Consider v(x) = textsgn(x-2) as defined above. Examine the graph of v(x) near x=2 shown below. Determine whether limx → 2 v(x) exists. If it exists, give its value; if it does not exist, explain why not, referencing the one-sided limits.

For x slightly less than 2, we have x - 2 0, so v(x) = 1 for all such x, giving limx → 2^+ v(x) = 1. Since the left-hand limit (-1) and the right-hand limit (1) are not equal, the two-sided limit limx→ 2 v(x) does not exist.

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Existence of a two-sided limit for the signum function

The signum function is defined as $v(x) = \text{sgn}(x-2) = \begin{cases} -1 & \text{if } x < 2 \\ 0 & \text{if } x = 2 \\ 1 & \text{if } x > 2 \end{cases}$

Consider $v(x) = \text{sgn}(x-2)$ as defined above. Examine the graph of $v(x)$ near $x=2$ shown below.

Existence of a two-sided limit for the signum function graph

Determine whether $\lim_{x \to 2} v(x)$ exists. If it exists, give its value; if it does not exist, explain why not, referencing the one-sided limits.

Hint

Compute the left-hand limit $\lim_{x\to 2^-} v(x)$ and the right-hand limit $\lim_{x\to 2^+} v(x)$ separately by considering values of $x$ just below and just above 2, then compare them.

Step-by-step solution

For $x$ slightly less than $2$ , we have $x - 2 < 0$ , so $v(x) = -1$ for all such $x$ . Therefore the left-hand limit is $\lim_{x \to 2^-} v(x) = -1$ . Note that the value $v(2) = 0$ plays no role in this computation since limits only depend on nearby values, not the value at $x=2$ itself.

For $x$ slightly greater than $2$ , we have $x-2 > 0$ , so $v(x) = 1$ for all such $x$ , giving $\lim_{x \to 2^+} v(x) = 1$ . Since the left-hand limit ( $-1$ ) and the right-hand limit ( $1$ ) are not equal, the two-sided limit $\lim_{x\to 2} v(x)$ does not exist.

Existence of a two-sided limit for the signum function

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