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. $$ f(x)=2\left|\frac{x}{x^{2}-4}\right| $$

Short answer

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

Step-by-step solution

Breaking down the absolute value function

The absolute value function breaks down into two cases: it equals its input when the input is greater than or equal zero, and it equals the negation when the input is less than zero. So, \(f(x)=2\left| \frac{x}{x^{2}-4} \right|\) becomes \(f(x)=2\frac{x}{x^{2}-4}\) when \(x>0\) and \(f(x)=-2\frac{x}{x^{2}-4}\) when \(x<0\).

Approaching -2 from the right

Since we're dealing with \(x\) approaching -2 from the right, \(x\) will be slightly more than -2 but still less than 0, hence, \(f(x)=-2\frac{x}{x^{2}-4}\) is considered. We substitute -2 into \(x\) and find \(f(x)\) approaches \(-\infty\) as this part of the function is negative.

Approaching -2 from the left

When \(x\) is approaching -2 from the left, \(x\) is slightly less than -2 and hence less than 0. Thus, \(f(x)=-2\frac{x}{x^{2}-4}\) is considered. Again, substituting -2 for \(x\) shows that \(f(x)\) approaches \(-\infty\), as this portion of the function is also negative.