Q. 40 - Calculus

Question

Calculate each of the limits in Exercises $29-70$.

$\underset{x\to \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\lim }\frac{6x^2+x-2}{3x+2}$.

Step-by-Step Solution

Answer

The value of $\underset{x\to \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\lim }\frac{6x^2+x-2}{3x+2}$ is, $-\frac{7}{3}$.

Step 1 . Given information

$\underset{x\to \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\lim }\frac{6x^2+x-2}{3x+2}$.

Step 2 . The limit is calculated as below:

$$\begin{gathered} \underset{x\to -2/3}{\lim }\frac{6x^2+x-2}{3x+2}=\underset{x\to -2/3}{\lim }\frac{6x^2+4x-3x-2}{3x+2} \\ =\underset{x\to -2/3}{\lim }\frac{2x(3x+2)-1(3x+2)}{3x+2} \\ =\underset{x\to -2/3}{\lim }\frac{(2x-1)(3x+2)}{(3x+2)} \\ =\underset{x\to -2/3}{\lim }(2x-1) \end{gathered}$$ 

Step 3 . Let us use the Sum rule in the obtained equation.

$$\begin{gathered} \underset{x\to -2/3}{\lim }(2x-1)=\underset{x\to -2/3}{\lim }\left(2x\right)-\underset{x\to -2/3}{\lim }\left(1\right) \\ =2\left(-\frac{2}{3}\right)-1 \\ =-\frac{4}{3}-1 \\ =-\frac{7}{3} \end{gathered}$$