Answer
$$\lim_{x \to -4}(\frac{1}{2}x-1)=-3$$For the proof by using the $\epsilon - \delta$ definition, see below.
Work Step by Step
$$\lim_{x \to -4} (\frac{1}{2}x-1)=\frac{1}{2}(-4)-1=-3$$ Now, we want to prove this limit by using $\epsilon - \delta$ definition; that is, we must show that for each $\epsilon >0$, there exists a $\delta >0$ such that $|(\frac{1}{2}x-1)-(-3)|< \epsilon$ whenever $|x-(-4)|< \delta$.
Now, we have$$|(\frac{1}{2}x-1)-(-3)|=|\frac{1}{2}x+2|=|\frac{1}{2}(x+4)|=\frac{1}{2}|x+4|.$$So, by choosing $\delta =2\epsilon$ we conclude that$$|x-(-4)|=|x+4|< \delta =2 \epsilon \quad \Rightarrow \quad |(\frac{1}{2}x-1)-(-3)|=\frac{1}{2}|x+4|$$.
