Question

a. evaluate \\(\lim_{x \to 1^+} \sqrt{x-1}\\).
b. explain why \\(\lim_{x \to 1^-} \sqrt{x-1}\\) does not exist.

a. \\(\lim_{x \to 1^+} \sqrt{x-1} = \square\\)

Explanation:

Evaluate the right-sided limit

$$ \lim_{x \to 1^{+}} \sqrt{x - 1} = \sqrt{1 - 1} = 0 $$

Analyze the left-sided limit domain

$$
LATEXBLOCK0
$$

Answer:

Question a

\(\lim_{x \to 1^{+}} \sqrt{x - 1} =\) <blank>0</blank>

Question b

The limit \(\lim_{x \to 1^{-}} \sqrt{x - 1}\) does not exist because the function \(f(x) = \sqrt{x - 1}\) is undefined for real numbers when \(x < 1\), meaning the expression inside the square root becomes negative as \(x\) approaches \(1\) from the left.