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Chapter 1: Problem 26

Find the \(x\) -values (if any) at which \(f\) is not continuous. Which of the discontinuities are removable? $$ f(x)=\frac{x}{x^{2}-1} $$

Short Answer

Expert verified
The function \(f(x) = \frac{x}{x^{2}-1}\) is not continuous at \(x = 1\) and \(x = -1\). The discontinuities at these points are not removable.

Step by step solution

01

Find Values That Make the Denominator Zero

To find where the function \(f(x) = \frac{x}{x^{2}-1}\) is not defined, we need to find the values of x that make the denominator zero. So, we solve the equation \(x^2 - 1 = 0\). The solutions are \(x = 1\) and \(x= -1\). Therefore, the function is not defined at \(x= -1\) and \(x= 1\). These are the points of discontinuity.
02

Determine If Discontinuities Are Removable

A discontinuity is removable if the limit as x approaches the point of discontinuity exists. The limits here would be \(\lim_{x \to 1} f(x)\) and \(\lim_{x \to -1} f(x)\). As these limits are \( \frac{1}{1-1} = \frac{1}{0}\) and \(\frac{-1}{-1-1} = \frac{-1}{0}\) respectively and they are undefined, the discontinuities at \(x = 1\) and \(x = -1\) are not removable.

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