Mixing
Check my proof that a monotone function has one-sided limits at any point
$begingroup$
Here $f:mathbb{R} rightarrow mathbb{R} $.
We consider the case when $f$ is an increasing function.
Let $A={f(x) | x<x_0}$. $A$ is bounded from above by $f(x_0)$,so there exists $alpha in mathbb{R} $ so that $sup A= alpha$.
Let $epsilon >0$. Since $alpha - epsilon < alpha => exists x_1 in mathbb{R}, x_1<x_0$ so that $f(x_1)>alpha - epsilon$.
$=>forall xin (x_1, x_0), alpha - epsilon < f(x_1) le f(x) <alpha + epsilon$
Let $delta_1=x_0 - x_1$.Then $forall xin mathbb{R}, |x-x_0|<delta_1,|f(x)-alpha|<epsilon$,so $lim_{xto x_0^{-}} f(x) =alpha$.
Let $B={f(x) | x>x_0}$. $B$ is bounded from below by $f(x_0)$,so there exists $beta in mathbb{R} $ so that $inf B= beta$.
Let $epsilon_1 >0$. Since $beta + epsilon_1 > beta => exists x_2 in mathbb{R}, x_2>x_0$ so that $f(x_2)<beta + epsilon_1$.
$=>forall xin (x_0, x_2), beta + epsilon_1 > f(x_2) ge f(x) >beta - epsilon_1$
Let $delta_2=x_2 - x_0$.Then $forall xin mathbb{R}, |x-x_0|<delta_2,|f(x)-beta|<epsilon_1$,so $lim_{xto x_0^{+}} f(x) =beta$.
limits functions proof-verification
asked Jan 13 at 16:15
JustAnAmateurJustAnAmateur
876
$endgroup$
add a comment |
$begingroup$
Here $f:mathbb{R} rightarrow mathbb{R} $.
We consider the case when $f$ is an increasing function.
Let $A={f(x) | x<x_0}$. $A$ is bounded from above by $f(x_0)$,so there exists $alpha in mathbb{R} $ so that $sup A= alpha$.
Let $epsilon >0$. Since $alpha - epsilon < alpha => exists x_1 in mathbb{R}, x_1<x_0$ so that $f(x_1)>alpha - epsilon$.
$=>forall xin (x_1, x_0), alpha - epsilon < f(x_1) le f(x) <alpha + epsilon$
Let $delta_1=x_0 - x_1$.Then $forall xin mathbb{R}, |x-x_0|<delta_1,|f(x)-alpha|<epsilon$,so $lim_{xto x_0^{-}} f(x) =alpha$.
Let $B={f(x) | x>x_0}$. $B$ is bounded from below by $f(x_0)$,so there exists $beta in mathbb{R} $ so that $inf B= beta$.
Let $epsilon_1 >0$. Since $beta + epsilon_1 > beta => exists x_2 in mathbb{R}, x_2>x_0$ so that $f(x_2)<beta + epsilon_1$.
$=>forall xin (x_0, x_2), beta + epsilon_1 > f(x_2) ge f(x) >beta - epsilon_1$
Let $delta_2=x_2 - x_0$.Then $forall xin mathbb{R}, |x-x_0|<delta_2,|f(x)-beta|<epsilon_1$,so $lim_{xto x_0^{+}} f(x) =beta$.
limits functions proof-verification
asked Jan 13 at 16:15
JustAnAmateurJustAnAmateur
876
$endgroup$
add a comment |
$begingroup$
Here $f:mathbb{R} rightarrow mathbb{R} $.
We consider the case when $f$ is an increasing function.
Let $A={f(x) | x<x_0}$. $A$ is bounded from above by $f(x_0)$,so there exists $alpha in mathbb{R} $ so that $sup A= alpha$.
Let $epsilon >0$. Since $alpha - epsilon < alpha => exists x_1 in mathbb{R}, x_1<x_0$ so that $f(x_1)>alpha - epsilon$.
$=>forall xin (x_1, x_0), alpha - epsilon < f(x_1) le f(x) <alpha + epsilon$
Let $delta_1=x_0 - x_1$.Then $forall xin mathbb{R}, |x-x_0|<delta_1,|f(x)-alpha|<epsilon$,so $lim_{xto x_0^{-}} f(x) =alpha$.
Let $B={f(x) | x>x_0}$. $B$ is bounded from below by $f(x_0)$,so there exists $beta in mathbb{R} $ so that $inf B= beta$.
Let $epsilon_1 >0$. Since $beta + epsilon_1 > beta => exists x_2 in mathbb{R}, x_2>x_0$ so that $f(x_2)<beta + epsilon_1$.
$=>forall xin (x_0, x_2), beta + epsilon_1 > f(x_2) ge f(x) >beta - epsilon_1$
Let $delta_2=x_2 - x_0$.Then $forall xin mathbb{R}, |x-x_0|<delta_2,|f(x)-beta|<epsilon_1$,so $lim_{xto x_0^{+}} f(x) =beta$.
limits functions proof-verification
asked Jan 13 at 16:15
JustAnAmateurJustAnAmateur
876
$endgroup$
Here $f:mathbb{R} rightarrow mathbb{R} $.
We consider the case when $f$ is an increasing function.
Let $A={f(x) | x<x_0}$. $A$ is bounded from above by $f(x_0)$,so there exists $alpha in mathbb{R} $ so that $sup A= alpha$.
Let $epsilon >0$. Since $alpha - epsilon < alpha => exists x_1 in mathbb{R}, x_1<x_0$ so that $f(x_1)>alpha - epsilon$.
$=>forall xin (x_1, x_0), alpha - epsilon < f(x_1) le f(x) <alpha + epsilon$
Let $delta_1=x_0 - x_1$.Then $forall xin mathbb{R}, |x-x_0|<delta_1,|f(x)-alpha|<epsilon$,so $lim_{xto x_0^{-}} f(x) =alpha$.
Let $B={f(x) | x>x_0}$. $B$ is bounded from below by $f(x_0)$,so there exists $beta in mathbb{R} $ so that $inf B= beta$.
Let $epsilon_1 >0$. Since $beta + epsilon_1 > beta => exists x_2 in mathbb{R}, x_2>x_0$ so that $f(x_2)<beta + epsilon_1$.
$=>forall xin (x_0, x_2), beta + epsilon_1 > f(x_2) ge f(x) >beta - epsilon_1$
Let $delta_2=x_2 - x_0$.Then $forall xin mathbb{R}, |x-x_0|<delta_2,|f(x)-beta|<epsilon_1$,so $lim_{xto x_0^{+}} f(x) =beta$.
limits functions proof-verification
limits functions proof-verification
asked Jan 13 at 16:15
JustAnAmateurJustAnAmateur
876
asked Jan 13 at 16:15
JustAnAmateurJustAnAmateur
876
asked Jan 13 at 16:15
JustAnAmateurJustAnAmateur
876
asked Jan 13 at 16:15
JustAnAmateurJustAnAmateur
876
asked Jan 13 at 16:15
JustAnAmateurJustAnAmateur
876
876
add a comment |
add a comment |
1 Answer
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$begingroup$
It is really good, there is just one small mistake. When you defined $delta_1$ you wrote the required inequality $|f(x)-alpha|<epsilon$ for all $xinmathbb{R}$ such that $|x-x_0|<delta_1$. But the inequality might not hold if $x>x_0$. This is a one sided limit so you needed to write for all $xinmathbb{R}$ such that $x_0-delta_1<x<x_0$. Same thing when you defined $delta_2$.
answered Jan 13 at 16:28
MarkMark
7,293417
$endgroup$
$begingroup$
Thank you very much for pointing this out!
$endgroup$
– JustAnAmateur
Jan 13 at 16:57
add a comment |
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1 Answer
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1 Answer
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$begingroup$
It is really good, there is just one small mistake. When you defined $delta_1$ you wrote the required inequality $|f(x)-alpha|<epsilon$ for all $xinmathbb{R}$ such that $|x-x_0|<delta_1$. But the inequality might not hold if $x>x_0$. This is a one sided limit so you needed to write for all $xinmathbb{R}$ such that $x_0-delta_1<x<x_0$. Same thing when you defined $delta_2$.
answered Jan 13 at 16:28
MarkMark
7,293417
$endgroup$
$begingroup$
Thank you very much for pointing this out!
$endgroup$
– JustAnAmateur
Jan 13 at 16:57
add a comment |
$begingroup$
It is really good, there is just one small mistake. When you defined $delta_1$ you wrote the required inequality $|f(x)-alpha|<epsilon$ for all $xinmathbb{R}$ such that $|x-x_0|<delta_1$. But the inequality might not hold if $x>x_0$. This is a one sided limit so you needed to write for all $xinmathbb{R}$ such that $x_0-delta_1<x<x_0$. Same thing when you defined $delta_2$.
answered Jan 13 at 16:28
MarkMark
7,293417
$endgroup$
$begingroup$
Thank you very much for pointing this out!
$endgroup$
– JustAnAmateur
Jan 13 at 16:57
add a comment |
$begingroup$
It is really good, there is just one small mistake. When you defined $delta_1$ you wrote the required inequality $|f(x)-alpha|<epsilon$ for all $xinmathbb{R}$ such that $|x-x_0|<delta_1$. But the inequality might not hold if $x>x_0$. This is a one sided limit so you needed to write for all $xinmathbb{R}$ such that $x_0-delta_1<x<x_0$. Same thing when you defined $delta_2$.
answered Jan 13 at 16:28
MarkMark
7,293417
$endgroup$
It is really good, there is just one small mistake. When you defined $delta_1$ you wrote the required inequality $|f(x)-alpha|<epsilon$ for all $xinmathbb{R}$ such that $|x-x_0|<delta_1$. But the inequality might not hold if $x>x_0$. This is a one sided limit so you needed to write for all $xinmathbb{R}$ such that $x_0-delta_1<x<x_0$. Same thing when you defined $delta_2$.
answered Jan 13 at 16:28
MarkMark
7,293417
answered Jan 13 at 16:28
MarkMark
7,293417
answered Jan 13 at 16:28
MarkMark
7,293417
answered Jan 13 at 16:28
MarkMark
7,293417
7,293417
$begingroup$
Thank you very much for pointing this out!
$endgroup$
– JustAnAmateur
Jan 13 at 16:57
add a comment |
$begingroup$
Thank you very much for pointing this out!
$endgroup$
– JustAnAmateur
Jan 13 at 16:57
$begingroup$
Thank you very much for pointing this out!
$endgroup$
– JustAnAmateur
Jan 13 at 16:57
$begingroup$
Thank you very much for pointing this out!
$endgroup$
– JustAnAmateur
Jan 13 at 16:57
add a comment |
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