Q. 31 - Calculus

Question

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

$\underset{x\to -1}{\lim } \left(x^2-\frac{3x}{x+2}\right)$.

Step-by-Step Solution

Answer

The value of $\underset{x\to -1}{\lim } \left(x^2-\frac{3x}{x+2}\right)$ is, $4$.

Step 1 . Given information

$\underset{x\to -1}{\lim } \left(x^2-\frac{3x}{x+2}\right)$.

Step 2 . The limit is calculated as below:

$\underset{x\to -1}{\lim } \left(x^2-\frac{3x}{x+2}\right)=\underset{x\to -1}{\lim } \left(x^2\right)-\underset{x\to -1}{\lim } \left(\frac{3x}{x+2}\right)$ [Sum rule]

                               $=\underset{x\to -1}{\lim } \left(x^2\right)-\frac{\underset{x\to -1}{\lim } \left(3x\right)}{\underset{x\to -1}{\lim } \left(x+2\right)}$ [Quotient rule]

                               $$\begin{gathered} ={\left(-1\right)}^2-\frac{\left(-3\right)}{\left(-1+2\right)} \\ =1-\frac{\left(-3\right)}{1} \\ =1+3 \\ =4 \end{gathered}$$