$$ \left.\begin{array}{l} ext { 14. Given } f(x)=\left\{\begin{array}{cc}2 x^{2}-x & ext { if } x

Question

$$ \left.\begin{array}{l}	ext { 14. Given } f(x)=\left\{\begin{array}{cc}2 x^{2}-x & 	ext { if } x<3 \ 3-x & 	ext { if } x \geq 3\end{array} 	ext {. Find } \lim _{x ightarrow 3} f(x) .ight. \ 	ext { 15. Given } f(x)=\left\{\begin{array}{cc}\frac{1}{x-1} & 	ext { if } x<-1 \ x^{2}+2 x & 	ext { if } x \geq-1\end{array} 	ext {. Find } \lim _{x ightarrow-1} f(x) .ight. \ 	ext { 16. Given } f(x)=\left\{\begin{array}{ll}x^{2}-3 x & 	ext { if } x<2 \ 4+2 x & 	ext { if } x \geq 2\end{array} 	ext { Is } f(x) 	ext { continuous at } 2 	ext { ? }ight. \ 	ext { 17. For what value(s) of } a 	ext { will the function } f(x)=\left\{\begin{array}{c}x^{2}+2 x 	ext { if } x \geq a \ -1\end{array} 	ext { if } x

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Let's solve each of the problems step by step.

### Problem 14
Given the piecewise function:
$$
f(x) = \begin{cases}
2x^2 - x & \text{if } x < 3 \
3 - x & \text{if } x \geq 3
\end{cases}
$$
We need to find $$\lim_{x \to 3} f(x)$$.

1. **Left-hand limit** as $$x$$ approaches 3:
   $$
   \lim_{x \to 3^-} f(x) = \lim_{x \to 3} (2x^2 - x) = 2(3^2) - 3 = 18 - 3 = 15
   $$

2. **Right-hand limit** as $$x$$ approaches 3:
   $$
   \lim_{x \to 3^+} f(x) = \lim_{x \to 3} (3 - x) = 3 - 3 = 0
   $$

Since the left-hand limit (15) and the right-hand limit (0) are not equal, the limit does not exist:
$$
\lim_{x \to 3} f(x) \text{ does not exist.}
$$

### Problem 15
Given the piecewise function:
$$
f(x) = \begin{cases}
\frac{1}{x-1} & \text{if } x < -1 \
x^2 + 2x & \text{if } x \geq -1
\end{cases}
$$
We need to find $$\lim_{x \to -1} f(x)$$.

1. **Left-hand limit** as $$x$$ approaches -1:
   $$
   \lim_{x \to -1^-} f(x) = \lim_{x \to -1} \frac{1}{x-1} = \frac{1}{-1-1} = \frac{1}{-2} = -\frac{1}{2}
   $$

2. **Right-hand limit** as $$x$$ approaches -1:
   $$
   \lim_{x \to -1^+} f(x) = \lim_{x \to -1} (x^2 + 2x) = (-1)^2 + 2(-1) = 1 - 2 = -1
   $$

Since the left-hand limit $$-\frac{1}{2}$$ and the right-hand limit $$-1$$ are not equal, the limit does not exist:
$$
\lim_{x \to -1} f(x) \text{ does not exist.}
$$

### Problem 16
Given the piecewise function:
$$
f(x) = \begin{cases}
x^2 - 3x & \text{if } x < 2 \
4 + 2x & \text{if } x \geq 2
\end{cases}
$$
We need to check if $$f(x)$$ is continuous at $$x = 2$$.

1. **Left-hand limit** as $$x$$ approaches 2:
   $$
   \lim_{x \to 2^-} f(x) = \lim_{x \to 2} (x^2 - 3x) = 2^2 - 3(2) = 4 - 6 = -2
   $$

2. **Right-hand limit** as $$x$$ approaches 2:
   $$
   \lim_{x \to 2^+} f(x) = \lim_{x \to 2} (4 + 2x) = 4 + 2(2) = 4 + 4 = 8
   $$

3. **Value of the function at $$x = 2$$**:
   $$
   f(2) = 4 + 2(2) = 8
   $$

Since the left-hand limit $$-2$$ and the right-hand limit $$8$$ are not equal, $$f(x)$$ is not continuous at $$x = 2$$:
$$
f(x) \text{ is not continuous at } 2.
$$

### Problem 17
Given the piecewise function:
$$
f(x) = \begin{cases}
x^2 + 2x & \text{if } x \geq a \
-1 & \text{if } x < a
\end{cases}
$$
We need to find the value(s) of $$a$$ such that $$f(x)$$ is continuous at $$x = a$$.

1. **Left-hand limit** as $$x$$ approaches $$a$$:
   $$
   \lim_{x \to a^-} f(x) = -1
   $$

2. **Right-hand limit** as $$x$$ approaches $$a$$:
   $$
   \lim_{x \to a^+} f(x) = a^2 + 2a
   $$

3. **Setting the limits equal for continuity**:
   $$
   -1 = a^2 + 2a
   $$
   Rearranging gives:
   $$
   a^2 + 2a + 1 = 0
   $$
   This factors to:
   $$
   (a + 1)^2 = 0
   $$
   Thus, $$a + 1 = 0$$ or $$a = -1$$.

The value of $$a$$ for which $$f(x)$$ is continuous at $$x = a$$ is:
$$
a = -1.
$$
- Problem 14: The limit does not exist. - Problem 15: The limit does not exist. - Problem 16: $$f(x)$$ is not continuous at $$x = 2$$. - Problem 17: $$a = -1$$.

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