Simpson Collins
04/10/2023 · Junior High School
Let \( f(x)=\left\{\begin{array}{ll}4-x-x^{2} & \text { if } x \leq 1 \\ 2 x-5 & \text { if } x>1\end{array}\right. \) Calculate the following limits. Enter "DNE" if the limit does not exist. \( \lim _{x \rightarrow 1^{-}} f(x)=\square \) \( \lim _{x \rightarrow 1^{+}} f(x)=\square \) \( \lim _{x \rightarrow 1} f(x)=\square \)
UpStudy ThothAI Solution
Tutor-Verified Answer
Step-by-step Solution
To calculate the limits of the piecewise function \( f(x) \), we will evaluate the left-hand limit, the right-hand limit, and the overall limit as \( x \) approaches 1.
1. **Left-hand limit** \( \lim_{x \rightarrow 1^{-}} f(x) \):
For \( x \leq 1 \), we use the first piece of the function:
\[
f(x) = 4 - x - x^2.
\]
We evaluate this as \( x \) approaches 1 from the left:
\[
\lim_{x \rightarrow 1^{-}} f(x) = 4 - 1 - 1^2 = 4 - 1 - 1 = 2.
\]
2. **Right-hand limit** \( \lim_{x \rightarrow 1^{+}} f(x) \):
For \( x > 1 \), we use the second piece of the function:
\[
f(x) = 2x - 5.
\]
We evaluate this as \( x \) approaches 1 from the right:
\[
\lim_{x \rightarrow 1^{+}} f(x) = 2(1) - 5 = 2 - 5 = -3.
\]
3. **Overall limit** \( \lim_{x \rightarrow 1} f(x) \):
The overall limit exists if the left-hand limit and the right-hand limit are equal. Since we found:
\[
\lim_{x \rightarrow 1^{-}} f(x) = 2 \quad \text{and} \quad \lim_{x \rightarrow 1^{+}} f(x) = -3,
\]
these two limits are not equal. Therefore, the overall limit does not exist:
\[
\lim_{x \rightarrow 1} f(x) = \text{DNE}.
\]
In summary, the limits are:
\[
\lim_{x \rightarrow 1^{-}} f(x) = 2,
\]
\[
\lim_{x \rightarrow 1^{+}} f(x) = -3,
\]
\[
\lim_{x \rightarrow 1} f(x) = \text{DNE}.
\]
Thus, the final answers are:
\[
\lim_{x \rightarrow 1^{-}} f(x) = 2,
\]
\[
\lim_{x \rightarrow 1^{+}} f(x) = -3,
\]
\[
\lim_{x \rightarrow 1} f(x) = \text{DNE}.
\]
Quick Answer
\[
\lim_{x \rightarrow 1^{-}} f(x) = 2,
\]
\[
\lim_{x \rightarrow 1^{+}} f(x) = -3,
\]
\[
\lim_{x \rightarrow 1} f(x) = \text{DNE}.
\]
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