To determine the limit [tex]\(\lim_{x \to -5} g(x)\)[/tex] for the piecewise function
[tex]\[ g(x)=\left\{\begin{array}{ll}
x^2 + 3x - 2, & x < -5 \\
9x + 4, & x = -5 \\
4x + 12, & x > -5
\end{array}\right. \][/tex]
we need to evaluate the left-hand limit (as [tex]\(x\)[/tex] approaches [tex]\(-5\)[/tex] from the left) and the right-hand limit (as [tex]\(x\)[/tex] approaches [tex]\(-5\)[/tex] from the right), and then compare these limits.
1. Calculate the left-hand limit as [tex]\(x\)[/tex] approaches [tex]\(-5\)[/tex]:
For [tex]\(x < -5\)[/tex], we use the expression [tex]\(x^2 + 3x - 2\)[/tex]. Plugging in [tex]\(x = -5\)[/tex],
[tex]\[
(-5)^2 + 3(-5) - 2 = 25 - 15 - 2 = 8
\][/tex]
2. Calculate the right-hand limit as [tex]\(x\)[/tex] approaches [tex]\(-5\)[/tex]:
For [tex]\(x > -5\)[/tex], we use the expression [tex]\(4x + 12\)[/tex]. Plugging in [tex]\(x = -5\)[/tex],
[tex]\[
4(-5) + 12 = -20 + 12 = -8
\][/tex]
3. Compare the limits:
- Left-hand limit as [tex]\(x\)[/tex] approaches [tex]\(-5\)[/tex]: [tex]\(8\)[/tex]
- Right-hand limit as [tex]\(x\)[/tex] approaches [tex]\(-5\)[/tex]: [tex]\(-8\)[/tex]
Since the left-hand limit ([tex]\(8\)[/tex]) and the right-hand limit ([tex]\(-8\)[/tex]) are not equal, the two-sided limit [tex]\(\lim_{x \to -5} g(x)\)[/tex] does not exist.
4. Evaluate the function at [tex]\(x = -5\)[/tex]:
For [tex]\(x = -5\)[/tex], we directly use the definition given in the piecewise function: [tex]\(g(x) = 9x + 4\)[/tex]. Plugging in [tex]\(x = -5\)[/tex],
[tex]\[
9(-5) + 4 = -45 + 4 = -41
\][/tex]
Thus, the limits from the left and right are:
- Left-hand limit: [tex]\(8\)[/tex]
- Right-hand limit: [tex]\(-8\)[/tex]
These limits are not equal, indicating that [tex]\(\lim_{x \to -5} g(x)\)[/tex] does not exist. Additionally, the value of the function at [tex]\(x = -5\)[/tex] is [tex]\(-41\)[/tex].
