I've got these two problems.

Any help would be appreciated!

Thanks!

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- Apr 1st 2008, 10:47 PMBACONATORFinding the limit
I've got these two problems.

Any help would be appreciated!

Thanks! - Apr 1st 2008, 10:51 PMTheEmptySet
- Apr 1st 2008, 10:58 PMTheEmptySet
$\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.