I've got these two problems.
Any help would be appreciated!
Thanks!
$\displaystyle f(x)=\begin{cases}
\frac{x^2+x-12}{x-3}=\frac{(x+4)(x-3)}{(x-3)}=(x+4) \mbox{ if } x < 3 \\
ae^{x-3} \mbox{ if } x \ge 3
\end{cases}$
so to be cont at x=3 we need
$\displaystyle (3+4)=ae^{3-3} \iff 7=a$
for part b use the limit defintion.