{2}{x}right) = xe^x cdot e^{-x}sinfrac{2}{x} cdot frac{sin left(e^{-x}sinfrac{2}{x} right)}{e^{-x}sinfrac{2}{x}} = 2cdot frac{sinfrac{2}{x}}{frac{2}{x}} cdot frac{sin left(e^{-x}sinfrac{2}{x} right)}{e^{-x}sinfrac{2}{x}}$$
$endgroup$
– trancelocation
Jan 7 at 13:11
$begingroup$
finally the given lim is equal to 2
$endgroup$
– brisk tuti
Jan 7 at 13:22
$begingroup$
Exactly. This is the limit.
$endgroup$
– trancelocation
Jan 7 at 13:26
$begingroup$
thanks sir, I didnt now the solve and another lim, can you help me please $$lim_{xto-infty}xe^{sin x}$$
$endgroup$
– brisk tuti
Jan 7 at 13:35
add a comment |
$begingroup$
$lim_{xto+infty}xe^xsinleft(e^{-x}sinfrac{2}{x}right)$ is not to equal for the second $cdot$
$endgroup$
– brisk tuti
Jan 7 at 13:06
$begingroup$
Just use the hints and write: $$xe^xsinleft(e^{-x}sinfrac{2}{x}right) = xe^x cdot e^{-x}sinfrac{2}{x} cdot frac{sin left(e^{-x}sinfrac{2}{x} right)}{e^{-x}sinfrac{2}{x}} = 2cdot frac{sinfrac{2}{x}}{frac{2}{x}} cdot frac{sin left(e^{-x}sinfrac{2}{x} right)}{e^{-x}sinfrac{2}{x}}$$
$endgroup$
– trancelocation
Jan 7 at 13:11
$begingroup$
finally the given lim is equal to 2
$endgroup$
– brisk tuti
Jan 7 at 13:22
$begingroup$
Exactly. This is the limit.
$endgroup$
– trancelocation
Jan 7 at 13:26
$begingroup$
thanks sir, I didnt now the solve and another lim, can you help me please $$lim_{xto-infty}xe^{sin x}$$
$endgroup$
– brisk tuti
Jan 7 at 13:35
$begingroup$
$lim_{xto+infty}xe^xsinleft(e^{-x}sinfrac{2}{x}right)$ is not to equal for the second $cdot$
$endgroup$
– brisk tuti
Jan 7 at 13:06
$begingroup$
$lim_{xto+infty}xe^xsinleft(e^{-x}sinfrac{2}{x}right)$ is not to equal for the second $cdot$
$endgroup$
– brisk tuti
Jan 7 at 13:06
$begingroup$
Just use the hints and write: $$xe^xsinleft(e^{-x}sinfrac{2}{x}right) = xe^x cdot e^{-x}sinfrac{2}{x} cdot frac{sin left(e^{-x}sinfrac{2}{x} right)}{e^{-x}sinfrac{2}{x}} = 2cdot frac{sinfrac{2}{x}}{frac{2}{x}} cdot frac{sin left(e^{-x}sinfrac{2}{x} right)}{e^{-x}sinfrac{2}{x}}$$
$endgroup$
– trancelocation
Jan 7 at 13:11
$begingroup$
Just use the hints and write: $$xe^xsinleft(e^{-x}sinfrac{2}{x}right) = xe^x cdot e^{-x}sinfrac{2}{x} cdot frac{sin left(e^{-x}sinfrac{2}{x} right)}{e^{-x}sinfrac{2}{x}} = 2cdot frac{sinfrac{2}{x}}{frac{2}{x}} cdot frac{sin left(e^{-x}sinfrac{2}{x} right)}{e^{-x}sinfrac{2}{x}}$$
$endgroup$
– trancelocation
Jan 7 at 13:11
$begingroup$
finally the given lim is equal to 2
$endgroup$
– brisk tuti
Jan 7 at 13:22
$begingroup$
finally the given lim is equal to 2
$endgroup$
– brisk tuti
Jan 7 at 13:22
$begingroup$
Exactly. This is the limit.
$endgroup$
– trancelocation
Jan 7 at 13:26
$begingroup$
Exactly. This is the limit.
$endgroup$
– trancelocation
Jan 7 at 13:26
$begingroup$
thanks sir, I didnt now the solve and another lim, can you help me please $$lim_{xto-infty}xe^{sin x}$$
$endgroup$
– brisk tuti
Jan 7 at 13:35
$begingroup$
thanks sir, I didnt now the solve and another lim, can you help me please $$lim_{xto-infty}xe^{sin x}$$
$endgroup$
– brisk tuti
Jan 7 at 13:35
add a comment |
2
$begingroup$
Please share how you have tried to solve by editing the post
$endgroup$
– Shubham Johri
Jan 7 at 12:39
2
$begingroup$
Please show your work. Do you have any ideas?
$endgroup$
– KM101
Jan 7 at 12:40
$begingroup$
It's unclear what you are allowed to use. How about Taylor expansion?
$endgroup$
– roman
Jan 7 at 12:41