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Limits & properties

mathmari

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Apr 14, 2013
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Hey!! :eek:

Could you give me a hint how to prove the following statements? (Wondering)

Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be differentiable (or twice differentiable).
  1. $\left.\begin{matrix}
    \displaystyle{\lim_{x\rightarrow +\infty}f(x)=\ell} \ (\text{or } \displaystyle{\lim_{x\rightarrow -\infty}f(x)=\ell}), \ell\in \mathbb{R} \\
    \text{and } f \text{ convex (or concave)}
    \end{matrix}\right\}\lim\limits_{\substack{x\rightarrow +\infty \\ (\text{or } x\rightarrow -\infty)}}f'(x)=0 $

    $\Rightarrow\ f$ is strictly monotone

  2. If $\displaystyle{\lim_{x\rightarrow +\infty}f''(x)}$ exists, then $\displaystyle{\lim_{x\rightarrow +\infty}f''(x)=\lim_{x\rightarrow +\infty}f'(x)=0}$ (i.e. we don't assume that $\displaystyle{\lim_{x\rightarrow +\infty}f'(x)}$ exists).
 

Klaas van Aarsen

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Mar 5, 2012
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Hey mathmari !!

Let's start with:

Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be differentiable.

1a.
$$\left.\begin{matrix}
\lim\limits_{x\to +\infty}f(x)=\ell,\,\ell\in \mathbb{R} \\
\text{and } f \text{ convex}
\end{matrix}\right\}\lim\limits_{x\to +\infty}f'(x)=0
\quad\implies\quad f{\text{ is strictly monotone}}$$
It follows from $\lim\limits_{x\rightarrow +\infty}f(x)=\ell$ that $\lim\limits_{x\to +\infty}f'(x)=0$ doesn't it? $f$ doesn't need to be convex for that, does it? (Wondering)


Anyway, from wiki:
A differentiable function of one variable is convex on an interval if and only if its graph lies above all of its tangents:
$$f(x)\geq f(y)+f'(y)(x-y)$$
for all x and y in the interval.

Can we use it to find if $f$ is strictly monotone?
What does strictly monotone mean exactly? (Wondering)
 

mathmari

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MHB Site Helper
Apr 14, 2013
4,028
Let's start with:

It follows from $\lim\limits_{x\rightarrow +\infty}f(x)=\ell$ that $\lim\limits_{x\to +\infty}f'(x)=0$ doesn't it? $f$ doesn't need to be convex for that, does it? (Wondering)
In a previous exercise I showed that $$\lim_{x\rightarrow +\infty}(f(x+1)-f(x))=L \Rightarrow \lim_{x\rightarrow +\infty}f'(x)=L$$ if we know that $\displaystyle{\lim_{x\rightarrow +\infty}f'(x)}$ exists.

When we know that the function is convex does it follow that $\displaystyle{\lim_{x\rightarrow +\infty}f'(x)}$ exists and so we can use this statement to show that $\displaystyle{\lim\limits_{x\to +\infty}f'(x)=0}$ ? (Wondering)


Or how does it follow from $\lim\limits_{x\rightarrow +\infty}f(x)=\ell$ that $\lim\limits_{x\to +\infty}f'(x)=0$ ? (Wondering)


Anyway, from wiki:
A differentiable function of one variable is convex on an interval if and only if its graph lies above all of its tangents:
$$f(x)\geq f(y)+f'(y)(x-y)$$
for all x and y in the interval.

Can we use it to find if $f$ is strictly monotone?
What does strictly monotone mean exactly? (Wondering)
Strictly monotone means that at the inequality the equality doesn't hold, i.e. strictly increasing is $x_1<x_2 \Rightarrow f(x_1)<f(x_2)$.
 

Klaas van Aarsen

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Mar 5, 2012
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In a previous exercise I showed that $$\lim_{x\rightarrow +\infty}(f(x+1)-f(x))=L \Rightarrow \lim_{x\rightarrow +\infty}f'(x)=L$$ if we know that $\displaystyle{\lim_{x\rightarrow +\infty}f'(x)}$ exists.

When we know that the function is convex does it follow that $\displaystyle{\lim_{x\rightarrow +\infty}f'(x)}$ exists and so we can use this statement to show that $\displaystyle{\lim\limits_{x\to +\infty}f'(x)=0}$ ?
Let's see, suppose we pick a convex function, say $f(x)=x^2$. It's convex isn't it?
Does $\lim\limits_{x\to +\infty}f'(x)$ exist? (Wondering)

Or how does it follow from $\lim\limits_{x\rightarrow +\infty}f(x)=\ell$ that $\lim\limits_{x\to +\infty}f'(x)=0$ ?
How about:
$$\lim_{x\to +\infty} f'(x) = \lim_{x\to +\infty} \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}
= \lim_{h\to 0}\lim_{x\to +\infty} \frac{f(x+h)-f(x)}{h}
= \lim_{h\to 0} \frac{\ell-\ell}{h} = 0
$$
(Thinking)

Strictly monotone means that at the inequality the equality doesn't hold, i.e. strictly increasing is $x_1<x_2 \Rightarrow f(x_1)<f(x_2)$.
Good.
Can we deduce that from the property that says when a differentiable function is convex? (Thinking)
 

mathmari

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Apr 14, 2013
4,028
Let's see, suppose we pick a convex function, say $f(x)=x^2$. It's convex isn't it?
Does $\lim\limits_{x\to +\infty}f'(x)$ exist? (Wondering)
At the previous exercise $L$ belong to $\mathbb{R}\cup \{\pm \infty\}$. So since $f'(x)=2x$ and the limit is equal to $+\infty$ and so it exists. (Wondering)


How about:
$$\lim_{x\to +\infty} f'(x) = \lim_{x\to +\infty} \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}
= \lim_{h\to 0}\lim_{x\to +\infty} \frac{f(x+h)-f(x)}{h}
= \lim_{h\to 0} \frac{\ell-\ell}{h} = 0
$$
(Thinking)
Ahh ok! I see!! (Malthe)


Can we deduce that from the property that says when a differentiable function is convex? (Thinking)
Since $f$ is convex we have that $f(x)\geq f(y)+f'(y)(x-y)$. Do we consider the limit as $x\rightarrow +\infty$ ? (Wondering)
 

Klaas van Aarsen

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Staff member
Mar 5, 2012
8,713
At the previous exercise $L$ belong to $\mathbb{R}\cup \{\pm \infty\}$. So since $f'(x)=2x$ and the limit is equal to $+\infty$ and so it exists.
Ah okay. But that is not the case now is it? (Wondering)

Since $f$ is convex we have that $f(x)\geq f(y)+f'(y)(x-y)$. Do we consider the limit as $x\rightarrow +\infty$ ?
Sounds like a plan. (Nod)
 

mathmari

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Apr 14, 2013
4,028
Sounds like a plan. (Nod)
So we have the following:
\begin{align*}f(x)\geq f(y)+f'(y)(x-y)&\Rightarrow \lim_{x\rightarrow +\infty}f(x)\geq \lim_{x\rightarrow +\infty}(f(y)+f'(y)(x-y)) \Rightarrow \ell\geq f(y)+f'(y)(\lim_{x\rightarrow +\infty}x-y) \\ & \Rightarrow \ell\geq f(y)+f'(y)(+\infty-y) \Rightarrow \ell\geq +\infty\end{align*}
Does this help us? (Wondering)
 

Klaas van Aarsen

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Mar 5, 2012
8,713
So we have the following:
\begin{align*}f(x)\geq f(y)+f'(y)(x-y)&\Rightarrow \lim_{x\rightarrow +\infty}f(x)\geq \lim_{x\rightarrow +\infty}(f(y)+f'(y)(x-y)) \Rightarrow \ell\geq f(y)+f'(y)(\lim_{x\rightarrow +\infty}x-y) \\ & \Rightarrow \ell\geq f(y)+f'(y)(+\infty-y) \Rightarrow \ell\geq +\infty\end{align*}
Does this help us?
We have an expression with $f(y)$, $f'(y)$, and $\ell$.
And we already know that $\ell\in\mathbb R$, don't we? So it can't be $\pm\infty$ either.

What can we say about $f'(y)$ for the statement to hold? (Wondering)
 

mathmari

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Apr 14, 2013
4,028
We have an expression with $f(y)$, $f'(y)$, and $\ell$.
And we already know that $\ell\in\mathbb R$, don't we? So it can't be $\pm\infty$ either.

What can we say about $f'(y)$ for the statement to hold? (Wondering)
So that we get that $\ell$ is not infinity we have to get an undefined form, one such form is when infinity is multiplied with zero, so $f'(y)$ must be equal to $0$.

Is this correct? (Wondering)
 

Klaas van Aarsen

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Mar 5, 2012
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So that we get that $\ell$ is not infinity we have to get an undefined form, one such form is when infinity is multiplied with zero, so $f'(y)$ must be equal to $0$.

Is this correct?
That is a possibility yes.
What happens if $f'(y)$ is negative? (Wondering)
 

mathmari

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Apr 14, 2013
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That is a possibility yes.
What happens if $f'(y)$ is negative? (Wondering)
Ah $(-\infty)\cdot (+\infty)$ is also an undefined form, isn't it?

So $f'(y)$ is either $0$ or $-\infty$, right? (Wondering)
 

Klaas van Aarsen

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Mar 5, 2012
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Ah $(-\infty)\cdot (+\infty)$ is also an undefined form, isn't it?

So $f'(y)$ is either $0$ or $-\infty$, right?
Then the inequality also holds yes.
What if fill in, say, $f'(y)=-1$ in the inequality? Would it satisfy it? (Wondering)
 

mathmari

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Apr 14, 2013
4,028
Then the inequality also holds yes.
What if fill in, say, $f'(y)=-1$ in the inequality? Would it satisfy it? (Wondering)
Yes, because then from $\ell\geq f(y)+f'(y)(+\infty-y)$ we get $-\infty$ at the right side of the inequality, and so $\ell$ can be real.

Therefore it must hold that $f'(y)\leq 0$, or not? (Wondering)
 

Klaas van Aarsen

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Mar 5, 2012
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Yes, because then from $\ell\geq f(y)+f'(y)(+\infty-y)$ we get $-\infty$ at the right side of the inequality, and so $\ell$ can be real.

Therefore it must hold that $f'(y)\leq 0$, or not?
Yep. (Nod)
 

mathmari

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Apr 14, 2013
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Ok! That means that $f$ is monotone and escpecially descreasing, right? To get that $f$ is strictly monotone, do we have to get $f'(x)<0$ instead of $f'(x)\leq 0$ ? (Wondering)
 

Klaas van Aarsen

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Mar 5, 2012
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Ok! That means that $f$ is monotone and escpecially descreasing, right? To get that $f$ is strictly monotone, do we have to get $f'(x)<0$ instead of $f'(x)\leq 0$ ?
Indeed. (Thinking)
 

mathmari

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Apr 14, 2013
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Klaas van Aarsen

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Mar 5, 2012
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I don't see how we get the strict inequality. Is maybe the result wrong and it should be monotone instead of strictly monotone? (Wondering)
I believe so yes.
Consider $f(x)=\ell$. It satisfies all conditions, doesn't it? (Wondering)
And it is monotone instead of strictly monotone.

To get strictly monotone we need $f$ to be strictly convex. (Thinking)
 

mathmari

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Apr 14, 2013
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I believe so yes.
Consider $f(x)=\ell$. It satisfies all conditions, doesn't it? (Wondering)
And it is monotone instead of strictly monotone.

To get strictly monotone we need $f$ to be strictly convex. (Thinking)
OK, so it's graph is a decreasing function, right? Or can we say something more specifically? (Wondering)
 

Klaas van Aarsen

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Mar 5, 2012
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OK, so it's graph is a decreasing function, right? Or can we say something more specifically?
Yep. (Nod)
And no, nothing more specific.
 

mathmari

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Apr 14, 2013
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Ok!!

As for the second question, why is the limit always zero?
(Wondering)

If $\displaystyle{\lim_{x\rightarrow +\infty}f''(x)}$ exists, then $\displaystyle{\lim_{x\rightarrow +\infty}f''(x)=\lim_{x\rightarrow +\infty}f'(x)=0}$ (i.e. we don't assume that $\displaystyle{\lim_{x\rightarrow +\infty}f'(x)}$ exists).
 

Klaas van Aarsen

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Mar 5, 2012
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Ok!!

As for the second question, why is the limit always zero?
Let's take a look at a couple of examples.

If $f(x)=x$, then $f'(x)=1$ and $f''(x)=0$.
So $\lim\limits_{x\to +\infty}f'(x)\ne 0$, isn't it? (Worried)

If $f(x)=x^2$, then $f'(x)=2x$ and $f''(x)=2$.
So $\lim\limits_{x\to +\infty}f''(x)$ exists, but it isn't $0$ is it? (Worried)

Is there something missing? (Wondering)
 

mathmari

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Apr 14, 2013
4,028
Let's take a look at a couple of examples.

If $f(x)=x$, then $f'(x)=1$ and $f''(x)=0$.
So $\lim\limits_{x\to +\infty}f'(x)\ne 0$, isn't it? (Worried)

If $f(x)=x^2$, then $f'(x)=2x$ and $f''(x)=2$.
So $\lim\limits_{x\to +\infty}f''(x)$ exists, but it isn't $0$ is it? (Worried)

Is there something missing? (Wondering)
Ahh ok!

And if we assume that $\displaystyle{\lim_{x\rightarrow +\infty}f(x)=\ell\in\mathbb{R}}$ ? (Wondering)
 

Klaas van Aarsen

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Mar 5, 2012
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Ahh ok!

And if we assume that $\displaystyle{\lim_{x\rightarrow +\infty}f(x)=\ell\in\mathbb{R}}$ ? (Wondering)
Then we have:
How about:
$$\lim_{x\to +\infty} f'(x) = \lim_{x\to +\infty} \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}
= \lim_{h\to 0}\lim_{x\to +\infty} \frac{f(x+h)-f(x)}{h}
= \lim_{h\to 0} \frac{\ell-\ell}{h} = 0
$$
don't we? (Thinking)

It follows that the limit of $f''(x)$ is also $0$, doesn't it?
Then there is no need to start with the assumption that it is, is it? (Bandit)