Mixing
Find limit of $frac{e^x cos x - 1}{x}$ [on hold]
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-1
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Is it possible to find
$$
lim_{x to 0}frac{e^x cos x - 1}{x}
$$
without using L'hospital's rule?
limits limits-without-lhopital
edited yesterday
gimusi
85.8k74294
asked yesterday
Nikrom
1033
put on hold as off-topic by Jyrki Lahtonen, amWhy, Rebellos, Paramanand Singh, jgon yesterday
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." – Jyrki Lahtonen, amWhy, Rebellos, Paramanand Singh, jgon
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
-1
down vote
favorite
Is it possible to find
$$
lim_{x to 0}frac{e^x cos x - 1}{x}
$$
without using L'hospital's rule?
limits limits-without-lhopital
edited yesterday
gimusi
85.8k74294
asked yesterday
Nikrom
1033
put on hold as off-topic by Jyrki Lahtonen, amWhy, Rebellos, Paramanand Singh, jgon yesterday
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." – Jyrki Lahtonen, amWhy, Rebellos, Paramanand Singh, jgon
If this question can be reworded to fit the rules in the help center, please edit the question.
1
Make the product of Taylor series (very very faw terms).
– Claude Leibovici
yesterday
2
It is possible.
– Chickenmancer
yesterday
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Is it possible to find
$$
lim_{x to 0}frac{e^x cos x - 1}{x}
$$
without using L'hospital's rule?
limits limits-without-lhopital
edited yesterday
gimusi
85.8k74294
asked yesterday
Nikrom
1033
Is it possible to find
$$
lim_{x to 0}frac{e^x cos x - 1}{x}
$$
without using L'hospital's rule?
limits limits-without-lhopital
limits limits-without-lhopital
edited yesterday
gimusi
85.8k74294
asked yesterday
Nikrom
1033
edited yesterday
gimusi
85.8k74294
asked yesterday
Nikrom
1033
edited yesterday
gimusi
85.8k74294
edited yesterday
gimusi
85.8k74294
edited yesterday
gimusi
85.8k74294
85.8k74294
asked yesterday
Nikrom
1033
asked yesterday
Nikrom
1033
asked yesterday
Nikrom
1033
1033
put on hold as off-topic by Jyrki Lahtonen, amWhy, Rebellos, Paramanand Singh, jgon yesterday
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." – Jyrki Lahtonen, amWhy, Rebellos, Paramanand Singh, jgon
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by Jyrki Lahtonen, amWhy, Rebellos, Paramanand Singh, jgon yesterday
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." – Jyrki Lahtonen, amWhy, Rebellos, Paramanand Singh, jgon
If this question can be reworded to fit the rules in the help center, please edit the question.
1
Make the product of Taylor series (very very faw terms).
– Claude Leibovici
yesterday
2
It is possible.
– Chickenmancer
yesterday
add a comment |
1
Make the product of Taylor series (very very faw terms).
– Claude Leibovici
yesterday
2
It is possible.
– Chickenmancer
yesterday
1
1
Make the product of Taylor series (very very faw terms).
– Claude Leibovici
yesterday
Make the product of Taylor series (very very faw terms).
– Claude Leibovici
yesterday
2
2
It is possible.
– Chickenmancer
yesterday
It is possible.
– Chickenmancer
yesterday
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
We have that
$$frac{e^x cos x - 1}{x}=frac{e^x cos x-cos x+cos x - 1}{x}=$$$$=cos xfrac{e^x- 1}{x}+xfrac{cos x - 1}{x^2} to 1cdot 1+0cdotleft(-frac12right)=1$$
indeed by standard limits
- $frac{e^x- 1}{x}to 1$
- $frac{1-cos x}{x^2} to frac12$
answered yesterday
gimusi
85.8k74294
My upvote for the first approach. IMO, using the definition of the derivative is L'Hospital in disguise.
– Yves Daoust
yesterday
@YvesDaoust Thanks for your kind appreciation! I also like much more the first approach.
– gimusi
yesterday
add a comment |
up vote
0
down vote
As $xto0$ you can make the following approximations: $e^x=1+x+O(x),quad cos x=1-x^2/2+O(x^2)$
$$lim_{xto0}frac{(1+x)(1-x^2/2)-1}{x}=lim_{xto0}frac{1-x^2/2+x-x^3/2-1}x=lim_{xto0}frac{x(-x/2+1-x^2/2)}{x}=1$$
answered yesterday
Lorenzo B.
1,6222419
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
We have that
$$frac{e^x cos x - 1}{x}=frac{e^x cos x-cos x+cos x - 1}{x}=$$$$=cos xfrac{e^x- 1}{x}+xfrac{cos x - 1}{x^2} to 1cdot 1+0cdotleft(-frac12right)=1$$
indeed by standard limits
- $frac{e^x- 1}{x}to 1$
- $frac{1-cos x}{x^2} to frac12$
answered yesterday
gimusi
85.8k74294
My upvote for the first approach. IMO, using the definition of the derivative is L'Hospital in disguise.
– Yves Daoust
yesterday
@YvesDaoust Thanks for your kind appreciation! I also like much more the first approach.
– gimusi
yesterday
add a comment |
up vote
1
down vote
accepted
We have that
$$frac{e^x cos x - 1}{x}=frac{e^x cos x-cos x+cos x - 1}{x}=$$$$=cos xfrac{e^x- 1}{x}+xfrac{cos x - 1}{x^2} to 1cdot 1+0cdotleft(-frac12right)=1$$
indeed by standard limits
- $frac{e^x- 1}{x}to 1$
- $frac{1-cos x}{x^2} to frac12$
answered yesterday
gimusi
85.8k74294
My upvote for the first approach. IMO, using the definition of the derivative is L'Hospital in disguise.
– Yves Daoust
yesterday
@YvesDaoust Thanks for your kind appreciation! I also like much more the first approach.
– gimusi
yesterday
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
We have that
$$frac{e^x cos x - 1}{x}=frac{e^x cos x-cos x+cos x - 1}{x}=$$$$=cos xfrac{e^x- 1}{x}+xfrac{cos x - 1}{x^2} to 1cdot 1+0cdotleft(-frac12right)=1$$
indeed by standard limits
- $frac{e^x- 1}{x}to 1$
- $frac{1-cos x}{x^2} to frac12$
answered yesterday
gimusi
85.8k74294
We have that
$$frac{e^x cos x - 1}{x}=frac{e^x cos x-cos x+cos x - 1}{x}=$$$$=cos xfrac{e^x- 1}{x}+xfrac{cos x - 1}{x^2} to 1cdot 1+0cdotleft(-frac12right)=1$$
indeed by standard limits
- $frac{e^x- 1}{x}to 1$
- $frac{1-cos x}{x^2} to frac12$
answered yesterday
gimusi
85.8k74294
edited yesterday
answered yesterday
gimusi
85.8k74294
answered yesterday
gimusi
85.8k74294
answered yesterday
gimusi
85.8k74294
85.8k74294
My upvote for the first approach. IMO, using the definition of the derivative is L'Hospital in disguise.
– Yves Daoust
yesterday
@YvesDaoust Thanks for your kind appreciation! I also like much more the first approach.
– gimusi
yesterday
add a comment |
My upvote for the first approach. IMO, using the definition of the derivative is L'Hospital in disguise.
– Yves Daoust
yesterday
@YvesDaoust Thanks for your kind appreciation! I also like much more the first approach.
– gimusi
yesterday
My upvote for the first approach. IMO, using the definition of the derivative is L'Hospital in disguise.
– Yves Daoust
yesterday
My upvote for the first approach. IMO, using the definition of the derivative is L'Hospital in disguise.
– Yves Daoust
yesterday
@YvesDaoust Thanks for your kind appreciation! I also like much more the first approach.
– gimusi
yesterday
@YvesDaoust Thanks for your kind appreciation! I also like much more the first approach.
– gimusi
yesterday
add a comment |
up vote
0
down vote
As $xto0$ you can make the following approximations: $e^x=1+x+O(x),quad cos x=1-x^2/2+O(x^2)$
$$lim_{xto0}frac{(1+x)(1-x^2/2)-1}{x}=lim_{xto0}frac{1-x^2/2+x-x^3/2-1}x=lim_{xto0}frac{x(-x/2+1-x^2/2)}{x}=1$$
answered yesterday
Lorenzo B.
1,6222419
add a comment |
up vote
0
down vote
As $xto0$ you can make the following approximations: $e^x=1+x+O(x),quad cos x=1-x^2/2+O(x^2)$
$$lim_{xto0}frac{(1+x)(1-x^2/2)-1}{x}=lim_{xto0}frac{1-x^2/2+x-x^3/2-1}x=lim_{xto0}frac{x(-x/2+1-x^2/2)}{x}=1$$
answered yesterday
Lorenzo B.
1,6222419
add a comment |
up vote
0
down vote
up vote
0
down vote
As $xto0$ you can make the following approximations: $e^x=1+x+O(x),quad cos x=1-x^2/2+O(x^2)$
$$lim_{xto0}frac{(1+x)(1-x^2/2)-1}{x}=lim_{xto0}frac{1-x^2/2+x-x^3/2-1}x=lim_{xto0}frac{x(-x/2+1-x^2/2)}{x}=1$$
answered yesterday
Lorenzo B.
1,6222419
As $xto0$ you can make the following approximations: $e^x=1+x+O(x),quad cos x=1-x^2/2+O(x^2)$
$$lim_{xto0}frac{(1+x)(1-x^2/2)-1}{x}=lim_{xto0}frac{1-x^2/2+x-x^3/2-1}x=lim_{xto0}frac{x(-x/2+1-x^2/2)}{x}=1$$
answered yesterday
Lorenzo B.
1,6222419
answered yesterday
Lorenzo B.
1,6222419
answered yesterday
Lorenzo B.
1,6222419
answered yesterday
Lorenzo B.
1,6222419
1,6222419
add a comment |
add a comment |
1
Make the product of Taylor series (very very faw terms).
– Claude Leibovici
yesterday
2
It is possible.
– Chickenmancer
yesterday