Let's carefully evaluate the given piecewise function:
[tex]\[
f(x) =
\begin{cases}
8x^2 - 10x + 5 & \text{if } x < 0 \\
-3 \sin(x) & \text{if } x \geq 0
\end{cases}
\][/tex]
We need to find the following limits:
1. [tex]\(\lim_{x \to 0^-} f(x)\)[/tex]
2. [tex]\(\lim_{x \to 0^+} f(x)\)[/tex]
3. [tex]\(\lim_{x \to 0} f(x)\)[/tex]
Step 1: Evaluate [tex]\(\lim_{x \to 0^-} f(x)\)[/tex]
For [tex]\( x < 0 \)[/tex], the function [tex]\( f(x) \)[/tex] is given by [tex]\( 8x^2 - 10x + 5 \)[/tex].
We need to find the limit of this expression as [tex]\( x \)[/tex] approaches [tex]\( 0 \)[/tex] from the left:
[tex]\[ \lim_{x \to 0^-} (8x^2 - 10x + 5) \][/tex]
Since this is a polynomial, we can substitute [tex]\( x = 0 \)[/tex] directly into the expression:
[tex]\[ 8(0)^2 - 10(0) + 5 = 5 \][/tex]
So,
[tex]\[ \lim_{x \to 0^-} f(x) = 5 \][/tex]
Step 2: Evaluate [tex]\(\lim_{x \to 0^+} f(x)\)[/tex]
For [tex]\( x \geq 0 \)[/tex], the function [tex]\( f(x) \)[/tex] is given by [tex]\( -3 \sin(x) \)[/tex].
We need to find the limit of this expression as [tex]\( x \)[/tex] approaches [tex]\( 0 \)[/tex] from the right:
[tex]\[ \lim_{x \to 0^+} (-3 \sin(x)) \][/tex]
Since [tex]\(\sin(x)\)[/tex] is continuous and [tex]\(\sin(0) = 0\)[/tex], we have:
[tex]\[ -3 \sin(0) = -3 \cdot 0 = 0 \][/tex]
So,
[tex]\[ \lim_{x \to 0^+} f(x) = 0 \][/tex]
Step 3: Evaluate [tex]\(\lim_{x \to 0} f(x)\)[/tex]
Now, we compare the left-hand limit and the right-hand limit as [tex]\( x \)[/tex] approaches [tex]\( 0 \)[/tex].
From Step 1:
[tex]\[ \lim_{x \to 0^-} f(x) = 5 \][/tex]
From Step 2:
[tex]\[ \lim_{x \to 0^+} f(x) = 0 \][/tex]
Since the left-hand limit and the right-hand limit are not equal, the overall limit [tex]\(\lim_{x \to 0} f(x)\)[/tex] does not exist.
Thus:
[tex]\[ \lim_{x \to 0} f(x) = \text{DNE (Does Not Exist)} \][/tex]
Summary:
[tex]\[
\lim_{x \rightarrow 0^{-}} f(x)= 5 \\
\lim_{x \rightarrow 0^{+}} f(x)= 0 \\
\lim_{x \rightarrow 0} f(x)= \text{DNE}
\][/tex]
