Chapter 11, Problem 7RE

Solution

To calculate: The limit of $f\left(x\right)=\frac{x^2-3x-10}{x+2}$ at $x=-2$ .

The limit of the given function at $x=-2$ is $-7$ .

Given information: The given function is $f\left(x\right)=\frac{x^2-3x-10}{x+2}$ .

It can be observed that on substituting $x=-2$ into the formula, the denominator will be 0. So, factor the numerator and cancel out the common factors.

  $\begin{array}{c}f\left(x\right)=\frac{x^2-5x+2x-10}{x+2}\\ =\frac{x\left(x-5\right)+2\left(x-5\right)}{x+2}\\ =\frac{\left(x-5\right)\left(x+2\right)}{x+2}\\ =x-5\end{array}$

Substitute $x=-2$ into the obtained function and simplify to obtain the limit of given function at $x=-2$ .

  $\begin{array}{c}\underset{x\to -2}{\lim }f\left(x\right)=\underset{x\to -2}{\lim }\left(x-5\right)\\ =-2-5\\ =-7\end{array}$

Thus, the limit of the given function at $x=-2$ is $-7$ .

Draw the graph of the function $f\left(x\right)=\frac{x^2-3x-10}{x+2}$ to verify the obtained limit.

  

Figure (1)

From the graph, it can be observed that the function approaches $-7$ at $x=-2$ . Hence, the required limit is $-7$ .