Answer
$a=-13$
$\lim\limits_{x \to -1} g(x)=6$
Work Step by Step
We are given the function:
$g(x)=\begin{cases}
x^2-5x,\text{ if }x\leq -1\\
ax^3-7,\text{ if }x>-1
\end{cases}$
For $\lim\limits_{x \to -1} g(x)$ to exist we must have:
$\lim\limits_{x \to -1^-} g(x)=\lim\limits_{x \to -1^+} g(x)$ (1)
Determine the left hand limit and the right hand limit:
$\lim\limits_{x \to -1^-} g(x)=\lim\limits_{x \to -1^-} (x^2-5x)=(-1)^2-5(-1)=1+5=6$
$\lim\limits_{x \to -1^+} g(x)=\lim\limits_{x \to -1^+} (ax^3-7)=a(-1)^3-7=-a-7$
For the condition (1) to be checked, we must have:
$6=-a-7\Rightarrow 6+7=-a\Rightarrow a=-13$
The value of the limit is:
$\lim\limits_{x \to -1} g(x)=6$
