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Let \( f(x)=\left\{\begin{array}{ll}10-x-x^{2} & \text { if } x \leq 2 \\ 2 x-7 & \text { if } x>2\end{array}\right. \) Calculate the following limits. Enter "DNE" if the limit does not exist. \( \lim _{x \rightarrow 2^{-}} f(x)=\square \) \( \lim _{x \rightarrow 2^{+}} f(x)=\square \) \( \lim _{x \rightarrow 2} f(x)=\square \)

Ask by Nunez Klein. in the United States
Jan 27,2025

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Tutor-Verified Answer

Answer

\[ \lim_{x \to 2^{-}} f(x)=4; \quad \lim_{x \to 2^{+}} f(x)=-3; \quad \lim_{x \to 2} f(x)=\text{DNE} \]

Solution

Let's evaluate the limits for the piecewise function: \[ f(x) = \begin{cases} 10 - x - x^2 & \text{if } x \leq 2 \\ 2x - 7 & \text{if } x > 2 \end{cases} \] 1. **Left-hand limit (\( \lim_{x \to 2^-} f(x) \))**: Since we approach 2 from the left, we use the first part of the function: \[ \lim_{x \to 2^-} f(x) = 10 - 2 - (2)^2 = 10 - 2 - 4 = 4 \] 2. **Right-hand limit (\( \lim_{x \to 2^+} f(x) \))**: Since we approach 2 from the right, we use the second part of the function: \[ \lim_{x \to 2^+} f(x) = 2(2) - 7 = 4 - 7 = -3 \] 3. **Overall limit (\( \lim_{x \to 2} f(x) \))**: The left-hand limit is 4 and the right-hand limit is -3. Since they are not equal, the overall limit does not exist. **Final Answers:** \[ \lim_{x \to 2^-} f(x) = 4 \] \[ \lim_{x \to 2^+} f(x) = -3 \] \[ \lim_{x \to 2} f(x) = \text{DNE} \] Answer: \[ \lim_{x \to 2^{-}} f(x)=4; \quad \lim_{x \to 2^{+}} f(x)=-3; \quad \lim_{x \to 2} f(x)=\text{DNE} \]

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The Deep Dive

To calculate the limits for the function \( f(x) \), we will evaluate the left-hand limit, right-hand limit, and the overall limit at \( x = 2 \). 1. **Left-hand limit** \( \lim _{x \rightarrow 2^{-}} f(x) \): Since we're approaching from the left, we use the first piece of the function \( f(x) = 10 - x - x^2 \): \[ \lim _{x \rightarrow 2^{-}} f(x) = 10 - 2 - 2^2 = 10 - 2 - 4 = 4 \] 2. **Right-hand limit** \( \lim _{x \rightarrow 2^{+}} f(x) \): As we approach from the right, we use the second piece of the function \( f(x) = 2x - 7 \): \[ \lim _{x \rightarrow 2^{+}} f(x) = 2(2) - 7 = 4 - 7 = -3 \] 3. **Overall limit** \( \lim _{x \rightarrow 2} f(x) \): Since the left-hand limit and the right-hand limit are not equal, the overall limit does not exist: \[ \lim _{x \rightarrow 2} f(x) = DNE \] Putting these results together, we have: \[ \lim _{x \rightarrow 2^{-}} f(x) = 4, \quad \lim _{x \rightarrow 2^{+}} f(x) = -3, \quad \lim _{x \rightarrow 2} f(x) = \text{DNE} \] So the final answers are: \[ \lim _{x \rightarrow 2^{-}} f(x) = 4 \] \[ \lim _{x \rightarrow 2^{+}} f(x) = -3 \] \[ \lim _{x \rightarrow 2} f(x) = DNE \]

Related Questions

Multiple Choice Identify the choice that best completes the statement or answers the question. Find any points of discontinuity for the rational function. 1. \( y=\frac{(x-7)(x+2)(x-9)}{(x-5)(x-2)} \) a. \( x=-5, x=-2 \) b. \( x=5, x=2 \) c. \( x=-7, x=2, x=-9 \) d. \( x=7, x=-2, x=9 \) 2. \( y=\frac{(x+7)(x+4)(x+2)}{(x+5)(x-3)} \) a. \( x=-5, x=3 \) b. \( x=7, x=4, x=2 \) c. \( x=-7, x=-4, x=-2 \) d. \( x=5, x=-3 \) 3. \( y=\frac{x+4}{x^{2}+8 x+15} \) a. \( x=-5, x=-3 \) b. \( x=-4 \) c. \( x=-5, x=3 \) d. \( x=5, x=3 \) 4. \( y=\frac{x-3}{x^{2}+3 x-10} \) a. \( x=-5, x=2 \) b. \( x=5, x=-2 \) c. \( x=3 \) d. \( x \) \( =-5, x=-2 \) 6. What are the points of discontinuity? Are they all removable? \[ y=\frac{(x-4)}{x^{2}-13 x+36} \] a. \( x=-9, x=-4, x=8 \); yes b. \( x=1, x=8, x= \) -8; no c. \( x=9, x=4 \); no d. \( x=-9, x=-4 \); no 7. Describe the vertical asymptote(s) and hole(s) for the graph of \( y=\frac{(x-2)(x-5)}{(x-5)(x+2)} \). a. asymptote: \( x=2 \) and hole: \( x=-5 \) b. asymptotes: \( x=-2 \) and hole: \( x=-5 \) c. asymptote: \( x=-2 \) and hole: \( x=5 \) d. asymptote: \( x=-2 \) and hole: \( x=-2 \) a. \( x=-3, x=-8 \); no b. \( x=5, x=-7, x=1 \); no c. \( x=-5, x=7, x=-1 \); yes d. \( x=3, x=8 \); yes 8. Describe the vertical asymptote(s) and hole(s) for the graph of \( y=\frac{x+1}{x^{2}+6 x+5} \). a. asymptote: \( x=-1 \) and hole: \( x=-1 \) b. asymptote: \( x=-1 \) and hole: \( x=-1 \) c. asymptotes: \( x=-1,-1 \) and hole: \( x=-1 \) d. asymptote: -5 and hole: \( x=-1 \), 9. Find the horizontal asymptote of the graph of \( y=\frac{7 x^{6}+7 x+3}{9 x^{5}+7 x+3} \). a. \( y=0 \) b. \( y=\frac{7}{9} \) c. no horizontal asymptote d. \( y=\frac{6}{5} \)
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