Q. 33 - Calculus

Question

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

$\underset{x\to 2}{\lim } \frac{4+2x}{x+2}$.

Step-by-Step Solution

Answer

The value of $\underset{x\to 2}{\lim } \frac{4+2x}{x+2}$ is, $2$.

Step 1 . Given information

$\underset{x\to 2}{\lim } \frac{4+2x}{x+2}$.

Step 2 . The limit is calculated as below:

$\underset{x\to 2}{\lim } \frac{4+2x}{x+2}= \frac{\underset{x\to 2}{\lim }(4+2x)}{\underset{x\to 2}{\lim }(x+2)}$ [Quotient rule]

                    $=\frac{\underset{x\to 2}{\lim }\left(4\right)+\underset{x\to 2}{\lim }\left(2x\right)}{\underset{x\to 2}{\lim }\left(x\right)+\underset{x\to 2}{\lim }\left(2\right)}$ [Sum rule]

                     $$\begin{gathered} =\frac{4+2\left(2\right)}{2+2} \\ =\frac{8}{4} \\ =2 \end{gathered}$$