Mixing
Find $limlimits_{ntoinfty}{e^n - e^{frac1n + n}}$
up vote
1
down vote
favorite
So far I've tried: $${e^n - e^{frac1n + n}} = e^n(1- e^{frac1n}).$$ Then appling l'Hopitals rule to $$limlimits_{ntoinfty} e^n(1- e^{frac1n}) = limlimits_{ntoinfty} dfrac{(1- e^{frac1n})}{e^{-n}},$$ I have not found success. Is there another way to manipulate the expression to be able to apply l'Hopitals? I know the limit should approach -$infty$. To clarify, I can use l'Hopital.
calculus limits
asked 2 days ago
t.perez
356
add a comment |
up vote
1
down vote
favorite
So far I've tried: $${e^n - e^{frac1n + n}} = e^n(1- e^{frac1n}).$$ Then appling l'Hopitals rule to $$limlimits_{ntoinfty} e^n(1- e^{frac1n}) = limlimits_{ntoinfty} dfrac{(1- e^{frac1n})}{e^{-n}},$$ I have not found success. Is there another way to manipulate the expression to be able to apply l'Hopitals? I know the limit should approach -$infty$. To clarify, I can use l'Hopital.
calculus limits
asked 2 days ago
t.perez
356
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
So far I've tried: $${e^n - e^{frac1n + n}} = e^n(1- e^{frac1n}).$$ Then appling l'Hopitals rule to $$limlimits_{ntoinfty} e^n(1- e^{frac1n}) = limlimits_{ntoinfty} dfrac{(1- e^{frac1n})}{e^{-n}},$$ I have not found success. Is there another way to manipulate the expression to be able to apply l'Hopitals? I know the limit should approach -$infty$. To clarify, I can use l'Hopital.
calculus limits
asked 2 days ago
t.perez
356
So far I've tried: $${e^n - e^{frac1n + n}} = e^n(1- e^{frac1n}).$$ Then appling l'Hopitals rule to $$limlimits_{ntoinfty} e^n(1- e^{frac1n}) = limlimits_{ntoinfty} dfrac{(1- e^{frac1n})}{e^{-n}},$$ I have not found success. Is there another way to manipulate the expression to be able to apply l'Hopitals? I know the limit should approach -$infty$. To clarify, I can use l'Hopital.
calculus limits
calculus limits
asked 2 days ago
t.perez
356
asked 2 days ago
t.perez
356
asked 2 days ago
t.perez
356
asked 2 days ago
t.perez
356
asked 2 days ago
t.perez
356
356
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
4
down vote
HINT
We have that
$${e^n - e^{frac1n + n}}=e^n left(1-e^{frac1n}right)=-frac{e^n}n frac{e^{frac1n}-1}{frac1n}$$
then use standard limits.
answered 2 days ago
gimusi
86.1k74392
add a comment |
up vote
0
down vote
The answer is $-infty$.
The trick is to factor out an $e^{n}$ term, and force L'Hopital's Rule by writing the expression as a fraction. We have
$$lim_{ntoinfty} e^{n} - e^{frac{1}{n} + n} = lim_{ntoinfty} e^{n}left(1 - e^{1/n}right)$$
$$= lim_{ntoinfty} frac{e^{n}left(1 - e^{1/n}right)left(1 + e^{1/n}right)}{(1 + e^{1/n})} $$
$$= lim_{ntoinfty}frac{e^{n} left(1 - e^{2/n}right)}{1 + e^{1/n}} $$
$$= lim_{ntoinfty} frac{e^{n} - e^{n + frac{2}{n}}}{1 + e^{1/n}}.$$
By L'Hopital's Rule, the above expression equals
$$lim_{ntoinfty} frac{e^{n} - e^{n + frac{2}{n}}left(1 - frac{2}{n^{2}}right)}{-frac{1}{n^{2}} cdot e^{1/n}} \[1em] $$
$$= lim_{ntoinfty} frac{-n^{2} left( e^{n} - e^{n + frac{2}{n}}left(1 - frac{2}{n}right)right)}{e^{1/n}}.$$
As $n rightarrow infty$, $e^{1/n}$ approaches $1$, and the numerator clearly approaches $-infty$.
Therefore, the answer is $-infty$.
answered 2 days ago
Ekesh
3715
New contributor
Ekesh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
This is very helpful, thank you! A question though -- why does the numerator approach $-infty$? $$limlimits_{ntoinfty} -n^2left(e^n-e^{n+frac{2}{n}}(1-frac{2}{n})right) = limlimits_{ntoinfty}-n^2(e^n)(1-e^{frac{2}{n}}) = limlimits_{ntoinfty}-n^2 (e^n)(1-1)= 0.$$ I 'm having trouble algebraically getting there
– t.perez
2 days ago
$$lim_{ntoinfty} -n^{2}left(e^{n}right)left(1 - e^{2/n}right) = lim_{ntoinfty} -n^{2} left(e^{n} - e^{2/n + n}right) = ldots$$
– Ekesh
2 days ago
My apologies, but I still don't see it. $limlimits_{ntoinfty}e^n - e^{frac{2}{n} + n}$ is similar to the limit I originally am tying to prove
– t.perez
yesterday
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
HINT
We have that
$${e^n - e^{frac1n + n}}=e^n left(1-e^{frac1n}right)=-frac{e^n}n frac{e^{frac1n}-1}{frac1n}$$
then use standard limits.
answered 2 days ago
gimusi
86.1k74392
add a comment |
up vote
4
down vote
HINT
We have that
$${e^n - e^{frac1n + n}}=e^n left(1-e^{frac1n}right)=-frac{e^n}n frac{e^{frac1n}-1}{frac1n}$$
then use standard limits.
answered 2 days ago
gimusi
86.1k74392
add a comment |
up vote
4
down vote
up vote
4
down vote
HINT
We have that
$${e^n - e^{frac1n + n}}=e^n left(1-e^{frac1n}right)=-frac{e^n}n frac{e^{frac1n}-1}{frac1n}$$
then use standard limits.
answered 2 days ago
gimusi
86.1k74392
HINT
We have that
$${e^n - e^{frac1n + n}}=e^n left(1-e^{frac1n}right)=-frac{e^n}n frac{e^{frac1n}-1}{frac1n}$$
then use standard limits.
answered 2 days ago
gimusi
86.1k74392
answered 2 days ago
gimusi
86.1k74392
answered 2 days ago
gimusi
86.1k74392
answered 2 days ago
gimusi
86.1k74392
86.1k74392
add a comment |
add a comment |
up vote
0
down vote
The answer is $-infty$.
The trick is to factor out an $e^{n}$ term, and force L'Hopital's Rule by writing the expression as a fraction. We have
$$lim_{ntoinfty} e^{n} - e^{frac{1}{n} + n} = lim_{ntoinfty} e^{n}left(1 - e^{1/n}right)$$
$$= lim_{ntoinfty} frac{e^{n}left(1 - e^{1/n}right)left(1 + e^{1/n}right)}{(1 + e^{1/n})} $$
$$= lim_{ntoinfty}frac{e^{n} left(1 - e^{2/n}right)}{1 + e^{1/n}} $$
$$= lim_{ntoinfty} frac{e^{n} - e^{n + frac{2}{n}}}{1 + e^{1/n}}.$$
By L'Hopital's Rule, the above expression equals
$$lim_{ntoinfty} frac{e^{n} - e^{n + frac{2}{n}}left(1 - frac{2}{n^{2}}right)}{-frac{1}{n^{2}} cdot e^{1/n}} \[1em] $$
$$= lim_{ntoinfty} frac{-n^{2} left( e^{n} - e^{n + frac{2}{n}}left(1 - frac{2}{n}right)right)}{e^{1/n}}.$$
As $n rightarrow infty$, $e^{1/n}$ approaches $1$, and the numerator clearly approaches $-infty$.
Therefore, the answer is $-infty$.
answered 2 days ago
Ekesh
3715
New contributor
Ekesh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
This is very helpful, thank you! A question though -- why does the numerator approach $-infty$? $$limlimits_{ntoinfty} -n^2left(e^n-e^{n+frac{2}{n}}(1-frac{2}{n})right) = limlimits_{ntoinfty}-n^2(e^n)(1-e^{frac{2}{n}}) = limlimits_{ntoinfty}-n^2 (e^n)(1-1)= 0.$$ I 'm having trouble algebraically getting there
– t.perez
2 days ago
$$lim_{ntoinfty} -n^{2}left(e^{n}right)left(1 - e^{2/n}right) = lim_{ntoinfty} -n^{2} left(e^{n} - e^{2/n + n}right) = ldots$$
– Ekesh
2 days ago
My apologies, but I still don't see it. $limlimits_{ntoinfty}e^n - e^{frac{2}{n} + n}$ is similar to the limit I originally am tying to prove
– t.perez
yesterday
add a comment |
up vote
0
down vote
The answer is $-infty$.
The trick is to factor out an $e^{n}$ term, and force L'Hopital's Rule by writing the expression as a fraction. We have
$$lim_{ntoinfty} e^{n} - e^{frac{1}{n} + n} = lim_{ntoinfty} e^{n}left(1 - e^{1/n}right)$$
$$= lim_{ntoinfty} frac{e^{n}left(1 - e^{1/n}right)left(1 + e^{1/n}right)}{(1 + e^{1/n})} $$
$$= lim_{ntoinfty}frac{e^{n} left(1 - e^{2/n}right)}{1 + e^{1/n}} $$
$$= lim_{ntoinfty} frac{e^{n} - e^{n + frac{2}{n}}}{1 + e^{1/n}}.$$
By L'Hopital's Rule, the above expression equals
$$lim_{ntoinfty} frac{e^{n} - e^{n + frac{2}{n}}left(1 - frac{2}{n^{2}}right)}{-frac{1}{n^{2}} cdot e^{1/n}} \[1em] $$
$$= lim_{ntoinfty} frac{-n^{2} left( e^{n} - e^{n + frac{2}{n}}left(1 - frac{2}{n}right)right)}{e^{1/n}}.$$
As $n rightarrow infty$, $e^{1/n}$ approaches $1$, and the numerator clearly approaches $-infty$.
Therefore, the answer is $-infty$.
answered 2 days ago
Ekesh
3715
New contributor
Ekesh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
This is very helpful, thank you! A question though -- why does the numerator approach $-infty$? $$limlimits_{ntoinfty} -n^2left(e^n-e^{n+frac{2}{n}}(1-frac{2}{n})right) = limlimits_{ntoinfty}-n^2(e^n)(1-e^{frac{2}{n}}) = limlimits_{ntoinfty}-n^2 (e^n)(1-1)= 0.$$ I 'm having trouble algebraically getting there
– t.perez
2 days ago
$$lim_{ntoinfty} -n^{2}left(e^{n}right)left(1 - e^{2/n}right) = lim_{ntoinfty} -n^{2} left(e^{n} - e^{2/n + n}right) = ldots$$
– Ekesh
2 days ago
My apologies, but I still don't see it. $limlimits_{ntoinfty}e^n - e^{frac{2}{n} + n}$ is similar to the limit I originally am tying to prove
– t.perez
yesterday
add a comment |
up vote
0
down vote
up vote
0
down vote
The answer is $-infty$.
The trick is to factor out an $e^{n}$ term, and force L'Hopital's Rule by writing the expression as a fraction. We have
$$lim_{ntoinfty} e^{n} - e^{frac{1}{n} + n} = lim_{ntoinfty} e^{n}left(1 - e^{1/n}right)$$
$$= lim_{ntoinfty} frac{e^{n}left(1 - e^{1/n}right)left(1 + e^{1/n}right)}{(1 + e^{1/n})} $$
$$= lim_{ntoinfty}frac{e^{n} left(1 - e^{2/n}right)}{1 + e^{1/n}} $$
$$= lim_{ntoinfty} frac{e^{n} - e^{n + frac{2}{n}}}{1 + e^{1/n}}.$$
By L'Hopital's Rule, the above expression equals
$$lim_{ntoinfty} frac{e^{n} - e^{n + frac{2}{n}}left(1 - frac{2}{n^{2}}right)}{-frac{1}{n^{2}} cdot e^{1/n}} \[1em] $$
$$= lim_{ntoinfty} frac{-n^{2} left( e^{n} - e^{n + frac{2}{n}}left(1 - frac{2}{n}right)right)}{e^{1/n}}.$$
As $n rightarrow infty$, $e^{1/n}$ approaches $1$, and the numerator clearly approaches $-infty$.
Therefore, the answer is $-infty$.
answered 2 days ago
Ekesh
3715
New contributor
Ekesh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The answer is $-infty$.
The trick is to factor out an $e^{n}$ term, and force L'Hopital's Rule by writing the expression as a fraction. We have
$$lim_{ntoinfty} e^{n} - e^{frac{1}{n} + n} = lim_{ntoinfty} e^{n}left(1 - e^{1/n}right)$$
$$= lim_{ntoinfty} frac{e^{n}left(1 - e^{1/n}right)left(1 + e^{1/n}right)}{(1 + e^{1/n})} $$
$$= lim_{ntoinfty}frac{e^{n} left(1 - e^{2/n}right)}{1 + e^{1/n}} $$
$$= lim_{ntoinfty} frac{e^{n} - e^{n + frac{2}{n}}}{1 + e^{1/n}}.$$
By L'Hopital's Rule, the above expression equals
$$lim_{ntoinfty} frac{e^{n} - e^{n + frac{2}{n}}left(1 - frac{2}{n^{2}}right)}{-frac{1}{n^{2}} cdot e^{1/n}} \[1em] $$
$$= lim_{ntoinfty} frac{-n^{2} left( e^{n} - e^{n + frac{2}{n}}left(1 - frac{2}{n}right)right)}{e^{1/n}}.$$
As $n rightarrow infty$, $e^{1/n}$ approaches $1$, and the numerator clearly approaches $-infty$.
Therefore, the answer is $-infty$.
answered 2 days ago
Ekesh
3715
New contributor
Ekesh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
Ekesh
3715
New contributor
Ekesh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
Ekesh
3715
answered 2 days ago
Ekesh
3715
3715
New contributor
Ekesh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ekesh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ekesh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
This is very helpful, thank you! A question though -- why does the numerator approach $-infty$? $$limlimits_{ntoinfty} -n^2left(e^n-e^{n+frac{2}{n}}(1-frac{2}{n})right) = limlimits_{ntoinfty}-n^2(e^n)(1-e^{frac{2}{n}}) = limlimits_{ntoinfty}-n^2 (e^n)(1-1)= 0.$$ I 'm having trouble algebraically getting there
– t.perez
2 days ago
$$lim_{ntoinfty} -n^{2}left(e^{n}right)left(1 - e^{2/n}right) = lim_{ntoinfty} -n^{2} left(e^{n} - e^{2/n + n}right) = ldots$$
– Ekesh
2 days ago
My apologies, but I still don't see it. $limlimits_{ntoinfty}e^n - e^{frac{2}{n} + n}$ is similar to the limit I originally am tying to prove
– t.perez
yesterday
add a comment |
This is very helpful, thank you! A question though -- why does the numerator approach $-infty$? $$limlimits_{ntoinfty} -n^2left(e^n-e^{n+frac{2}{n}}(1-frac{2}{n})right) = limlimits_{ntoinfty}-n^2(e^n)(1-e^{frac{2}{n}}) = limlimits_{ntoinfty}-n^2 (e^n)(1-1)= 0.$$ I 'm having trouble algebraically getting there
– t.perez
2 days ago
$$lim_{ntoinfty} -n^{2}left(e^{n}right)left(1 - e^{2/n}right) = lim_{ntoinfty} -n^{2} left(e^{n} - e^{2/n + n}right) = ldots$$
– Ekesh
2 days ago
My apologies, but I still don't see it. $limlimits_{ntoinfty}e^n - e^{frac{2}{n} + n}$ is similar to the limit I originally am tying to prove
– t.perez
yesterday
This is very helpful, thank you! A question though -- why does the numerator approach $-infty$? $$limlimits_{ntoinfty} -n^2left(e^n-e^{n+frac{2}{n}}(1-frac{2}{n})right) = limlimits_{ntoinfty}-n^2(e^n)(1-e^{frac{2}{n}}) = limlimits_{ntoinfty}-n^2 (e^n)(1-1)= 0.$$ I 'm having trouble algebraically getting there
– t.perez
2 days ago
This is very helpful, thank you! A question though -- why does the numerator approach $-infty$? $$limlimits_{ntoinfty} -n^2left(e^n-e^{n+frac{2}{n}}(1-frac{2}{n})right) = limlimits_{ntoinfty}-n^2(e^n)(1-e^{frac{2}{n}}) = limlimits_{ntoinfty}-n^2 (e^n)(1-1)= 0.$$ I 'm having trouble algebraically getting there
– t.perez
2 days ago
$$lim_{ntoinfty} -n^{2}left(e^{n}right)left(1 - e^{2/n}right) = lim_{ntoinfty} -n^{2} left(e^{n} - e^{2/n + n}right) = ldots$$
– Ekesh
2 days ago
$$lim_{ntoinfty} -n^{2}left(e^{n}right)left(1 - e^{2/n}right) = lim_{ntoinfty} -n^{2} left(e^{n} - e^{2/n + n}right) = ldots$$
– Ekesh
2 days ago
My apologies, but I still don't see it. $limlimits_{ntoinfty}e^n - e^{frac{2}{n} + n}$ is similar to the limit I originally am tying to prove
– t.perez
yesterday
My apologies, but I still don't see it. $limlimits_{ntoinfty}e^n - e^{frac{2}{n} + n}$ is similar to the limit I originally am tying to prove
– t.perez
yesterday
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3005568%2ffind-lim-limits-n-to-inftyen-e-frac1n-n%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
