Exercise 5.19

For (a) and (b), we need to find an algebraic expression that relates $D_n$ to the difference quotients found in the definition of $f^{\prime }(0)$, $\left(f({\alpha }_n)-f(0)\right)∕{\alpha }_n$ and $\left(f({\beta }_n)-f(0)\right)∕{\beta }_n$ in such a way that we can safely let $n\to \infty$. To simplify the algebra, replace $f$ with $F=f-f(0)$ so that $F(0)=0$. We start with

To get $D_n$, we need to multiply by ${\beta }_n∕({\beta }_n-{\alpha }_n)$, and this gives us what we need:

Rearranging and substituting back $f-f(0)$ for $F$, we get

For (b), the ${\beta }_n∕({\beta }_n-{\alpha }_n)$ factor is assumed to be bounded, and for part (a) it is bounded by 1. In either case, we can pass to a subsequence where the factor converges to a finite limit, so letting $n\to \infty$ we get $\lim <!--\; FUNCTION\; APPLICATION\; -->D_n=f^{\prime }(0)$.

For part (c), we can apply Theorem 5.10 to get $D_n=f^{\prime }({\gamma }_n)$ for some ${\gamma }_n$ between ${\alpha }_n$ and ${\beta }_n$. Since $\lim <!--\; FUNCTION\; APPLICATION\; -->{\gamma }_n=0$ and $f^{\prime }$ is continuous, we get $\lim <!--\; FUNCTION\; APPLICATION\; -->D_n=f^{\prime }(0)$.

Let $f$ be the function in Example 5.6(b), $f(x)=x^2\sin <!--\; FUNCTION\; APPLICATION\; -->(1∕x)$ for $x\ne 0$ and $f(0)=0$. It was shown that $f^{\prime }(0)=0$, although $f^{\prime }$ is not continuous at 0. Let

Then $\sin <!--\; FUNCTION\; APPLICATION\; -->(1∕{\alpha }_n)=-1$ and $\sin <!--\; FUNCTION\; APPLICATION\; -->(1∕{\beta }_n)=1$, so that $f({\alpha }_n)=-{\alpha }_n^2$ and $f({\beta }_n)={\beta }_n^2$. Hence

which tends to $2∕\pi$ as $n\to \infty$.