Proof. Let $h_{n} \rightarrow 0, h_{n} \neq 0$. Since $E$ is continuous, $E\left(h_{n}\right) \rightarrow E(0)=$ 1. Let $x_{n}=E\left(h_{n}\right)$; then $x_{n}>0, x_{n} \rightarrow 1$ and $x_{n} \neq 1$ (because $\left.h_{n} \neq 0\right)$, so by $9.5 .10$
$$
\frac{L\left(x_{n}\right)}{x_{n}-1} \rightarrow 1
$$
That is,
$$
\frac{h_{n}}{E\left(h_{n}\right)-1} \rightarrow 1
$$
whence the lemmn.
5CCM221A COURSE NOTES ：
(i) Show that $A$ is an odd function: $A(-x)=-A(x)$ for all $x \in \mathbb{R}$.
{Hint: $f$ is an even function.}
(ii) $A$ is strictly increasing. {Hint: $\left.A^{\prime}=f .\right}$
(iii) For every positive integer $k$,
$$
\frac{1}{1+k^{2}} \leq A(k)-A(k-1) \leq \frac{1}{1+(k-1)^{2}} .
$$
{Hint: Integrate $f$ over the interval
(iv) Let
$$
s_{n}=\sum_{k=1}^{n} \frac{1}{1+k^{2}} .
$$