Show $lim_{x to x_0^+} f(x)(x-x_0) =0$ when $f(mathbb{R}) subset mathbb{R}^+$ & monotone increasing.
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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
795216
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
795216
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
795216
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
795216
asked Nov 18 at 1:16
Moreblue
795216
asked Nov 18 at 1:16
Moreblue
795216
asked Nov 18 at 1:16
Moreblue
795216
asked Nov 18 at 1:16
Moreblue
795216
795216
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
17.3k21027
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
2589
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
17.3k21027
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
17.3k21027
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
17.3k21027
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
17.3k21027
answered Nov 18 at 1:42
Jacky Chong
17.3k21027
answered Nov 18 at 1:42
Jacky Chong
17.3k21027
answered Nov 18 at 1:42
Jacky Chong
17.3k21027
17.3k21027
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
2589
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
2589
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
2589
Use $M=f(x_0+1)$ and cosider $delta=min{frac{1}{2},frac{epsilon}{2M}}$.
answered Nov 18 at 1:50
Offlaw
2589
answered Nov 18 at 1:50
Offlaw
2589
answered Nov 18 at 1:50
Offlaw
2589
answered Nov 18 at 1:50
Offlaw
2589
2589
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 |
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