Q. 39 - Calculus

Question

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

$\underset{x\to \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\lim }\frac{4x-2}{6x^2+x-2}$.

Step-by-Step Solution

Answer

The value of $\underset{x\to \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\lim }\frac{4x-2}{6x^2+x-2}$ is, $\frac{4}{7}$.

Step 1 . Given information

$\underset{x\to \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\lim }\frac{4x-2}{6x^2+x-2}$.

Step 2 . The limit is calculated as below:

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

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

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