Answer
$-1$
Work Step by Step
For the function to be continuous at $x=4$, the left and right side limits need to be equal at this point, we have:
$\lim_{x\to4^-}\frac{x^2-9}{x+3}=\lim_{x\to4^-}(x-3)=1$
$\lim_{x\to4^+}(kx+5)=4k+5$
Let $4k+5=1$, we get $k=-1$
