Limits & properties

mathmari

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Hey!!

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

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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Hey mathmari !!

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?

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?

mathmari

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MHB Site Helper

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?
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 $\displaystyle{\lim\limits_{x\to +\infty}f'(x)=0}$ ?

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

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?
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

MHB Seeker
Staff member
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?

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

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?

mathmari

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MHB Site Helper
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?
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.

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

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

Klaas van Aarsen

MHB Seeker
Staff member
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?

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.

mathmari

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MHB Site Helper
Sounds like a plan.
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?

Klaas van Aarsen

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Staff member
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?

mathmari

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MHB Site Helper
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?
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?

Klaas van Aarsen

MHB Seeker
Staff member
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?

mathmari

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

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

Klaas van Aarsen

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Staff member
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?

mathmari

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MHB Site Helper
Then the inequality also holds yes.
What if fill in, say, $f'(y)=-1$ in the inequality? Would it satisfy it?
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?

Klaas van Aarsen

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Staff member
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.

mathmari

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MHB Site Helper
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 $f'(x)\leq 0$ ?

Klaas van Aarsen

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Staff member
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.

mathmari

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MHB Site Helper
I don't see how we get the strict inequality. Is maybe the result wrong and it should be monotone instead of strictly monotone?

Klaas van Aarsen

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

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

mathmari

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

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

Klaas van Aarsen

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Staff member
OK, so it's graph is a decreasing function, right? Or can we say something more specifically?
Yep.
And no, nothing more specific.

mathmari

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Ok!!

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

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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Staff member
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?

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?

Is there something missing?

mathmari

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MHB Site Helper
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?

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?

Is there something missing?
Ahh ok!

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

Klaas van Aarsen

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Staff member
Ahh ok!

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

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?