Question

1. let $g(x) = \

$$\begin{cases} 3x - 5 & \\text{if } 0 \\leq x < 1 \\\\ 1 & \\text{if } x = 1 \\\\ x^2 - 2x - 1 & \\text{if } x > 1 \\end{cases}$$

$
(5pts) evaluate the following limits algebraically:
(a) $\lim\limits_{x \to 1^+} g(x)$
(b) $\lim\limits_{x \to 1^-} g(x)$
(c) $\lim\limits_{x \to 1} g(x)$

Explanation:

Step1: Identify right-hand limit formula

For $x \to 1^+$, use $g(x)=x^2-2x-1$

Step2: Substitute $x=1$ into the formula

$\lim_{x \to 1^+} g(x) = 1^2 - 2(1) - 1 = 1 - 2 - 1$

Step3: Calculate right-hand limit value

$\lim_{x \to 1^+} g(x) = -2$

Step4: Identify left-hand limit formula

For $x \to 1^-$, use $g(x)=3x-5$

Step5: Substitute $x=1$ into the formula

$\lim_{x \to 1^-} g(x) = 3(1) - 5 = 3 - 5$

Step6: Calculate left-hand limit value

$\lim_{x \to 1^-} g(x) = -2$

Step7: Compare left/right limits for overall limit

Since $\lim_{x \to 1^+} g(x) = \lim_{x \to 1^-} g(x) = -2$, the two-sided limit exists and equals this value.

Answer:

(a) $\boldsymbol{-2}$
(b) $\boldsymbol{-2}$
(c) $\boldsymbol{-2}$