Sandwich Theorem not working?
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1
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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
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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
5
You cannot 'divide' infinity by infinity. It is not defined.
– thedilated
Nov 19 at 10:48
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
Nov 19 at 10:56
2
Your "simplification" is also not correct. You "trade" $n^5$ for $n^4$.
– maxmilgram
Nov 19 at 10:59
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
Nov 19 at 11:16
And hence the simplification: $lim_{n to infty} frac{2(4 cos(n) - 3n)^2}{(6n^3 + 5n sin(n))}$
– Leon Vladimirov
Nov 19 at 11:17
|
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
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 Nov 19 at 11:16
Robert Z
91.2k1058129
91.2k1058129
asked Nov 19 at 10:47
Leon Vladimirov
113
113
5
You cannot 'divide' infinity by infinity. It is not defined.
– thedilated
Nov 19 at 10:48
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
Nov 19 at 10:56
2
Your "simplification" is also not correct. You "trade" $n^5$ for $n^4$.
– maxmilgram
Nov 19 at 10:59
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
Nov 19 at 11:16
And hence the simplification: $lim_{n to infty} frac{2(4 cos(n) - 3n)^2}{(6n^3 + 5n sin(n))}$
– Leon Vladimirov
Nov 19 at 11:17
|
show 2 more comments
5
You cannot 'divide' infinity by infinity. It is not defined.
– thedilated
Nov 19 at 10:48
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
Nov 19 at 10:56
2
Your "simplification" is also not correct. You "trade" $n^5$ for $n^4$.
– maxmilgram
Nov 19 at 10:59
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
Nov 19 at 11:16
And hence the simplification: $lim_{n to infty} frac{2(4 cos(n) - 3n)^2}{(6n^3 + 5n sin(n))}$
– Leon Vladimirov
Nov 19 at 11:17
5
5
You cannot 'divide' infinity by infinity. It is not defined.
– thedilated
Nov 19 at 10:48
You cannot 'divide' infinity by infinity. It is not defined.
– thedilated
Nov 19 at 10:48
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
Nov 19 at 10:56
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
Nov 19 at 10:56
2
2
Your "simplification" is also not correct. You "trade" $n^5$ for $n^4$.
– maxmilgram
Nov 19 at 10:59
Your "simplification" is also not correct. You "trade" $n^5$ for $n^4$.
– maxmilgram
Nov 19 at 10:59
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
Nov 19 at 11:16
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
Nov 19 at 11:16
And hence the simplification: $lim_{n to infty} frac{2(4 cos(n) - 3n)^2}{(6n^3 + 5n sin(n))}$
– Leon Vladimirov
Nov 19 at 11:17
And hence the simplification: $lim_{n to infty} frac{2(4 cos(n) - 3n)^2}{(6n^3 + 5n sin(n))}$
– Leon Vladimirov
Nov 19 at 11:17
|
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?
Thank you for the answer! So if we just get rid of the infitesimals the answer is $frac{-6}{6} = -1$?
– Leon Vladimirov
Nov 19 at 11:01
Yes, that's it! Factoring out the main powers of $n$ is the key point.
– Robert Z
Nov 19 at 11:02
Thank you! I now understand.
– Leon Vladimirov
Nov 19 at 11:08
@LeonVladimirov Thanks for appreciating.
– Robert Z
Nov 19 at 11:14
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
Nov 19 at 11:20
|
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.
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?
Thank you for the answer! So if we just get rid of the infitesimals the answer is $frac{-6}{6} = -1$?
– Leon Vladimirov
Nov 19 at 11:01
Yes, that's it! Factoring out the main powers of $n$ is the key point.
– Robert Z
Nov 19 at 11:02
Thank you! I now understand.
– Leon Vladimirov
Nov 19 at 11:08
@LeonVladimirov Thanks for appreciating.
– Robert Z
Nov 19 at 11:14
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
Nov 19 at 11:20
|
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?
Thank you for the answer! So if we just get rid of the infitesimals the answer is $frac{-6}{6} = -1$?
– Leon Vladimirov
Nov 19 at 11:01
Yes, that's it! Factoring out the main powers of $n$ is the key point.
– Robert Z
Nov 19 at 11:02
Thank you! I now understand.
– Leon Vladimirov
Nov 19 at 11:08
@LeonVladimirov Thanks for appreciating.
– Robert Z
Nov 19 at 11:14
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
Nov 19 at 11:20
|
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?
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?
edited Nov 19 at 11:15
answered Nov 19 at 10:56
Robert Z
91.2k1058129
91.2k1058129
Thank you for the answer! So if we just get rid of the infitesimals the answer is $frac{-6}{6} = -1$?
– Leon Vladimirov
Nov 19 at 11:01
Yes, that's it! Factoring out the main powers of $n$ is the key point.
– Robert Z
Nov 19 at 11:02
Thank you! I now understand.
– Leon Vladimirov
Nov 19 at 11:08
@LeonVladimirov Thanks for appreciating.
– Robert Z
Nov 19 at 11:14
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
Nov 19 at 11:20
|
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
Nov 19 at 11:01
Yes, that's it! Factoring out the main powers of $n$ is the key point.
– Robert Z
Nov 19 at 11:02
Thank you! I now understand.
– Leon Vladimirov
Nov 19 at 11:08
@LeonVladimirov Thanks for appreciating.
– Robert Z
Nov 19 at 11:14
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
Nov 19 at 11:20
Thank you for the answer! So if we just get rid of the infitesimals the answer is $frac{-6}{6} = -1$?
– Leon Vladimirov
Nov 19 at 11:01
Thank you for the answer! So if we just get rid of the infitesimals the answer is $frac{-6}{6} = -1$?
– Leon Vladimirov
Nov 19 at 11:01
Yes, that's it! Factoring out the main powers of $n$ is the key point.
– Robert Z
Nov 19 at 11:02
Yes, that's it! Factoring out the main powers of $n$ is the key point.
– Robert Z
Nov 19 at 11:02
Thank you! I now understand.
– Leon Vladimirov
Nov 19 at 11:08
Thank you! I now understand.
– Leon Vladimirov
Nov 19 at 11:08
@LeonVladimirov Thanks for appreciating.
– Robert Z
Nov 19 at 11:14
@LeonVladimirov Thanks for appreciating.
– Robert Z
Nov 19 at 11:14
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
Nov 19 at 11:20
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
Nov 19 at 11:20
|
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.
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.
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.
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 Nov 19 at 11:05
gimusi
88.8k74394
88.8k74394
add a comment |
add a comment |
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5
You cannot 'divide' infinity by infinity. It is not defined.
– thedilated
Nov 19 at 10:48
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
Nov 19 at 10:56
2
Your "simplification" is also not correct. You "trade" $n^5$ for $n^4$.
– maxmilgram
Nov 19 at 10:59
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
Nov 19 at 11:16
And hence the simplification: $lim_{n to infty} frac{2(4 cos(n) - 3n)^2}{(6n^3 + 5n sin(n))}$
– Leon Vladimirov
Nov 19 at 11:17