## Calculus: Early Transcendentals (2nd Edition)

$a=-13$ $\lim\limits_{x \to -1} g(x)=6$
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: $\lim\limits_{x \to -1} g(x)=6$