Mixing
Sandwich Theorem not working?
up vote
1
down vote
favorite
This is the limit I need to solve:
$$lim_{n to infty} frac{(4 cos(n) - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4}$$
I simplified it to this:
$$lim_{n to infty} frac{2(4 cos(n) - 3n^2)}{(6n^3 + 5n sin(n))}.$$
At this point I want to use the Sandwich Theorem on the Numerator and Denominator to evaluate the limit.
I use the fact that $lim_{n to infty} frac{a}{b} = frac{lim_{n to infty} a}{lim_{n to infty} b}$ when $bne 0$.
By the Sandwich Theorem both the Numerator and Denominator is $infty$.
Hence the answer is 1.
But if I calculate the limit whole without splitting it into two I get $frac{3}{2}$. Which answer is correct? Please Help!
calculus limits limits-without-lhopital
edited 19 hours ago
Robert Z
89.8k1056128
asked 20 hours ago
Leon Vladimirov
113
New contributor
Leon Vladimirov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
|
show 2 more comments
up vote
1
down vote
favorite
This is the limit I need to solve:
$$lim_{n to infty} frac{(4 cos(n) - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4}$$
I simplified it to this:
$$lim_{n to infty} frac{2(4 cos(n) - 3n^2)}{(6n^3 + 5n sin(n))}.$$
At this point I want to use the Sandwich Theorem on the Numerator and Denominator to evaluate the limit.
I use the fact that $lim_{n to infty} frac{a}{b} = frac{lim_{n to infty} a}{lim_{n to infty} b}$ when $bne 0$.
By the Sandwich Theorem both the Numerator and Denominator is $infty$.
Hence the answer is 1.
But if I calculate the limit whole without splitting it into two I get $frac{3}{2}$. Which answer is correct? Please Help!
calculus limits limits-without-lhopital
edited 19 hours ago
Robert Z
89.8k1056128
asked 20 hours ago
Leon Vladimirov
113
New contributor
Leon Vladimirov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
5
You cannot 'divide' infinity by infinity. It is not defined.
– thedilated
20 hours ago
Thank you! I completely missed that I did that! The fraction is definitely undefined. So we need to apply the Theorem to the whole expression.
– Leon Vladimirov
20 hours ago
2
Your "simplification" is also not correct. You "trade" $n^5$ for $n^4$.
– maxmilgram
20 hours ago
I'm sorry. I just noticed that I wrote the wrong LaTeX formula. It should actually be: $lim_{n to infty} frac{(4 cos(n) - 3n)^2(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4)}$
– Leon Vladimirov
19 hours ago
And hence the simplification: $lim_{n to infty} frac{2(4 cos(n) - 3n)^2}{(6n^3 + 5n sin(n))}$
– Leon Vladimirov
19 hours ago
|
show 2 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This is the limit I need to solve:
$$lim_{n to infty} frac{(4 cos(n) - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4}$$
I simplified it to this:
$$lim_{n to infty} frac{2(4 cos(n) - 3n^2)}{(6n^3 + 5n sin(n))}.$$
At this point I want to use the Sandwich Theorem on the Numerator and Denominator to evaluate the limit.
I use the fact that $lim_{n to infty} frac{a}{b} = frac{lim_{n to infty} a}{lim_{n to infty} b}$ when $bne 0$.
By the Sandwich Theorem both the Numerator and Denominator is $infty$.
Hence the answer is 1.
But if I calculate the limit whole without splitting it into two I get $frac{3}{2}$. Which answer is correct? Please Help!
calculus limits limits-without-lhopital
edited 19 hours ago
Robert Z
89.8k1056128
asked 20 hours ago
Leon Vladimirov
113
New contributor
Leon Vladimirov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
This is the limit I need to solve:
$$lim_{n to infty} frac{(4 cos(n) - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4}$$
I simplified it to this:
$$lim_{n to infty} frac{2(4 cos(n) - 3n^2)}{(6n^3 + 5n sin(n))}.$$
At this point I want to use the Sandwich Theorem on the Numerator and Denominator to evaluate the limit.
I use the fact that $lim_{n to infty} frac{a}{b} = frac{lim_{n to infty} a}{lim_{n to infty} b}$ when $bne 0$.
By the Sandwich Theorem both the Numerator and Denominator is $infty$.
Hence the answer is 1.
But if I calculate the limit whole without splitting it into two I get $frac{3}{2}$. Which answer is correct? Please Help!
calculus limits limits-without-lhopital
calculus limits limits-without-lhopital
edited 19 hours ago
Robert Z
89.8k1056128
asked 20 hours ago
Leon Vladimirov
113
New contributor
Leon Vladimirov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 19 hours ago
Robert Z
89.8k1056128
asked 20 hours ago
Leon Vladimirov
113
New contributor
Leon Vladimirov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 19 hours ago
Robert Z
89.8k1056128
edited 19 hours ago
Robert Z
89.8k1056128
edited 19 hours ago
Robert Z
89.8k1056128
89.8k1056128
asked 20 hours ago
Leon Vladimirov
113
New contributor
Leon Vladimirov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 20 hours ago
Leon Vladimirov
113
asked 20 hours ago
Leon Vladimirov
113
113
New contributor
Leon Vladimirov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Leon Vladimirov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Leon Vladimirov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
5
You cannot 'divide' infinity by infinity. It is not defined.
– thedilated
20 hours ago
Thank you! I completely missed that I did that! The fraction is definitely undefined. So we need to apply the Theorem to the whole expression.
– Leon Vladimirov
20 hours ago
2
Your "simplification" is also not correct. You "trade" $n^5$ for $n^4$.
– maxmilgram
20 hours ago
I'm sorry. I just noticed that I wrote the wrong LaTeX formula. It should actually be: $lim_{n to infty} frac{(4 cos(n) - 3n)^2(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4)}$
– Leon Vladimirov
19 hours ago
And hence the simplification: $lim_{n to infty} frac{2(4 cos(n) - 3n)^2}{(6n^3 + 5n sin(n))}$
– Leon Vladimirov
19 hours ago
|
show 2 more comments
5
You cannot 'divide' infinity by infinity. It is not defined.
– thedilated
20 hours ago
Thank you! I completely missed that I did that! The fraction is definitely undefined. So we need to apply the Theorem to the whole expression.
– Leon Vladimirov
20 hours ago
2
Your "simplification" is also not correct. You "trade" $n^5$ for $n^4$.
– maxmilgram
20 hours ago
I'm sorry. I just noticed that I wrote the wrong LaTeX formula. It should actually be: $lim_{n to infty} frac{(4 cos(n) - 3n)^2(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4)}$
– Leon Vladimirov
19 hours ago
And hence the simplification: $lim_{n to infty} frac{2(4 cos(n) - 3n)^2}{(6n^3 + 5n sin(n))}$
– Leon Vladimirov
19 hours ago
5
5
You cannot 'divide' infinity by infinity. It is not defined.
– thedilated
20 hours ago
You cannot 'divide' infinity by infinity. It is not defined.
– thedilated
20 hours ago
Thank you! I completely missed that I did that! The fraction is definitely undefined. So we need to apply the Theorem to the whole expression.
– Leon Vladimirov
20 hours ago
Thank you! I completely missed that I did that! The fraction is definitely undefined. So we need to apply the Theorem to the whole expression.
– Leon Vladimirov
20 hours ago
2
2
Your "simplification" is also not correct. You "trade" $n^5$ for $n^4$.
– maxmilgram
20 hours ago
Your "simplification" is also not correct. You "trade" $n^5$ for $n^4$.
– maxmilgram
20 hours ago
I'm sorry. I just noticed that I wrote the wrong LaTeX formula. It should actually be: $lim_{n to infty} frac{(4 cos(n) - 3n)^2(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4)}$
– Leon Vladimirov
19 hours ago
I'm sorry. I just noticed that I wrote the wrong LaTeX formula. It should actually be: $lim_{n to infty} frac{(4 cos(n) - 3n)^2(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4)}$
– Leon Vladimirov
19 hours ago
And hence the simplification: $lim_{n to infty} frac{2(4 cos(n) - 3n)^2}{(6n^3 + 5n sin(n))}$
– Leon Vladimirov
19 hours ago
And hence the simplification: $lim_{n to infty} frac{2(4 cos(n) - 3n)^2}{(6n^3 + 5n sin(n))}$
– Leon Vladimirov
19 hours ago
|
show 2 more comments
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
You should revise your work. My advice is to apply the Sandwich Theorem in a different way.
Note that the given limit can be written as
$$lim_{n to infty} frac{n^2cdot (frac{4 cos(n)}{n^2} - 3)cdot n^5cdot(2 - frac{1}{n^2} + frac{1}{n^5})}{n^3cdot (6 + frac{5sin(n)}{n^2})cdot n^4cdot (1 + frac{2}{n})^4}$$
Simplify the powers of $n$ and recall that, just by the Sandwich Theorem, if $a_nto 0$ and $b_n$ is bounded then $lim_{nto infty}(a_ncdot b_n)=0$.
What is the final answer?
answered 20 hours ago
Robert Z
89.8k1056128
Thank you for the answer! So if we just get rid of the infitesimals the answer is $frac{-6}{6} = -1$?
– Leon Vladimirov
19 hours ago
Yes, that's it! Factoring out the main powers of $n$ is the key point.
– Robert Z
19 hours ago
Thank you! I now understand.
– Leon Vladimirov
19 hours ago
@LeonVladimirov Thanks for appreciating.
– Robert Z
19 hours ago
I'm terribly sorry. I just noticed that I wrote the wrong LaTeX formula. The limit I actually have is $lim_{n to infty} frac{(4 cos(n) - 3n)^2(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4)}$. Would the answer in this case be $frac{3}{2}$?
– Leon Vladimirov
19 hours ago
|
show 4 more comments
up vote
0
down vote
We have that
$$frac{(-4 - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4)}le frac{(4 cos(n) - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4)}le frac{(4 - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4)}$$
and we can conclude by squeeze theorem since for both bounds
$$frac{(pm4 - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4)}sim frac{-6n^7}{6n^7} = -1$$
as already noticed by RobertZ, in a simpler way, we can directly use the same argument for the original limit.
answered 19 hours ago
gimusi
85.5k74294
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
You should revise your work. My advice is to apply the Sandwich Theorem in a different way.
Note that the given limit can be written as
$$lim_{n to infty} frac{n^2cdot (frac{4 cos(n)}{n^2} - 3)cdot n^5cdot(2 - frac{1}{n^2} + frac{1}{n^5})}{n^3cdot (6 + frac{5sin(n)}{n^2})cdot n^4cdot (1 + frac{2}{n})^4}$$
Simplify the powers of $n$ and recall that, just by the Sandwich Theorem, if $a_nto 0$ and $b_n$ is bounded then $lim_{nto infty}(a_ncdot b_n)=0$.
What is the final answer?
answered 20 hours ago
Robert Z
89.8k1056128
Thank you for the answer! So if we just get rid of the infitesimals the answer is $frac{-6}{6} = -1$?
– Leon Vladimirov
19 hours ago
Yes, that's it! Factoring out the main powers of $n$ is the key point.
– Robert Z
19 hours ago
Thank you! I now understand.
– Leon Vladimirov
19 hours ago
@LeonVladimirov Thanks for appreciating.
– Robert Z
19 hours ago
I'm terribly sorry. I just noticed that I wrote the wrong LaTeX formula. The limit I actually have is $lim_{n to infty} frac{(4 cos(n) - 3n)^2(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4)}$. Would the answer in this case be $frac{3}{2}$?
– Leon Vladimirov
19 hours ago
|
show 4 more comments
up vote
1
down vote
accepted
You should revise your work. My advice is to apply the Sandwich Theorem in a different way.
Note that the given limit can be written as
$$lim_{n to infty} frac{n^2cdot (frac{4 cos(n)}{n^2} - 3)cdot n^5cdot(2 - frac{1}{n^2} + frac{1}{n^5})}{n^3cdot (6 + frac{5sin(n)}{n^2})cdot n^4cdot (1 + frac{2}{n})^4}$$
Simplify the powers of $n$ and recall that, just by the Sandwich Theorem, if $a_nto 0$ and $b_n$ is bounded then $lim_{nto infty}(a_ncdot b_n)=0$.
What is the final answer?
answered 20 hours ago
Robert Z
89.8k1056128
Thank you for the answer! So if we just get rid of the infitesimals the answer is $frac{-6}{6} = -1$?
– Leon Vladimirov
19 hours ago
Yes, that's it! Factoring out the main powers of $n$ is the key point.
– Robert Z
19 hours ago
Thank you! I now understand.
– Leon Vladimirov
19 hours ago
@LeonVladimirov Thanks for appreciating.
– Robert Z
19 hours ago
I'm terribly sorry. I just noticed that I wrote the wrong LaTeX formula. The limit I actually have is $lim_{n to infty} frac{(4 cos(n) - 3n)^2(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4)}$. Would the answer in this case be $frac{3}{2}$?
– Leon Vladimirov
19 hours ago
|
show 4 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You should revise your work. My advice is to apply the Sandwich Theorem in a different way.
Note that the given limit can be written as
$$lim_{n to infty} frac{n^2cdot (frac{4 cos(n)}{n^2} - 3)cdot n^5cdot(2 - frac{1}{n^2} + frac{1}{n^5})}{n^3cdot (6 + frac{5sin(n)}{n^2})cdot n^4cdot (1 + frac{2}{n})^4}$$
Simplify the powers of $n$ and recall that, just by the Sandwich Theorem, if $a_nto 0$ and $b_n$ is bounded then $lim_{nto infty}(a_ncdot b_n)=0$.
What is the final answer?
answered 20 hours ago
Robert Z
89.8k1056128
You should revise your work. My advice is to apply the Sandwich Theorem in a different way.
Note that the given limit can be written as
$$lim_{n to infty} frac{n^2cdot (frac{4 cos(n)}{n^2} - 3)cdot n^5cdot(2 - frac{1}{n^2} + frac{1}{n^5})}{n^3cdot (6 + frac{5sin(n)}{n^2})cdot n^4cdot (1 + frac{2}{n})^4}$$
Simplify the powers of $n$ and recall that, just by the Sandwich Theorem, if $a_nto 0$ and $b_n$ is bounded then $lim_{nto infty}(a_ncdot b_n)=0$.
What is the final answer?
answered 20 hours ago
Robert Z
89.8k1056128
edited 19 hours ago
answered 20 hours ago
Robert Z
89.8k1056128
answered 20 hours ago
Robert Z
89.8k1056128
answered 20 hours ago
Robert Z
89.8k1056128
89.8k1056128
Thank you for the answer! So if we just get rid of the infitesimals the answer is $frac{-6}{6} = -1$?
– Leon Vladimirov
19 hours ago
Yes, that's it! Factoring out the main powers of $n$ is the key point.
– Robert Z
19 hours ago
Thank you! I now understand.
– Leon Vladimirov
19 hours ago
@LeonVladimirov Thanks for appreciating.
– Robert Z
19 hours ago
I'm terribly sorry. I just noticed that I wrote the wrong LaTeX formula. The limit I actually have is $lim_{n to infty} frac{(4 cos(n) - 3n)^2(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4)}$. Would the answer in this case be $frac{3}{2}$?
– Leon Vladimirov
19 hours ago
|
show 4 more comments
Thank you for the answer! So if we just get rid of the infitesimals the answer is $frac{-6}{6} = -1$?
– Leon Vladimirov
19 hours ago
Yes, that's it! Factoring out the main powers of $n$ is the key point.
– Robert Z
19 hours ago
Thank you! I now understand.
– Leon Vladimirov
19 hours ago
@LeonVladimirov Thanks for appreciating.
– Robert Z
19 hours ago
I'm terribly sorry. I just noticed that I wrote the wrong LaTeX formula. The limit I actually have is $lim_{n to infty} frac{(4 cos(n) - 3n)^2(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4)}$. Would the answer in this case be $frac{3}{2}$?
– Leon Vladimirov
19 hours ago
Thank you for the answer! So if we just get rid of the infitesimals the answer is $frac{-6}{6} = -1$?
– Leon Vladimirov
19 hours ago
Thank you for the answer! So if we just get rid of the infitesimals the answer is $frac{-6}{6} = -1$?
– Leon Vladimirov
19 hours ago
Yes, that's it! Factoring out the main powers of $n$ is the key point.
– Robert Z
19 hours ago
Yes, that's it! Factoring out the main powers of $n$ is the key point.
– Robert Z
19 hours ago
Thank you! I now understand.
– Leon Vladimirov
19 hours ago
Thank you! I now understand.
– Leon Vladimirov
19 hours ago
@LeonVladimirov Thanks for appreciating.
– Robert Z
19 hours ago
@LeonVladimirov Thanks for appreciating.
– Robert Z
19 hours ago
I'm terribly sorry. I just noticed that I wrote the wrong LaTeX formula. The limit I actually have is $lim_{n to infty} frac{(4 cos(n) - 3n)^2(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4)}$. Would the answer in this case be $frac{3}{2}$?
– Leon Vladimirov
19 hours ago
I'm terribly sorry. I just noticed that I wrote the wrong LaTeX formula. The limit I actually have is $lim_{n to infty} frac{(4 cos(n) - 3n)^2(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4)}$. Would the answer in this case be $frac{3}{2}$?
– Leon Vladimirov
19 hours ago
|
show 4 more comments
up vote
0
down vote
We have that
$$frac{(-4 - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4)}le frac{(4 cos(n) - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4)}le frac{(4 - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4)}$$
and we can conclude by squeeze theorem since for both bounds
$$frac{(pm4 - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4)}sim frac{-6n^7}{6n^7} = -1$$
as already noticed by RobertZ, in a simpler way, we can directly use the same argument for the original limit.
answered 19 hours ago
gimusi
85.5k74294
add a comment |
up vote
0
down vote
We have that
$$frac{(-4 - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4)}le frac{(4 cos(n) - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4)}le frac{(4 - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4)}$$
and we can conclude by squeeze theorem since for both bounds
$$frac{(pm4 - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4)}sim frac{-6n^7}{6n^7} = -1$$
as already noticed by RobertZ, in a simpler way, we can directly use the same argument for the original limit.
answered 19 hours ago
gimusi
85.5k74294
add a comment |
up vote
0
down vote
up vote
0
down vote
We have that
$$frac{(-4 - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4)}le frac{(4 cos(n) - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4)}le frac{(4 - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4)}$$
and we can conclude by squeeze theorem since for both bounds
$$frac{(pm4 - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4)}sim frac{-6n^7}{6n^7} = -1$$
as already noticed by RobertZ, in a simpler way, we can directly use the same argument for the original limit.
answered 19 hours ago
gimusi
85.5k74294
We have that
$$frac{(-4 - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4)}le frac{(4 cos(n) - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4)}le frac{(4 - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4)}$$
and we can conclude by squeeze theorem since for both bounds
$$frac{(pm4 - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4)}sim frac{-6n^7}{6n^7} = -1$$
as already noticed by RobertZ, in a simpler way, we can directly use the same argument for the original limit.
answered 19 hours ago
gimusi
85.5k74294
answered 19 hours ago
gimusi
85.5k74294
answered 19 hours ago
gimusi
85.5k74294
answered 19 hours ago
gimusi
85.5k74294
85.5k74294
add a comment |
add a comment |
Leon Vladimirov is a new contributor. Be nice, and check out our Code of Conduct.
Leon Vladimirov is a new contributor. Be nice, and check out our Code of Conduct.
Leon Vladimirov is a new contributor. Be nice, and check out our Code of Conduct.
Leon Vladimirov is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3004776%2fsandwich-theorem-not-working%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
5
You cannot 'divide' infinity by infinity. It is not defined.
– thedilated
20 hours ago
Thank you! I completely missed that I did that! The fraction is definitely undefined. So we need to apply the Theorem to the whole expression.
– Leon Vladimirov
20 hours ago
2
Your "simplification" is also not correct. You "trade" $n^5$ for $n^4$.
– maxmilgram
20 hours ago
I'm sorry. I just noticed that I wrote the wrong LaTeX formula. It should actually be: $lim_{n to infty} frac{(4 cos(n) - 3n)^2(2n^5 - n^3 + 1)}{(6n^3 + 5n sin(n))(n + 2)^4)}$
– Leon Vladimirov
19 hours ago
And hence the simplification: $lim_{n to infty} frac{2(4 cos(n) - 3n)^2}{(6n^3 + 5n sin(n))}$
– Leon Vladimirov
19 hours ago