Q. 38 - Calculus

Question

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

$\underset{x\to -3}{\lim } \frac{3x^2+8x-3}{x+3}$.

Step-by-Step Solution

Answer

The value of $\underset{x\to -3}{\lim } \frac{3x^2+8x-3}{x+3}$ is $-10$.

Step 1 . Given information

$\underset{x\to -3}{\lim } \frac{3x^2+8x-3}{x+3}$.

Step 2 . The limit is calculated as below:

$$\begin{gathered} \underset{x\to -3}{\lim } \frac{3x^2+8x-3}{x+3}=\underset{x\to -3}{\lim } \frac{3x^2+9x-x-3}{x+3} \\ =\underset{x\to -3}{\lim } \frac{3x\left(x+3\right)-1\left(x+3\right)}{x+3} \\ =\underset{x\to -3}{\lim } \frac{\left(x+3\right)\left(3x-1\right)}{(x+3)} \\ =\underset{x\to -3}{\lim } \left(3x-1\right) \end{gathered}$$

                                $=\underset{x\to -3}{\lim } \left(3x\right)-\underset{x\to -3}{\lim }\left(1\right)$  [Sum rule]

                                $$\begin{gathered} =3\left(-3\right)-1 \\ =-9-1 \\ =-10 \end{gathered}$$