Chapter 2 - Limits - 2.5 Evaluating Limits Algebraically - Exercises - Page 73: 47

$$\lim _{t \rightarrow-1}(4 t-2 a t+3 a) =5a-4$$

Given $$\lim _{t \rightarrow-1}(4 t-2 a t+3 a), \ \ \ a \ \ \text{is constant}$$ let $$ f(t) = (4 t-2 a t+3 a)$$ Since, we have $$ f(-1)= (-4+2 a+3 a)=5a-4$$ So, we get \begin{aligned} L&=\lim _{t \rightarrow-1}(4 t-2 a t+3 a)\\ &= (-4+2 a+3 a)\\ &=5a-4 \end{aligned}