Answer
$$\lim_{x \to 3}(\frac{3}{4}x+1)=\frac{13}{4}$$For the proof by using the $\epsilon - \delta$ definition, see below.
Work Step by Step
$$\lim_{x \to 3} (\frac{3}{4}x+1)=\frac{3}{4}(3)+1=\frac{13}{4}$$ 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 $|(\frac{3}{4}x+1)-\frac{13}{4}|< \epsilon$ whenever $|x-3|< \delta$.
Now, we have$$|(\frac{3}{4}x+1)-\frac{13}{4}|=|\frac{3}{4}(x-3)|=\frac{3}{4}|x-3|.$$So, by choosing $\delta =\frac{4}{3}\epsilon$ we conclude that$$|x-3|< \delta =\frac{4}{3} \epsilon \quad \Rightarrow \quad |(\frac{3}{4}x+1)-\frac{13}{4}|=\frac{3}{4}|x-3|$$.
