Proving of limits
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- Let $a>0$ and $n$ be a positive integer. Prove that $$lim_xto a x^1/n = a^1/n$$
For this question, I'm not entirely sure whether I should proceed with the $epsilon$-$delta$ definition of a limit, or simply prove the power law for limits, since the question did not specify the need to use the precise definition of a limit in this proof. I'm thinking that I could try to prove that the function is continuous at $x=a$, but proving continuity requires $lim_xto a x^1/n = a^1/n$. Could someone hint me in the right direction for this?
Let $a$ be a real number. Prove that
(a) $lim_xto a^- frac1(x-a)^3 = -infty$ and
(b)$lim_xto a^+ frac1(x-a)^3 = infty$
My attempt at the second question is as follows:
(a) We want to prove $forall N<0$, $exists delta >0$ such that if $a - delta < x < a$, then $frac1(x-a)^3 < N$. $$frac1(x-a)^3<N$$ $$frac1N > (x-a)^3$$ so $x-a > sqrt[3]frac1N$, i.e. $x > sqrt[3]frac1N$, so we choose $delta = -sqrt[3]frac1N$. Note that $sqrt[3]frac1N + a < x < a$, so $frac1(x-a)^3 < N$ (Q.E.D)
(b) Similarly, we want to prove that $forall M > 0$, $exists delta >0$ such that if $a<x<a+delta$ then $frac1(x-a)^3 > M$. $$frac1M > (x-a)^3$$ $$x < sqrt[3]frac1M + a$$, so we choose $delta = sqrt[3]frac1M$. Note that $a<x<a+ sqrt[3]frac1M$, so $frac1(x-a)^3 > M$ (Q.E.D)
I feel like this proof I produced is missing something necessary to conclude the respective limits, or it could be completely wrong, so any criticism for the proof would be appreciated!
Thanks in advance!
calculus limits
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up vote
2
down vote
favorite
- Let $a>0$ and $n$ be a positive integer. Prove that $$lim_xto a x^1/n = a^1/n$$
For this question, I'm not entirely sure whether I should proceed with the $epsilon$-$delta$ definition of a limit, or simply prove the power law for limits, since the question did not specify the need to use the precise definition of a limit in this proof. I'm thinking that I could try to prove that the function is continuous at $x=a$, but proving continuity requires $lim_xto a x^1/n = a^1/n$. Could someone hint me in the right direction for this?
Let $a$ be a real number. Prove that
(a) $lim_xto a^- frac1(x-a)^3 = -infty$ and
(b)$lim_xto a^+ frac1(x-a)^3 = infty$
My attempt at the second question is as follows:
(a) We want to prove $forall N<0$, $exists delta >0$ such that if $a - delta < x < a$, then $frac1(x-a)^3 < N$. $$frac1(x-a)^3<N$$ $$frac1N > (x-a)^3$$ so $x-a > sqrt[3]frac1N$, i.e. $x > sqrt[3]frac1N$, so we choose $delta = -sqrt[3]frac1N$. Note that $sqrt[3]frac1N + a < x < a$, so $frac1(x-a)^3 < N$ (Q.E.D)
(b) Similarly, we want to prove that $forall M > 0$, $exists delta >0$ such that if $a<x<a+delta$ then $frac1(x-a)^3 > M$. $$frac1M > (x-a)^3$$ $$x < sqrt[3]frac1M + a$$, so we choose $delta = sqrt[3]frac1M$. Note that $a<x<a+ sqrt[3]frac1M$, so $frac1(x-a)^3 > M$ (Q.E.D)
I feel like this proof I produced is missing something necessary to conclude the respective limits, or it could be completely wrong, so any criticism for the proof would be appreciated!
Thanks in advance!
calculus limits
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
- Let $a>0$ and $n$ be a positive integer. Prove that $$lim_xto a x^1/n = a^1/n$$
For this question, I'm not entirely sure whether I should proceed with the $epsilon$-$delta$ definition of a limit, or simply prove the power law for limits, since the question did not specify the need to use the precise definition of a limit in this proof. I'm thinking that I could try to prove that the function is continuous at $x=a$, but proving continuity requires $lim_xto a x^1/n = a^1/n$. Could someone hint me in the right direction for this?
Let $a$ be a real number. Prove that
(a) $lim_xto a^- frac1(x-a)^3 = -infty$ and
(b)$lim_xto a^+ frac1(x-a)^3 = infty$
My attempt at the second question is as follows:
(a) We want to prove $forall N<0$, $exists delta >0$ such that if $a - delta < x < a$, then $frac1(x-a)^3 < N$. $$frac1(x-a)^3<N$$ $$frac1N > (x-a)^3$$ so $x-a > sqrt[3]frac1N$, i.e. $x > sqrt[3]frac1N$, so we choose $delta = -sqrt[3]frac1N$. Note that $sqrt[3]frac1N + a < x < a$, so $frac1(x-a)^3 < N$ (Q.E.D)
(b) Similarly, we want to prove that $forall M > 0$, $exists delta >0$ such that if $a<x<a+delta$ then $frac1(x-a)^3 > M$. $$frac1M > (x-a)^3$$ $$x < sqrt[3]frac1M + a$$, so we choose $delta = sqrt[3]frac1M$. Note that $a<x<a+ sqrt[3]frac1M$, so $frac1(x-a)^3 > M$ (Q.E.D)
I feel like this proof I produced is missing something necessary to conclude the respective limits, or it could be completely wrong, so any criticism for the proof would be appreciated!
Thanks in advance!
calculus limits
- Let $a>0$ and $n$ be a positive integer. Prove that $$lim_xto a x^1/n = a^1/n$$
For this question, I'm not entirely sure whether I should proceed with the $epsilon$-$delta$ definition of a limit, or simply prove the power law for limits, since the question did not specify the need to use the precise definition of a limit in this proof. I'm thinking that I could try to prove that the function is continuous at $x=a$, but proving continuity requires $lim_xto a x^1/n = a^1/n$. Could someone hint me in the right direction for this?
Let $a$ be a real number. Prove that
(a) $lim_xto a^- frac1(x-a)^3 = -infty$ and
(b)$lim_xto a^+ frac1(x-a)^3 = infty$
My attempt at the second question is as follows:
(a) We want to prove $forall N<0$, $exists delta >0$ such that if $a - delta < x < a$, then $frac1(x-a)^3 < N$. $$frac1(x-a)^3<N$$ $$frac1N > (x-a)^3$$ so $x-a > sqrt[3]frac1N$, i.e. $x > sqrt[3]frac1N$, so we choose $delta = -sqrt[3]frac1N$. Note that $sqrt[3]frac1N + a < x < a$, so $frac1(x-a)^3 < N$ (Q.E.D)
(b) Similarly, we want to prove that $forall M > 0$, $exists delta >0$ such that if $a<x<a+delta$ then $frac1(x-a)^3 > M$. $$frac1M > (x-a)^3$$ $$x < sqrt[3]frac1M + a$$, so we choose $delta = sqrt[3]frac1M$. Note that $a<x<a+ sqrt[3]frac1M$, so $frac1(x-a)^3 > M$ (Q.E.D)
I feel like this proof I produced is missing something necessary to conclude the respective limits, or it could be completely wrong, so any criticism for the proof would be appreciated!
Thanks in advance!
calculus limits
calculus limits
edited Sep 9 at 11:07
asked Sep 9 at 11:00
uznam
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2 Answers
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1
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For question 1, use the following geometric series identity:
$$fracp^n - q^np - q = p^n-1 + p^n - 2 q + p^n - 3 q^2 + ldots + q^n - 1$$
with $p = x^frac1n$ and $q = a^frac1n$.
As for question 2, you're doing fine. There is what I think is an inconsequential typo under your first two inequalities, where you say $x > sqrt[3]frac1N$, where I think you may mean $x > a + sqrt[3]frac1N$, but it doesn't cause any problems down the road.
It's hard for me to gauge exactly what is bothering you; I can only guess. However, what slightly bothers me is that the proof is presented slightly backwards. It's presented in the order of your working out, rather than in logical order.
You're supposed to be able to take a number $N < 0$, and instantly produce a $delta > 0$ with the given property. So, your proof should probably take the form:
Suppose $N < 0$, and let $delta = -sqrt[3]frac1N > 0$. Then $a - delta < x < a implies ldots implies frac1(x - a)^3 < N.$
The working you've written is valuable working behind the scenes, but it's nicer to present the proof in the proper order.
EDIT: In response to your comment, that's not too hard to do, and I think it is worth covering. Anything you feel is shaky is worth delving into. A proof is supposed to be convincing, and if it can't even convince the person who wrote it, then it's not working.
To prove this, suppose that $a + sqrt[3]frac1N < x < a$. Then
$$x - a > sqrt[3]frac1N.$$
Then, because the function $x mapsto x^3$ is increasing, we have
$$(x - a)^3 > left(sqrt[3]frac1Nright)^3 = frac1N.$$
Now, $x < a$, so $x - a < 0$. For negative numbers, the map $x mapsto frac1x$ is decreasing, so
$$frac1(x - a)^3 < N,$$
which is what we wanted to prove.
Thank you for your response. To clarify what exactly was bothering me about my proof: I feel like there could be some way to explain how we conclude that $frac1(x-a)^3 < N$ from the premise that $sqrt[3]frac1N + a < x < a$ for part (a), and likewise for part (b). Or is it unnecessary to explain this since it is just a simple fact from mere observation?
â uznam
Sep 9 at 13:36
1
@uznam Edited. Hope it helps.
â Theo Bendit
Sep 9 at 14:37
add a comment |Â
up vote
1
down vote
For $(1)$ you can use this lemma:
Let $x > y ge 0$ and $n in mathbbN$. Then $$x^1/n - y^1/n le (x-y)^1/n$$
Proof:
Let $t = left(fracyxright)^1/n in [0,1rangle$. Then $$(1-t)^n le 1-t le 1-t^n$$
so $1-t le (1-t^n)^1/n$. Multiplying with $x^1/n$ gives the desired result.
Now you should be able to finish with
$$left|a^1/n - x^1/nright| le |a-x|^1/n$$
Your attempt at $(2)$ looks good.
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
For question 1, use the following geometric series identity:
$$fracp^n - q^np - q = p^n-1 + p^n - 2 q + p^n - 3 q^2 + ldots + q^n - 1$$
with $p = x^frac1n$ and $q = a^frac1n$.
As for question 2, you're doing fine. There is what I think is an inconsequential typo under your first two inequalities, where you say $x > sqrt[3]frac1N$, where I think you may mean $x > a + sqrt[3]frac1N$, but it doesn't cause any problems down the road.
It's hard for me to gauge exactly what is bothering you; I can only guess. However, what slightly bothers me is that the proof is presented slightly backwards. It's presented in the order of your working out, rather than in logical order.
You're supposed to be able to take a number $N < 0$, and instantly produce a $delta > 0$ with the given property. So, your proof should probably take the form:
Suppose $N < 0$, and let $delta = -sqrt[3]frac1N > 0$. Then $a - delta < x < a implies ldots implies frac1(x - a)^3 < N.$
The working you've written is valuable working behind the scenes, but it's nicer to present the proof in the proper order.
EDIT: In response to your comment, that's not too hard to do, and I think it is worth covering. Anything you feel is shaky is worth delving into. A proof is supposed to be convincing, and if it can't even convince the person who wrote it, then it's not working.
To prove this, suppose that $a + sqrt[3]frac1N < x < a$. Then
$$x - a > sqrt[3]frac1N.$$
Then, because the function $x mapsto x^3$ is increasing, we have
$$(x - a)^3 > left(sqrt[3]frac1Nright)^3 = frac1N.$$
Now, $x < a$, so $x - a < 0$. For negative numbers, the map $x mapsto frac1x$ is decreasing, so
$$frac1(x - a)^3 < N,$$
which is what we wanted to prove.
Thank you for your response. To clarify what exactly was bothering me about my proof: I feel like there could be some way to explain how we conclude that $frac1(x-a)^3 < N$ from the premise that $sqrt[3]frac1N + a < x < a$ for part (a), and likewise for part (b). Or is it unnecessary to explain this since it is just a simple fact from mere observation?
â uznam
Sep 9 at 13:36
1
@uznam Edited. Hope it helps.
â Theo Bendit
Sep 9 at 14:37
add a comment |Â
up vote
1
down vote
accepted
For question 1, use the following geometric series identity:
$$fracp^n - q^np - q = p^n-1 + p^n - 2 q + p^n - 3 q^2 + ldots + q^n - 1$$
with $p = x^frac1n$ and $q = a^frac1n$.
As for question 2, you're doing fine. There is what I think is an inconsequential typo under your first two inequalities, where you say $x > sqrt[3]frac1N$, where I think you may mean $x > a + sqrt[3]frac1N$, but it doesn't cause any problems down the road.
It's hard for me to gauge exactly what is bothering you; I can only guess. However, what slightly bothers me is that the proof is presented slightly backwards. It's presented in the order of your working out, rather than in logical order.
You're supposed to be able to take a number $N < 0$, and instantly produce a $delta > 0$ with the given property. So, your proof should probably take the form:
Suppose $N < 0$, and let $delta = -sqrt[3]frac1N > 0$. Then $a - delta < x < a implies ldots implies frac1(x - a)^3 < N.$
The working you've written is valuable working behind the scenes, but it's nicer to present the proof in the proper order.
EDIT: In response to your comment, that's not too hard to do, and I think it is worth covering. Anything you feel is shaky is worth delving into. A proof is supposed to be convincing, and if it can't even convince the person who wrote it, then it's not working.
To prove this, suppose that $a + sqrt[3]frac1N < x < a$. Then
$$x - a > sqrt[3]frac1N.$$
Then, because the function $x mapsto x^3$ is increasing, we have
$$(x - a)^3 > left(sqrt[3]frac1Nright)^3 = frac1N.$$
Now, $x < a$, so $x - a < 0$. For negative numbers, the map $x mapsto frac1x$ is decreasing, so
$$frac1(x - a)^3 < N,$$
which is what we wanted to prove.
Thank you for your response. To clarify what exactly was bothering me about my proof: I feel like there could be some way to explain how we conclude that $frac1(x-a)^3 < N$ from the premise that $sqrt[3]frac1N + a < x < a$ for part (a), and likewise for part (b). Or is it unnecessary to explain this since it is just a simple fact from mere observation?
â uznam
Sep 9 at 13:36
1
@uznam Edited. Hope it helps.
â Theo Bendit
Sep 9 at 14:37
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
For question 1, use the following geometric series identity:
$$fracp^n - q^np - q = p^n-1 + p^n - 2 q + p^n - 3 q^2 + ldots + q^n - 1$$
with $p = x^frac1n$ and $q = a^frac1n$.
As for question 2, you're doing fine. There is what I think is an inconsequential typo under your first two inequalities, where you say $x > sqrt[3]frac1N$, where I think you may mean $x > a + sqrt[3]frac1N$, but it doesn't cause any problems down the road.
It's hard for me to gauge exactly what is bothering you; I can only guess. However, what slightly bothers me is that the proof is presented slightly backwards. It's presented in the order of your working out, rather than in logical order.
You're supposed to be able to take a number $N < 0$, and instantly produce a $delta > 0$ with the given property. So, your proof should probably take the form:
Suppose $N < 0$, and let $delta = -sqrt[3]frac1N > 0$. Then $a - delta < x < a implies ldots implies frac1(x - a)^3 < N.$
The working you've written is valuable working behind the scenes, but it's nicer to present the proof in the proper order.
EDIT: In response to your comment, that's not too hard to do, and I think it is worth covering. Anything you feel is shaky is worth delving into. A proof is supposed to be convincing, and if it can't even convince the person who wrote it, then it's not working.
To prove this, suppose that $a + sqrt[3]frac1N < x < a$. Then
$$x - a > sqrt[3]frac1N.$$
Then, because the function $x mapsto x^3$ is increasing, we have
$$(x - a)^3 > left(sqrt[3]frac1Nright)^3 = frac1N.$$
Now, $x < a$, so $x - a < 0$. For negative numbers, the map $x mapsto frac1x$ is decreasing, so
$$frac1(x - a)^3 < N,$$
which is what we wanted to prove.
For question 1, use the following geometric series identity:
$$fracp^n - q^np - q = p^n-1 + p^n - 2 q + p^n - 3 q^2 + ldots + q^n - 1$$
with $p = x^frac1n$ and $q = a^frac1n$.
As for question 2, you're doing fine. There is what I think is an inconsequential typo under your first two inequalities, where you say $x > sqrt[3]frac1N$, where I think you may mean $x > a + sqrt[3]frac1N$, but it doesn't cause any problems down the road.
It's hard for me to gauge exactly what is bothering you; I can only guess. However, what slightly bothers me is that the proof is presented slightly backwards. It's presented in the order of your working out, rather than in logical order.
You're supposed to be able to take a number $N < 0$, and instantly produce a $delta > 0$ with the given property. So, your proof should probably take the form:
Suppose $N < 0$, and let $delta = -sqrt[3]frac1N > 0$. Then $a - delta < x < a implies ldots implies frac1(x - a)^3 < N.$
The working you've written is valuable working behind the scenes, but it's nicer to present the proof in the proper order.
EDIT: In response to your comment, that's not too hard to do, and I think it is worth covering. Anything you feel is shaky is worth delving into. A proof is supposed to be convincing, and if it can't even convince the person who wrote it, then it's not working.
To prove this, suppose that $a + sqrt[3]frac1N < x < a$. Then
$$x - a > sqrt[3]frac1N.$$
Then, because the function $x mapsto x^3$ is increasing, we have
$$(x - a)^3 > left(sqrt[3]frac1Nright)^3 = frac1N.$$
Now, $x < a$, so $x - a < 0$. For negative numbers, the map $x mapsto frac1x$ is decreasing, so
$$frac1(x - a)^3 < N,$$
which is what we wanted to prove.
edited Sep 9 at 14:45
answered Sep 9 at 11:15
Theo Bendit
13.7k12045
13.7k12045
Thank you for your response. To clarify what exactly was bothering me about my proof: I feel like there could be some way to explain how we conclude that $frac1(x-a)^3 < N$ from the premise that $sqrt[3]frac1N + a < x < a$ for part (a), and likewise for part (b). Or is it unnecessary to explain this since it is just a simple fact from mere observation?
â uznam
Sep 9 at 13:36
1
@uznam Edited. Hope it helps.
â Theo Bendit
Sep 9 at 14:37
add a comment |Â
Thank you for your response. To clarify what exactly was bothering me about my proof: I feel like there could be some way to explain how we conclude that $frac1(x-a)^3 < N$ from the premise that $sqrt[3]frac1N + a < x < a$ for part (a), and likewise for part (b). Or is it unnecessary to explain this since it is just a simple fact from mere observation?
â uznam
Sep 9 at 13:36
1
@uznam Edited. Hope it helps.
â Theo Bendit
Sep 9 at 14:37
Thank you for your response. To clarify what exactly was bothering me about my proof: I feel like there could be some way to explain how we conclude that $frac1(x-a)^3 < N$ from the premise that $sqrt[3]frac1N + a < x < a$ for part (a), and likewise for part (b). Or is it unnecessary to explain this since it is just a simple fact from mere observation?
â uznam
Sep 9 at 13:36
Thank you for your response. To clarify what exactly was bothering me about my proof: I feel like there could be some way to explain how we conclude that $frac1(x-a)^3 < N$ from the premise that $sqrt[3]frac1N + a < x < a$ for part (a), and likewise for part (b). Or is it unnecessary to explain this since it is just a simple fact from mere observation?
â uznam
Sep 9 at 13:36
1
1
@uznam Edited. Hope it helps.
â Theo Bendit
Sep 9 at 14:37
@uznam Edited. Hope it helps.
â Theo Bendit
Sep 9 at 14:37
add a comment |Â
up vote
1
down vote
For $(1)$ you can use this lemma:
Let $x > y ge 0$ and $n in mathbbN$. Then $$x^1/n - y^1/n le (x-y)^1/n$$
Proof:
Let $t = left(fracyxright)^1/n in [0,1rangle$. Then $$(1-t)^n le 1-t le 1-t^n$$
so $1-t le (1-t^n)^1/n$. Multiplying with $x^1/n$ gives the desired result.
Now you should be able to finish with
$$left|a^1/n - x^1/nright| le |a-x|^1/n$$
Your attempt at $(2)$ looks good.
add a comment |Â
up vote
1
down vote
For $(1)$ you can use this lemma:
Let $x > y ge 0$ and $n in mathbbN$. Then $$x^1/n - y^1/n le (x-y)^1/n$$
Proof:
Let $t = left(fracyxright)^1/n in [0,1rangle$. Then $$(1-t)^n le 1-t le 1-t^n$$
so $1-t le (1-t^n)^1/n$. Multiplying with $x^1/n$ gives the desired result.
Now you should be able to finish with
$$left|a^1/n - x^1/nright| le |a-x|^1/n$$
Your attempt at $(2)$ looks good.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
For $(1)$ you can use this lemma:
Let $x > y ge 0$ and $n in mathbbN$. Then $$x^1/n - y^1/n le (x-y)^1/n$$
Proof:
Let $t = left(fracyxright)^1/n in [0,1rangle$. Then $$(1-t)^n le 1-t le 1-t^n$$
so $1-t le (1-t^n)^1/n$. Multiplying with $x^1/n$ gives the desired result.
Now you should be able to finish with
$$left|a^1/n - x^1/nright| le |a-x|^1/n$$
Your attempt at $(2)$ looks good.
For $(1)$ you can use this lemma:
Let $x > y ge 0$ and $n in mathbbN$. Then $$x^1/n - y^1/n le (x-y)^1/n$$
Proof:
Let $t = left(fracyxright)^1/n in [0,1rangle$. Then $$(1-t)^n le 1-t le 1-t^n$$
so $1-t le (1-t^n)^1/n$. Multiplying with $x^1/n$ gives the desired result.
Now you should be able to finish with
$$left|a^1/n - x^1/nright| le |a-x|^1/n$$
Your attempt at $(2)$ looks good.
answered Sep 9 at 11:34
mechanodroid
24.7k62245
24.7k62245
add a comment |Â
add a comment |Â
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