Show $lim_{x to x_0^+} f(x)(x-x_0) =0$ when $f(mathbb{R}) subset mathbb{R}^+$ & monotone increasing.
up vote
1
down vote
favorite
Show $lim_{x to x_0^+} f(x)(x-x_0) =0$ when $f(mathbb{R}) subset mathbb{R}^+$ & monotone increasing.
Try
I need to show,
$$
forall epsilon >0, exists delta >0 : x in (x_0, x_0 + delta) Rightarrow |f(x) (x-x_0)| < epsilon
$$
I think I could find some upper bound $M >0$ such that $|f(x) (x-x_0)| le M |x - x_0|$.
Let $M = f(x_0 + epsilon)$, and let $delta = frac{epsilon}{max {2M, 2 }}$, then clearly $f(x) le f(x_0 + epsilon) = M$
But I'm not sure $|f(x) (x-x_0)| le M |x - x_0|$.
Any hint about how I should proceed?
real-analysis
asked Nov 18 at 1:16
Moreblue
800216
add a comment |
up vote
1
down vote
favorite
Show $lim_{x to x_0^+} f(x)(x-x_0) =0$ when $f(mathbb{R}) subset mathbb{R}^+$ & monotone increasing.
Try
I need to show,
$$
forall epsilon >0, exists delta >0 : x in (x_0, x_0 + delta) Rightarrow |f(x) (x-x_0)| < epsilon
$$
I think I could find some upper bound $M >0$ such that $|f(x) (x-x_0)| le M |x - x_0|$.
Let $M = f(x_0 + epsilon)$, and let $delta = frac{epsilon}{max {2M, 2 }}$, then clearly $f(x) le f(x_0 + epsilon) = M$
But I'm not sure $|f(x) (x-x_0)| le M |x - x_0|$.
Any hint about how I should proceed?
real-analysis
asked Nov 18 at 1:16
Moreblue
800216
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Show $lim_{x to x_0^+} f(x)(x-x_0) =0$ when $f(mathbb{R}) subset mathbb{R}^+$ & monotone increasing.
Try
I need to show,
$$
forall epsilon >0, exists delta >0 : x in (x_0, x_0 + delta) Rightarrow |f(x) (x-x_0)| < epsilon
$$
I think I could find some upper bound $M >0$ such that $|f(x) (x-x_0)| le M |x - x_0|$.
Let $M = f(x_0 + epsilon)$, and let $delta = frac{epsilon}{max {2M, 2 }}$, then clearly $f(x) le f(x_0 + epsilon) = M$
But I'm not sure $|f(x) (x-x_0)| le M |x - x_0|$.
Any hint about how I should proceed?
real-analysis
asked Nov 18 at 1:16
Moreblue
800216
Show $lim_{x to x_0^+} f(x)(x-x_0) =0$ when $f(mathbb{R}) subset mathbb{R}^+$ & monotone increasing.
Try
I need to show,
$$
forall epsilon >0, exists delta >0 : x in (x_0, x_0 + delta) Rightarrow |f(x) (x-x_0)| < epsilon
$$
I think I could find some upper bound $M >0$ such that $|f(x) (x-x_0)| le M |x - x_0|$.
Let $M = f(x_0 + epsilon)$, and let $delta = frac{epsilon}{max {2M, 2 }}$, then clearly $f(x) le f(x_0 + epsilon) = M$
But I'm not sure $|f(x) (x-x_0)| le M |x - x_0|$.
Any hint about how I should proceed?
real-analysis
real-analysis
asked Nov 18 at 1:16
Moreblue
800216
asked Nov 18 at 1:16
Moreblue
800216
asked Nov 18 at 1:16
Moreblue
800216
asked Nov 18 at 1:16
Moreblue
800216
asked Nov 18 at 1:16
Moreblue
800216
800216
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
Hint: Observe
begin{align}
|f(x)(x-x_0)|leq |f(x_0)||x-x_0|
end{align}
for all $xleq x_0$.
answered Nov 18 at 1:42
Jacky Chong
17.3k21128
add a comment |
up vote
1
down vote
Use $M=f(x_0+1)$ and cosider $delta=min{frac{1}{2},frac{epsilon}{2M}}$.
answered Nov 18 at 1:50
Offlaw
2649
add a comment |
up vote
1
down vote
Fix $varepsilon>0$. Let $M=f(x_0+1)$ and choose $delta=mathrm{min}{1,frac{varepsilon}{M}}$. For each $xin(x_0,x_0+delta)$, $|f(x)|leq M$ since $f$ is strictly increasing. Thus, $|f(x)(x-x_0)|leq M|x-x_0|<Mdeltaleqvarepsilon$.
answered Nov 18 at 2:07
gHem
583
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Hint: Observe
begin{align}
|f(x)(x-x_0)|leq |f(x_0)||x-x_0|
end{align}
for all $xleq x_0$.
answered Nov 18 at 1:42
Jacky Chong
17.3k21128
add a comment |
up vote
1
down vote
accepted
Hint: Observe
begin{align}
|f(x)(x-x_0)|leq |f(x_0)||x-x_0|
end{align}
for all $xleq x_0$.
answered Nov 18 at 1:42
Jacky Chong
17.3k21128
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Hint: Observe
begin{align}
|f(x)(x-x_0)|leq |f(x_0)||x-x_0|
end{align}
for all $xleq x_0$.
answered Nov 18 at 1:42
Jacky Chong
17.3k21128
Hint: Observe
begin{align}
|f(x)(x-x_0)|leq |f(x_0)||x-x_0|
end{align}
for all $xleq x_0$.
answered Nov 18 at 1:42
Jacky Chong
17.3k21128
answered Nov 18 at 1:42
Jacky Chong
17.3k21128
answered Nov 18 at 1:42
Jacky Chong
17.3k21128
answered Nov 18 at 1:42
Jacky Chong
17.3k21128
17.3k21128
add a comment |
add a comment |
up vote
1
down vote
Use $M=f(x_0+1)$ and cosider $delta=min{frac{1}{2},frac{epsilon}{2M}}$.
answered Nov 18 at 1:50
Offlaw
2649
add a comment |
up vote
1
down vote
Use $M=f(x_0+1)$ and cosider $delta=min{frac{1}{2},frac{epsilon}{2M}}$.
answered Nov 18 at 1:50
Offlaw
2649
add a comment |
up vote
1
down vote
up vote
1
down vote
Use $M=f(x_0+1)$ and cosider $delta=min{frac{1}{2},frac{epsilon}{2M}}$.
answered Nov 18 at 1:50
Offlaw
2649
Use $M=f(x_0+1)$ and cosider $delta=min{frac{1}{2},frac{epsilon}{2M}}$.
answered Nov 18 at 1:50
Offlaw
2649
answered Nov 18 at 1:50
Offlaw
2649
answered Nov 18 at 1:50
Offlaw
2649
answered Nov 18 at 1:50
Offlaw
2649
2649
add a comment |
add a comment |
up vote
1
down vote
Fix $varepsilon>0$. Let $M=f(x_0+1)$ and choose $delta=mathrm{min}{1,frac{varepsilon}{M}}$. For each $xin(x_0,x_0+delta)$, $|f(x)|leq M$ since $f$ is strictly increasing. Thus, $|f(x)(x-x_0)|leq M|x-x_0|<Mdeltaleqvarepsilon$.
answered Nov 18 at 2:07
gHem
583
add a comment |
up vote
1
down vote
Fix $varepsilon>0$. Let $M=f(x_0+1)$ and choose $delta=mathrm{min}{1,frac{varepsilon}{M}}$. For each $xin(x_0,x_0+delta)$, $|f(x)|leq M$ since $f$ is strictly increasing. Thus, $|f(x)(x-x_0)|leq M|x-x_0|<Mdeltaleqvarepsilon$.
answered Nov 18 at 2:07
gHem
583
add a comment |
up vote
1
down vote
up vote
1
down vote
Fix $varepsilon>0$. Let $M=f(x_0+1)$ and choose $delta=mathrm{min}{1,frac{varepsilon}{M}}$. For each $xin(x_0,x_0+delta)$, $|f(x)|leq M$ since $f$ is strictly increasing. Thus, $|f(x)(x-x_0)|leq M|x-x_0|<Mdeltaleqvarepsilon$.
answered Nov 18 at 2:07
gHem
583
Fix $varepsilon>0$. Let $M=f(x_0+1)$ and choose $delta=mathrm{min}{1,frac{varepsilon}{M}}$. For each $xin(x_0,x_0+delta)$, $|f(x)|leq M$ since $f$ is strictly increasing. Thus, $|f(x)(x-x_0)|leq M|x-x_0|<Mdeltaleqvarepsilon$.
answered Nov 18 at 2:07
gHem
583
answered Nov 18 at 2:07
gHem
583
answered Nov 18 at 2:07
gHem
583
answered Nov 18 at 2:07
gHem
583
583
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
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%2f3003033%2fshow-lim-x-to-x-0-fxx-x-0-0-when-f-mathbbr-subset-mathbbr%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
