# Math Help - Prove this derivative is continuous

1. ## Prove this derivative is continuous

Let $U$ be a neighborhood of 0 in $\mathbb {R}$.

Suppose that $f:U \rightarrow \mathbb {R}$ is twice differentiable , f' and f'' are continuous, and $f(0)=0$.

Define $g: U \rightarrow \mathbb {R}$ by $g(x)= \frac {f(x)}{x}$ if $x \neq 0$ and $\lim _{ h \rightarrow 0 } \frac {f(h)}{h}$ if $x = 0$

Show that g' exists and is continuous on U.

Proof so far.

Claim: g' exists on U.

By the division theorem, we know that $f(x)=xg(x)$, implies that for $x \neq 0$, we would have $f'(x)=xg'(x)+g(x)$, so we have $g'(x)= \frac {f'(x)- \frac {f(x)}{x} }{x}$.

For $x=0$, we have $g'(0)= \lim _{h \rightarrow 0 } \frac {g(h)-g(0)}{h} = \lim _{h \rightarrow 0 } \frac { \frac {f(h)}{h} - \frac {f(h)}{h}}{h} = 0$

Claim: g' is continuous on U.

Write $f(x)=f(0)+f'(0)x+f'(t)x^2 \ \ \ \ \ 0, so we have $f(x)=f'(0)x+f''(xh)x^2 \ \ \ \ \ 0

In the case that $x \neq 0$:

Given $\epsilon > 0$, pick $\delta > 0$, then for each $x \in U$ and $|x-x_0|< \delta$, we have $|g'(x)-g'(x_0)|=| \frac {f'(x)- \frac {f(x)}{x}}{x} - \frac {f'(x_0)- \frac {f(x_0)}{x}}{x}|$ $= \frac {|f'(x)-f'(x_0)|}{x} + \frac {| \frac {f(x)}{x}- \frac {f(x_0)}{x_0}|}{x}$

I have a feeling I'm very wrong as I'm stuck at this point, any hints? Thank you.

Let $U$ be a neighborhood of 0 in $\mathbb {R}$.

Suppose that $f:U \rightarrow \mathbb {R}$ is twice differentiable , f' and f'' are continuous, and $f(0)=0$.

Define $g: U \rightarrow \mathbb {R}$ by $g(x)= \frac {f(x)}{x}$ if $x \neq 0$ and $\lim _{ h \rightarrow 0 } \frac {f(h)}{h}$ if $x = 0$

Show that g' exists and is continuous on U.

Proof so far.

Claim: g' exists on U.

By the division theorem, we know that $f(x)=xg(x)$, implies that for $x \neq 0$, we would have $f'(x)=xg'(x)+g(x)$, so we have $g'(x)= \frac {f'(x)- \frac {f(x)}{x} }{x}$.

For $x=0$, we have $g'(0)= \lim _{h \rightarrow 0 } \frac {g(h)-g(0)}{h} = \lim _{h \rightarrow 0 } \frac { \frac {f(h)}{h} - \frac {f(h)}{h}}{h} = 0$
- Note that in fact $g(0)=f'(0)$ (the limit defining $g(0)$ exists because $f$ is differentiable at 0).
- Your last limit is incorrect. You can't just replace $g(0)$ by $\frac{f(h)}{h}$. You could write instead: for any $h\neq 0$ in $U$,
$\frac{g(h)-g(0)}{h}=\frac{\frac{f(h)}{h}-f'(0)}{h}=\frac{f(h)-hf'(0)}{h^2}$,
and use Taylor's theorem to show that the previous quantity converges to $\frac{f''(0)}{2}$.

Claim: g' is continuous on U.
First, you wrote that $g'(x)= \frac {f'(x)- \frac {f(x)}{x} }{x}$, so it should be clear (from the hypotheses) that $g'$ is continuous on $U\setminus\{0\}$. The only thing to check is that $\lim_{h\to 0}g'(h)=g'(0)$ ( $=\frac{f''(0)}{2}$ because of the previous question) .

For any $h\neq 0$ (and $h\in U$), $g'(h)=\frac{hf'(h)-f(h)}{h^2}$, so you have to apply Taylor's theorem to $f$ and to $f'$.

3. My work so far:

1. Use the Taylor's Theorem to show that $\frac{g(h)-g(0)}{h}=\frac{\frac{f(h)}{h}-f'(0)}{h}=\frac{f(h)-hf'(0)}{h^2}$ converges to $
\frac{f''(0)}{2}
$

Proof.

By the Taylor's Theorem, write $f(x)=f(0)+f'(0)x+f''(t)x^2 \ \ \ \ \ 0, since $f(0)=0$, we have $f(x)=f'(0)x+f''(t)x^2$, implies that $f(h)=f'(0)h+f''(t)h^2 \ \ \ \ \ 0.

Then $g'(0)= \lim _{h \rightarrow 0} \frac {f(h)-hf''(0)}{h^2}$
$= \lim _{h \rightarrow 0} \frac {f'(0)h+f''(t)h^2-f''(0)h}{h^2}= \lim _{h \rightarrow 0}f''(t)$

Now, I know that $t \rightarrow 0$ as h approaches to 0.

So I have $g'(0) = f''(0)$.

But I don't have $\frac {f''(0)}{2}$, what did I do wrong?

2. Show that g' is continuous on U.

First, since
$
g'(x)= \frac {f'(x)- \frac {f(x)}{x} }{x}
$
, both f and f' are continuous, and $x \neq 0$, so g' must be continuous on $
U\setminus\{0\}
$
since the sum of two continuous functions is also continuous. (Is this reason enough?)

Now, to show that g' is continuous on $U \setminus \{ 0 \}$.

By the Taylor's Theorem, write $f'(h)=f'(0)+hf''(s), \ \ \ \ \ 0

Then $\lim _{h \rightarrow 0 } g'(h) = \lim _{h \rightarrow 0 } \frac {hf'(h)-f(h)}{h^2}$ $= \lim _{h \rightarrow 0 } \frac {h[f'(0)+hf''(s)]-[f(0)+hf'(0)+h^2f''(t)]}{h^2}$
$= \lim _{h \rightarrow 0 } f''(s)-f''(t)- \frac {f(0)}{h^2}$

So as h approaches 0, both s and t approaches 0 as they are sandwiched in between 0 and h, so $f''(s)-f''(t) \rightarrow f''(0)-f''(0) =0$

So all I have left is: $\lim _{h \rightarrow 0 } g'(h) = \lim _{h \rightarrow 0 } \frac {f(0)}{h^2}$

I must have done something wrong again, because this expression doesn't look converging to g'(0).

Thank you very much for your help!!!

4. The only problem is a $\frac{1}{2}$ factor in the Taylor's theorem (cf. wikipedia or whereever):

My work so far:

1. Use the Taylor's Theorem to show that $\frac{g(h)-g(0)}{h}=\frac{\frac{f(h)}{h}-f'(0)}{h}=\frac{f(h)-hf'(0)}{h^2}$ converges to $
\frac{f''(0)}{2}
$

Proof.

By the Taylor's Theorem, write $f(x)=f(0)+f'(0)x+{\color{red}\frac{1}{2}}f''(t)x^2 \ \ \ \ \ 0, since $f(0)=0$, we have $f(x)=f'(0)x+{\color{red}\frac{1}{2}}f''(t)x^2$, implies that $f(h)=f'(0)h+{\color{red}\frac{1}{2}}f''(t)h^2 \ \ \ \ \ 0.

Then $g'(0)= \lim _{h \rightarrow 0} \frac {f(h)-hf'(0)}{h^2}$
$= \lim _{h \rightarrow 0} \frac {f'(0)h+{\color{red}\frac{1}{2}}f''(t)h^2-f'(0)h}{h^2}= \lim _{h \rightarrow 0}{\color{red}\frac{1}{2}}f''(t)$

Now, I know that $t \rightarrow 0$ as h approaches to 0.

So I have $g'(0) = {\color{red}\frac{1}{2}}f''(0)$ because $\color{red}f''$ is assumed to be continuous.
2. Show that g' is continuous on U.

First, since
$
g'(x)= \frac {f'(x)- \frac {f(x)}{x} }{x}
$
, both f and f' are continuous, and $x \neq 0$, so g' must be continuous on $
U\setminus\{0\}
$
since the sum of two continuous functions is also continuous. (Is this reason enough?)

Now, to show that g' is continuous at 0.

By the Taylor's Theorem, write $f'(h)=f'(0)+hf''(s), \ \ \ \ \ 0

Then $\lim _{h \rightarrow 0 } g'(h) = \lim _{h \rightarrow 0 } \frac {hf'(h)-f(h)}{h^2}$ $= \lim _{h \rightarrow 0 } \frac {h[f'(0)+hf''(s)]-[f(0)+hf'(0)+{\color{red}\frac{1}{2}}h^2f''(t)]}{h^2}$
$= \lim _{h \rightarrow 0 } f''(s)-{\color{red}\frac{1}{2}}f''(t)- \frac {f(0)}{h^2}$
then, don't forget $f(0)=0$, and you'll be done.