Question

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

Short answer

The function \(f(x)=\frac{|x+2|}{x+2}\) is not continuous at \(x = -2\) and the discontinuity is non-removable.

Step-by-step solution

Identifying discontinuity

First, let's look for the values at which \( f(x) = \frac{|x+2|}{x+2} \) is not defined. This function is undefined at \( x = -2 \), as the denominator becomes 0 at this point. Hence, \( -2 \) is a point of discontinuity.

Analyzing the discontinuity

Now, let's analyze the discontinuity at \( x = -2 \). In order to do this, we find the left and right limit as \( x \) approaches -2.\n\nAs \(x\) approaches -2 from the left (i.e., \(x < -2\)), the expression |x+2| evaluates to - (x + 2) (because for negative values inside absolute value, the result is the negative of those values), hence \(f(x) = \frac{- (x+2)}{x+2} = -1.\) \n\nAs \(x\) approaches -2 from the right (i.e., \(x > -2\)), the expression |x+2| evaluates to x+2 (because for non-negative values inside absolute value, the result is just those values), hence \(f(x) = \frac{x+2}{x+2} = 1.\) \n\nSince the left limit and the right limit of \(f(x)\) at \(x=-2\) are not the same, we conclude that the function is not continuous at \(x=-2.\)

Classifying the discontinuity

Finally, let's classify the discontinuity. Given that the left and right limits exist but are not equal, the discontinuity at \(x = -2\) is non-removable. Removable discontinuities occur when a factor in the denominator cancels out with a factor in the numerator, meaning the function could be made continuous at that point by a minor adjustment. This is not the case here, so the discontinuity is non-removable.