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$g(x)={[e^(x)," for "-5 < x < -1],[(x)/(e)," for "-1 <= x < 0]:} Find lim_(x rarr-1^(-))g(x). Choose 1 answer: (A) -(1)/(e) (B) 0 (C) (1)/(e) (D) The limit doesn't exist.$

\[ g(x)=\left\{\begin{array}{ll} e^{x} & \text { for }-5

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Full solution

Q.

g(x)=\left\{\begin{array}{ll} e^{x} & \text { for }-5<x<-1 \\ \frac{x}{e} & \text { for }-1 \leq x<0 \end{array}\right.

\newline

Find

\lim _{x \rightarrow-1^{-}} g(x)

.

\newline

Choose

1

answer:

\newline

(A)

-\frac{1}{e}

\newline

(B)

0

\newline

(C)

\frac{1}{e}

\newline

(D) The limit doesn't exist.

Problem statement: We are asked to find the limit of $g(x)$ as $x$ approaches $-1$ from the left side. This means we are looking at the interval where $x$ is between $-5$ and $-1$ , and for this interval, the function $g(x)$ is defined as $e^{x}$ .
Substituting $x$ in $g(x)$ : To find the limit as $x$ approaches $-1$ from the left, we substitute $x$ with $-1$ in the expression for $g(x)$ in the given interval, which is $e^{x}$ . $\newline$ $\lim_{x \to -1^{-}}g(x) = e^{-1}$
Finding the limit: We know that $e^{-1}$ is the reciprocal of $e$ , which is $1/e$ . $\newline$ $\lim_{x \rightarrow -1^{-}}g(x) = \frac{1}{e}$
Final result: Therefore, the limit of $g(x)$ as $x$ approaches $-1$ from the left is $\frac{1}{e}$ , which corresponds to answer choice (C).

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