Chapter 1.3, Problem 36E

To determine

To Show: The given function is continuous by using theorem $7$ .

Answer to Problem 36E

The given function is continuous at the domain $R-\frac{\pi }{2}$ .

Explanation of Solution

Given:

  $f\left(x\right)=\tan \left(\frac{x^2}{x^2+4}\right)$

Calculation:

Theorem 7:

If $f\left(x\right)$ is continuous at $c$ and $g\left(x\right)$ is continuous at $f\left(c\right)$ , then the composite function $g°f\left(x\right)$ is continuous.

Given function, $f\left(x\right)=\tan \left(\frac{x^2}{x^2+4}\right)$ is a composite function where $f\left(x\right)=\frac{x^2}{x^2+4}$ and $g\left(x\right)=\tan x$ .

Show$f\left(x\right)=\frac{x^2}{x^2+4}$is continuous:

Set the denominator in $\frac{x^2}{x^2+4}$ equal to $0$ to find where the expression is undefined.

  $x^2+4=0$

Solve for $x$ :

  $x^2=-4$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

  $\begin{array}{l}x=\pm \sqrt{-4}\\ x=+2i,-2i\end{array}$

The domain is all real numbers.

Interval Notation: $\left(-\infty ,\infty \right)$

Set-Builder Notation: $\left\{x|x\in ℝ\right\}$

Since the domain is all real numbers, $f\left(x\right)=\frac{x^2}{x^2+4}$ is continuous over all real numbers.

Show$g\left(x\right)=\tan x$is continuous:

Set the argument in $\tan \left(x\right)$ equal to $\frac{\pi }{2}+\pi n$ to find where the expression is undefined.

  $x=\frac{\pi }{2}+\pi n$ for any integer $n$

The domain is all values of $x$ that make the expression defined.

Set-Builder Notation: $\left\{x\mid x\ne \frac{\pi }{2}+\pi n\right\},$ for any integer $n$ . So, $\tan x$ is a continuous function except at $x=\frac{\pi }{2}$ .

Hence, the given function is continuous at the domain $R-\frac{\pi }{2}$ .