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$h(x)={[2x+1," for "x < -2],[(1)/(x-1)," for "-2 <= x < 1]:} Find lim_(x rarr-3)h(x). Choose 1 answer: (A) -5 (B) -3 (C) -(1)/(4) (D) The limit doesn't exist.$

$h(x)=\left\{\begin{array}{ll} 2 x+1 & \text { for } x<-2 \\ \frac{1}{x-1} & \text { for }-2 \leq x<1 \end{array}\right.$ $\newline$ Find $\lim _{x \rightarrow-3} h(x)$ . $\newline$ Choose $1$ answer: $\newline$ (A) $-5$ $\newline$ (B) $-3$ $\newline$ (C) $-\frac{1}{4}$ $\newline$ (D) The limit doesn't exist.

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Full solution

Q.

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

\newline

Find

\lim _{x \rightarrow-3} h(x)

.

\newline

Choose

1

answer:

\newline

(A)

-5

\newline

(B)

-3

\newline

(C)

-\frac{1}{4}

\newline

(D) The limit doesn't exist.

Given function and limit: We are given a piecewise function $h(x)$ and need to find the limit as $x$ approaches $-3$ . Since $-3$ is less than $-2$ , we will use the first part of the piecewise function, which is $2x + 1$ .
Substituting $x$ into the function: Now we will substitute $x$ with $-3$ into the first part of the function to find the limit. $\lim_{x \to -3} h(x) = 2(-3) + 1$
Performing the calculation: Perform the calculation: $2(-3) + 1 = -6 + 1 = -5$
Final result and answer choice: The limit of $h(x)$ as $x$ approaches $-3$ is $-5$ , which corresponds to answer choice (A).

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