Question

In Exercises 1 and \(2,\) determine whether \(f(x)\) approaches \(\infty\) or \(-\infty\) as \(x\) approaches -2 from the left and from the right. $$ f(x)=\sec \frac{\pi x}{4} $$

Short answer

As \(x\) approaches -2 from the left, \(f(x)\) approaches \(-\infty\), and as \(x\) approaches -2 from the right, \(f(x)\) approaches \(\infty\).

Step-by-step solution

Calculate limit as \(x\) approaches -2 from left

Evaluate the function \(f(x) = \sec\left(\frac{\pi x}{4}\right)\) at a value slightly less than -2, such as -2.001. When \(x=-2.001,\) then \(f(x) = \sec\left(-\frac{\pi (2.001)}{4}\right)\). This is a large negative number, meaning that as \(x\) approaches -2 from the left, \(f(x)\) approaches \(-\infty\).\n

Calculate limit as \(x\) approaches -2 from right

Evaluate the function \(f(x) = \sec\left(\frac{\pi x}{4}\right)\) at a value slightly more than -2, such as -1.999. When \(x=-1.999,\) then \(f(x) = \sec\left(-\frac{\pi (1.999)}{4}\right)\). This results in a large positive number, implying that as \(x\) approaches -2 from the right, \(f(x)\) approaches \(\infty\).\n

Function behavior summary

As we derived in the step by step process, when \(x\) approaches -2 from the left, \(f(x)\) approaches \(-\infty\) and when \(x\) approaches -2 from the right, \(f(x)\) approaches \(\infty\). This indicates that there a vertical asymptote at \(x=-2\).