Q. 37 - Calculus

Question

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

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

Step-by-Step Solution

Answer

The value of $\underset{x\to -3}{\lim } \frac{x+3}{3x^2+8x-3}$ is, $-\frac{1}{10}$.

Step 1 . Given information

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

Step 2 . The limit is calculated as below:

$$\begin{gathered} \underset{x\to -3}{\lim } \frac{x+3}{3x^2+8x-3}=\underset{x\to -3}{\lim } \frac{x+3}{3x^2+9x-x-3} \\ =\underset{x\to -3}{\lim } \frac{x+3}{3x\left(x+3\right)-1\left(x+3\right)} \\ =\underset{x\to -3}{\lim } \frac{x+3}{\left(3x-1\right)\left(x+3\right)} \\ =\underset{x\to -3}{\lim } \frac{1}{\left(3x-1\right)} \\ =\frac{1}{3\left(-3\right)-1} \\ =\frac{1}{-9-1} \\ \underset{x\to -3}{\lim } \frac{x+3}{3x^2+8x-3}=-\frac{1}{10} \end{gathered}$$