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Quick question about bound on certain function while computing limit
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I'm given the limit: $$lim_ntoinftysqrt[n]n^2+n^3+n^4+2^n+3^n+4^n$$
The solution says:
Because exponentials grow faster than polynomials, there is some $n_0$ such that $forall ninBbbN,ngeq n_0:n^2leq n^3leq n^4 leq 2^n leq 3^n leq 4^n$ so that $$forall ninBbbN,ngeq n_0:4leqsqrt[n]n^2+n^3+n^4+2^n+3^n+4^nleqsqrt[n]6cdot4^n=4sqrt[n]6$$ and by squeeze lemma it follows that the limit is $4$. I'm quite confused, where did we get the bound that the function is always greater than $4$ i.e. the first inequality.
limits
edited Jul 22 at 13:19
Bernard
110k635103
asked Jul 22 at 13:11
Michal Dvořák
48912
add a comment |Â
up vote
1
down vote
favorite
I'm given the limit: $$lim_ntoinftysqrt[n]n^2+n^3+n^4+2^n+3^n+4^n$$
The solution says:
Because exponentials grow faster than polynomials, there is some $n_0$ such that $forall ninBbbN,ngeq n_0:n^2leq n^3leq n^4 leq 2^n leq 3^n leq 4^n$ so that $$forall ninBbbN,ngeq n_0:4leqsqrt[n]n^2+n^3+n^4+2^n+3^n+4^nleqsqrt[n]6cdot4^n=4sqrt[n]6$$ and by squeeze lemma it follows that the limit is $4$. I'm quite confused, where did we get the bound that the function is always greater than $4$ i.e. the first inequality.
limits
edited Jul 22 at 13:19
Bernard
110k635103
asked Jul 22 at 13:11
Michal Dvořák
48912
4
Each of the terms before $4^n$ are positive. You get something smaller if you replace them all by $0$. When you do, you get $sqrt[n]0+0+0+0+0+4^n=4$.
– user574889
Jul 22 at 13:12
@cactus. You can copy my text and post it as your answer if you are so inclined. I would think that your comment should be the answer to this question.
– Mason
Jul 22 at 15:22
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm given the limit: $$lim_ntoinftysqrt[n]n^2+n^3+n^4+2^n+3^n+4^n$$
The solution says:
Because exponentials grow faster than polynomials, there is some $n_0$ such that $forall ninBbbN,ngeq n_0:n^2leq n^3leq n^4 leq 2^n leq 3^n leq 4^n$ so that $$forall ninBbbN,ngeq n_0:4leqsqrt[n]n^2+n^3+n^4+2^n+3^n+4^nleqsqrt[n]6cdot4^n=4sqrt[n]6$$ and by squeeze lemma it follows that the limit is $4$. I'm quite confused, where did we get the bound that the function is always greater than $4$ i.e. the first inequality.
limits
edited Jul 22 at 13:19
Bernard
110k635103
asked Jul 22 at 13:11
Michal Dvořák
48912
I'm given the limit: $$lim_ntoinftysqrt[n]n^2+n^3+n^4+2^n+3^n+4^n$$
The solution says:
Because exponentials grow faster than polynomials, there is some $n_0$ such that $forall ninBbbN,ngeq n_0:n^2leq n^3leq n^4 leq 2^n leq 3^n leq 4^n$ so that $$forall ninBbbN,ngeq n_0:4leqsqrt[n]n^2+n^3+n^4+2^n+3^n+4^nleqsqrt[n]6cdot4^n=4sqrt[n]6$$ and by squeeze lemma it follows that the limit is $4$. I'm quite confused, where did we get the bound that the function is always greater than $4$ i.e. the first inequality.
limits
edited Jul 22 at 13:19
Bernard
110k635103
asked Jul 22 at 13:11
Michal Dvořák
48912
edited Jul 22 at 13:19
Bernard
110k635103
edited Jul 22 at 13:19
Bernard
110k635103
edited Jul 22 at 13:19
Bernard
110k635103
110k635103
asked Jul 22 at 13:11
Michal Dvořák
48912
asked Jul 22 at 13:11
Michal Dvořák
48912
asked Jul 22 at 13:11
Michal Dvořák
48912
48912
4
Each of the terms before $4^n$ are positive. You get something smaller if you replace them all by $0$. When you do, you get $sqrt[n]0+0+0+0+0+4^n=4$.
– user574889
Jul 22 at 13:12
@cactus. You can copy my text and post it as your answer if you are so inclined. I would think that your comment should be the answer to this question.
– Mason
Jul 22 at 15:22
add a comment |Â
4
Each of the terms before $4^n$ are positive. You get something smaller if you replace them all by $0$. When you do, you get $sqrt[n]0+0+0+0+0+4^n=4$.
– user574889
Jul 22 at 13:12
@cactus. You can copy my text and post it as your answer if you are so inclined. I would think that your comment should be the answer to this question.
– Mason
Jul 22 at 15:22
4
4
Each of the terms before $4^n$ are positive. You get something smaller if you replace them all by $0$. When you do, you get $sqrt[n]0+0+0+0+0+4^n=4$.
– user574889
Jul 22 at 13:12
Each of the terms before $4^n$ are positive. You get something smaller if you replace them all by $0$. When you do, you get $sqrt[n]0+0+0+0+0+4^n=4$.
– user574889
Jul 22 at 13:12
@cactus. You can copy my text and post it as your answer if you are so inclined. I would think that your comment should be the answer to this question.
– Mason
Jul 22 at 15:22
@cactus. You can copy my text and post it as your answer if you are so inclined. I would think that your comment should be the answer to this question.
– Mason
Jul 22 at 15:22
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
As is written in the comments we can see this through some straightforward replacements:
$$4$$
$$=sqrt[n]0+0+0+0+0+4^n$$
$$leqsqrt[n]n^2+n^3+n^4+2^n+3^n+4^n$$
$$leqsqrt[n]4^n+4^n+4^n+4^n+4^n+4^n$$
$$=sqrt[n]6cdot4^n$$
$$=4sqrt[n]6$$
Where for the last inequality we need $n>3$, since $n^4leq 4^n$, fails for some small values of $n$, for $n=3$.
answered Jul 22 at 15:20
Mason
1,2071224
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
As is written in the comments we can see this through some straightforward replacements:
$$4$$
$$=sqrt[n]0+0+0+0+0+4^n$$
$$leqsqrt[n]n^2+n^3+n^4+2^n+3^n+4^n$$
$$leqsqrt[n]4^n+4^n+4^n+4^n+4^n+4^n$$
$$=sqrt[n]6cdot4^n$$
$$=4sqrt[n]6$$
Where for the last inequality we need $n>3$, since $n^4leq 4^n$, fails for some small values of $n$, for $n=3$.
answered Jul 22 at 15:20
Mason
1,2071224
add a comment |Â
up vote
1
down vote
accepted
As is written in the comments we can see this through some straightforward replacements:
$$4$$
$$=sqrt[n]0+0+0+0+0+4^n$$
$$leqsqrt[n]n^2+n^3+n^4+2^n+3^n+4^n$$
$$leqsqrt[n]4^n+4^n+4^n+4^n+4^n+4^n$$
$$=sqrt[n]6cdot4^n$$
$$=4sqrt[n]6$$
Where for the last inequality we need $n>3$, since $n^4leq 4^n$, fails for some small values of $n$, for $n=3$.
answered Jul 22 at 15:20
Mason
1,2071224
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
As is written in the comments we can see this through some straightforward replacements:
$$4$$
$$=sqrt[n]0+0+0+0+0+4^n$$
$$leqsqrt[n]n^2+n^3+n^4+2^n+3^n+4^n$$
$$leqsqrt[n]4^n+4^n+4^n+4^n+4^n+4^n$$
$$=sqrt[n]6cdot4^n$$
$$=4sqrt[n]6$$
Where for the last inequality we need $n>3$, since $n^4leq 4^n$, fails for some small values of $n$, for $n=3$.
answered Jul 22 at 15:20
Mason
1,2071224
As is written in the comments we can see this through some straightforward replacements:
$$4$$
$$=sqrt[n]0+0+0+0+0+4^n$$
$$leqsqrt[n]n^2+n^3+n^4+2^n+3^n+4^n$$
$$leqsqrt[n]4^n+4^n+4^n+4^n+4^n+4^n$$
$$=sqrt[n]6cdot4^n$$
$$=4sqrt[n]6$$
Where for the last inequality we need $n>3$, since $n^4leq 4^n$, fails for some small values of $n$, for $n=3$.
answered Jul 22 at 15:20
Mason
1,2071224
edited Jul 22 at 15:32
user574889
answered Jul 22 at 15:20
Mason
1,2071224
answered Jul 22 at 15:20
Mason
1,2071224
answered Jul 22 at 15:20
Mason
1,2071224
1,2071224
add a comment |Â
add a comment |Â
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4
Each of the terms before $4^n$ are positive. You get something smaller if you replace them all by $0$. When you do, you get $sqrt[n]0+0+0+0+0+4^n=4$.
– user574889
Jul 22 at 13:12
@cactus. You can copy my text and post it as your answer if you are so inclined. I would think that your comment should be the answer to this question.
– Mason
Jul 22 at 15:22