187,188c187,188
< It follows that $g$ is concave up on $(- \infty, \, - 2)$,
< concave down on $( - 2, 4)$, and concave up on $(4, \infty)$.
---
> It follows that $g$ is concave up on $(- \infty, \, 0)$,
> concave down on $(0, 4)$, and concave up on $(4, \infty)$.
300,301c300,308
< This limit can be evaluated by substituting $x = 0$,
< and the result is~$k$.
---
> The last limit can be evaluated by substituting $x = 0$.
> That is,
> \[
> \lim_{x \to 0} \frac{e^{k \sin (x)} - 1}{\sin (x)}
>  = \lim_{x \to 0} \frac{e^{k \sin (x)} \cdot k \cos (x)}{\cos (x)}
>  = \frac{e^{k \sin (0)} \cdot k \cos (0)}{\cos (0)}
>  = \frac{e^{0} \cdot k \cdot1}{1}
>  = k.
> \]