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$g(x)={[(x)/(3)-2," for "0 < x < 6],[cos(x*pi)," for "6 <= x <= 10]:} Find lim_(x rarr6)g(x). Choose 1 answer: (A) -1 (B) 0 (C) 1 (D) The limit doesn't exist.$

\( g(x)=\left\{\begin{array}{ll}\frac{x}{3}-2 & \text { for } 0

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Full solution

g(x)=\left\{\begin{array}{ll}\frac{x}{3}-2 & \text { for } 0<x<6 \\ \cos (x \cdot \pi) & \text { for } 6 \leq x \leq 10\end{array}\right.

\newline

Find

\lim _{x \rightarrow 6} g(x)

\newline

Choose

1

answer:

\newline

(A)

-1

\newline

(B)

0

\newline

(C)

1

\newline

(D) The limit doesn't exist.

Define $g(x)$ : To find the limit of $g(x)$ as $x$ approaches $6$ , we need to consider the definition of $g(x)$ on both sides of $6$ . For values of $x$ less than $6$ , $g(x)$ is defined as $\frac{x}{3} - 2$ . For values of $x$ greater than or equal to $6$ , $g(x)$ is defined as $g(x)$ $3$ . We need to check the limit from both sides to see if they match.
Left-hand limit: First, let's find the limit of $g(x)$ as $x$ approaches $6$ from the left, which means we are looking at the piece of the function where $g(x) = \frac{x}{3} - 2$ . $\newline$ $\lim_{x \to 6^-} g(x) = \lim_{x \to 6^-} \left(\frac{x}{3} - 2\right)$
Substitute $x$ with $6$ : Now we substitute $x$ with $6$ in the expression $\frac{x}{3} - 2$ to find the left-hand limit. $\newline$ $\lim_{x \to 6^-} \left(\frac{x}{3} - 2\right) = \frac{6}{3} - 2 = 2 - 2 = 0$
Right-hand limit: Next, we find the limit of $g(x)$ as $x$ approaches $6$ from the right, which means we are looking at the piece of the function where $g(x) = \cos(x\pi)$ . $\lim_{x \to 6^+} g(x) = \lim_{x \to 6^+} \cos(x\pi)$
Substitute $x$ with $6$ : We substitute $x$ with $6$ in the expression $\cos(x\pi)$ to find the right-hand limit. $\lim_{x \to 6^+} \cos(x\pi) = \cos(6\pi) = \cos(2\pi\cdot3) = 1$
Conclusion: Since the left-hand limit as $x$ approaches $6$ is $0$ and the right-hand limit as $x$ approaches $6$ is $1$ , the two one-sided limits are not equal. Therefore, the limit of $g(x)$ as $x$ approaches $6$ does not exist.