# Thread: Piecewise function with Taylor's Theorem

1. ## Piecewise function with Taylor's Theorem

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 $g(0)$ exists and that g' exist and is continuous on U. Find $g(0),g'(0)$

Proof so far.

Find $g(0)$:

By the Taylor's Theorem, I can write $f(x)=f(0)+f'(0)x+f''(t)x^2=f'(0)x+f''(t)x^2$

Now, $g(0)= \lim _{h \rightarrow 0 } \frac {f(h)}{h} = \lim _{h \rightarrow 0 } \frac {f(h)-f(0)}{h} = f'(0)$

Is that right so far?

And how do I find g'(0)? 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 $g(0)$ exists and that g' exist and is continuous on U. Find $g(0),g'(0)$

Proof so far.

Find $g(0)$:

By the Taylor's Theorem, I can write $f(x)=f(0)+f'(0)x+f''(t)x^2=f'(0)x+f''(t)x^2$
No you can't, there has to be a remainder term.

Now, $g(0)= \lim _{h \rightarrow 0 } \frac {f(h)}{h} = \lim _{h \rightarrow 0 } \frac {f(h)-f(0)}{h} = f'(0)$

Is that right so far?
Since $f(0)=0$:

$
g(0)=\lim_{h \to 0} \frac{f(h)}{h}=\lim_{h \to 0} \frac{f(h)-f(0)}{h}=f'(0)
$

CB

3. Do I find g'(0) the same way then?

$g'(0) = \lim _{h \rightarrow 0 } \frac {g(h)-g(0)}{h} = \lim _{h \rightarrow 0 } \frac { \frac {f(h)}{h}-f'(0)}{h} = \frac {g(0)-g(0)}{h} = 0$

Is this right?