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$h(x)={[(1)/(2x)," for "x <= -2],[2^(x)," for "-2 < x <= 0]:} Find lim_(x rarr-2)h(x). Choose 1 answer: (A) -2 (B) -(1)/(4) (C) (1)/(4) (D) The limit doesn't exist.$

\[ h(x)=\left\{\begin{array}{ll} \frac{1}{2 x} & \text { for } x \leq-2 \\ 2^{x} & \text { for }-2

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Full solution

Q.

h(x)=\left\{\begin{array}{ll} \frac{1}{2 x} & \text { for } x \leq-2 \\ 2^{x} & \text { for }-2<x \leq 0 \end{array}\right.

\newline

Find

\lim _{x \rightarrow-2} h(x)

.

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Choose

1

answer:

\newline

(A)

-2

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(B)

-\frac{1}{4}

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(C)

\frac{1}{4}

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(D) The limit doesn't exist.

Consider the function definition: To find the limit of $h(x)$ as $x$ approaches $-2$ , we need to consider the definition of the function for values of $x$ that are less than or equal to $-2$ . According to the given piecewise function, $h(x) = \frac{1}{2x}$ for $x \leq -2$ .
Calculate the limit from the left side: We will calculate the limit from the left side, as $x$ approaches $-2$ . Since the function is continuous for $x \leq -2$ , the limit as $x$ approaches $-2$ from the left is simply the value of the function at $x = -2$ .
Substitute $x = -2$ into the function: Substitute $x = -2$ into the function $h(x) = \frac{1}{2x}$ to find the limit. $\lim_{x \to -2} h(x) = \frac{1}{2(-2)} = \frac{1}{-4} = -\frac{1}{4}$
Find the limit of $h(x)$ as $x$ approaches $-2$ : The limit of $h(x)$ as $x$ approaches $-2$ is $-\frac{1}{4}$ . Therefore, the correct answer is (B) $-\frac{1}{4}$ .

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