Question
Let \( F(x)=\left\{\begin{array}{cc}\sin (x-3) & x<3 \\ e^{x}+3 & 3 \leq x\end{array}\right. \) Then \( \lim _{x \rightarrow 3} F(x)=\square \)
Ask by Lynch Hilton. in the United States
Mar 08,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The limit as \( x \) approaches 3 does not exist because the left-hand limit is 0 and the right-hand limit is \( e^3 + 3 \), which are not equal.
Solution
1. For \( x < 3 \), the function is given by \( F(x) = \sin(x - 3) \). Evaluating the left-hand limit:
\[
\lim_{x \to 3^-} F(x) = \lim_{x \to 3^-} \sin(x - 3) = \sin(3 - 3) = \sin(0) = 0.
\]
2. For \( x \geq 3 \), the function is given by \( F(x) = e^x + 3 \). Evaluating the right-hand limit:
\[
\lim_{x \to 3^+} F(x) = \lim_{x \to 3^+} (e^x + 3) = e^3 + 3.
\]
3. Since the left-hand limit and right-hand limit are not equal:
\[
0 \neq e^3 + 3,
\]
the overall limit as \( x \to 3 \) does not exist.
\[
\lim_{x \to 3} F(x) \text{ does not exist.}
\]
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Simplify this solution Bonus Knowledge
To find \( \lim_{x \rightarrow 3} F(x) \), we need to evaluate the limit from both sides of \( x = 3 \). 1. From the left side, where \( x < 3 \): \[ \lim_{x \rightarrow 3^-} F(x) = \lim_{x \rightarrow 3} \sin(x - 3) = \sin(3 - 3) = \sin(0) = 0. \] 2. From the right side, where \( x \geq 3 \): \[ \lim_{x \rightarrow 3^+} F(x) = \lim_{x \rightarrow 3} (e^x + 3) = e^3 + 3. \] As the left-hand limit is \( 0 \) and the right-hand limit is \( e^3 + 3 \), since \( e^3 + 3 \) is greater than \( 0 \), we conclude that: \[ \lim_{x \rightarrow 3} F(x) \text{ does not exist.} \] This is due to the left-hand limit and the right-hand limit not being equal.
