Analyze the behavior of q(x) = dfracx-1|x-1| as x approaches 1. Find limx → 1^- q(x) and limx → 1^+ q(x), and use these to determine whether limx → 1 q(x) exists. If it exists, give its value; if not, explain why not.

When x 1, we have x - 1 > 0, so |x-1| = x-1, giving q(x) = dfracx-1x-1 = 1 for x near 1 from the right. Thus limx → 1^+ q(x) = 1. Since the left-hand limit (-1) and right-hand limit (1) are not equal, the two-sided limit limx → 1 q(x) does not exist.

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One-sided limits and existence of a two-sided limit

Consider the function $q(x) = \dfrac{x-1}{|x-1|}$ , which is defined for all $x \neq 1$ .

Analyze the behavior of $q(x) = \dfrac{x-1}{|x-1|}$ as $x$ approaches $1$ .

One-sided limits and existence of a two-sided limit graph

Find $\lim_{x \to 1^-} q(x)$ and $\lim_{x \to 1^+} q(x)$ , and use these to determine whether $\lim_{x \to 1} q(x)$ exists. If it exists, give its value; if not, explain why not.

Hint

Consider what $x - 1$ and $|x-1|$ look like separately when $x$ is slightly less than $1$ versus slightly greater than $1$ . The sign of the quotient depends entirely on the sign of $x-1$ .

Step-by-step solution

When $x < 1$ , we have $x - 1 < 0$ , so $|x-1| = -(x-1)$ . Thus for $x$ near $1$ from the left, $q(x) = \dfrac{x-1}{-(x-1)} = -1$ . Therefore $\lim_{x \to 1^-} q(x) = -1$ .

When $x > 1$ , we have $x - 1 > 0$ , so $|x-1| = x-1$ , giving $q(x) = \dfrac{x-1}{x-1} = 1$ for $x$ near $1$ from the right. Thus $\lim_{x \to 1^+} q(x) = 1$ . Since the left-hand limit ( $-1$ ) and right-hand limit ( $1$ ) are not equal, the two-sided limit $\lim_{x \to 1} q(x)$ does not exist.

One-sided limits and existence of a two-sided limit

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