Assertions about the derivative how do I show the following assertions:


*

*Let $f\colon I\to \mathbb{R}$ differentiable, let $a_n,b_n$ be sequences of real numbers such that $a_n \leq c \leq b_n$, then $f'(c)= \lim \frac{f(b_n)-f(a_n)}{b_n - a_n}$

*Let $f\colon[a,b]\to \mathbb{R}$ differentiable in $(a,b)$ with $f'$ bounded,if $f$ has the intermediate value property then $f$ is continuous in $[a,b]$.

*Let $f(x)=\sin\left(\frac 1x\right) ,x\neq 0$ and $f(0)=0$, then $f$ has the intermediate value property.


thanks.
 A: *

*We can write 
\begin{align*}
\frac{f(b_n)-f(a_n)}{b_n-a_n}&=\frac{f(b_n)-f(c)}{b_n-c}\frac{b_n-c}{b_n-a_n}+\frac{f(c)-f(a_n)}{c-a_n}\frac{c-a_n}{b_n-a_n}\\
&=\frac{f(b_n)-f(c)}{b_n-c}\left(1+\frac{a_n-c}{b_n-a_n}\right)+\frac{f(c)-f(a_n)}{c-a_n}\frac{c-a_n}{b_n-a_n}\\
&=\frac{f(b_n)-f(c)}{b_n-c}+\frac{c-a_n}{b_n-a_n}\left(\frac{f(c)-f(a_n)}{c-a_n}-\frac{f(b_n)-f(c)}{b_n-c}\right),
\end{align*}
and you can conclude since $\frac{c-a_n}{b_n-a_n}$ is bounded. 

*We only have to show that $f$ is continuous at $a$ and $b$. I will do it for $a$. Let $\{x_n\}$ a sequence which converges to $a$. If $f(x_n)$ doesn't converge to $a$ then we can find a $\delta>0$ and a subsequence $\{x_{n_k}\}$ such that $|f(x_{n_k})-f(a)|\geq \delta$. We have two cases: $f(x_{n_k})-f(a)\geq \delta$ for infinitely many $k$ or $f(x_{n_k})-f(a)<\delta$. In the first case, after taking a subsequence denoted by the same way, we can find $a\leq y_{n_k}\leq x_{n_k}$ such that $f(y_{n_k})=f(x_{n_k})-\delta$, since $f(a)\leq f(x_{n_k})-\delta\leq f(x_{n_k})$. We have 
$$\delta=|f(x_{n_k})-f(y_{n_k})|\leq \int_{x_{n_k}}^{y_{n_k}}f'(t)dt\leq \sup |f'||x_{n_k}-y_{n_k}|\leq \sup |f'||x_{n_k}-a|,$$
which is a contradiction. The argument is quite the same for for the second case and the continuity at $b$.

*Let $a<b$. If $a<b<0$ or $0<a<b$ the property is true since $f$ is continuous on $[a,b]$. If $a<0<b$ each value between $-1$ and $1$ is reached between $a$ and $b$ and if $a=0<b$, then $\left[0,\frac 1{2n\pi}\right]\subset [a,b]$ for some 
$n$ and $f\left(\left[0,\frac 1{2n\pi}\right]\right)=[-1,1]$. The case $a<b=0$ is the same since $f$ is odd.

