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$f(x)={[sin(x*pi)," for "-8 < x < 0],[(x)/(5)," for "0 <= x <= 10]:} Find lim_(x rarr-5)f(x). Choose 1 answer: (A) -1 (B) 0 (C) 1 (D) The limit doesn't exist.$

\[ f(x)=\left\{\begin{array}{ll} \sin (x \cdot \pi) & \text { for }-8

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Math Problems

Grade 8

Write variable expressions for arithmetic sequences

Full solution

f(x)=\left\{\begin{array}{ll} \sin (x \cdot \pi) & \text { for }-8<x<0 \\ \frac{x}{5} & \text { for } 0 \leq x \leq 10 \end{array}\right.

\newline

Find

\lim _{x \rightarrow-5} f(x)

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Choose

1

answer:

\newline

(A)

-1

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(B)

0

\newline

(C)

1

\newline

(D) The limit doesn't exist.

Determine applicable part: First, we need to determine which part of the piecewise function applies to the value $x = -5$ . The function $f(x)$ is defined as $\sin(x\pi)$ for -8 < x < 0 and as $\frac{x}{5}$ for $0 \leq x \leq 10$ . Since $-5$ falls within the first interval, we will use the $\sin(x\pi)$ part of the function to find the limit.
Calculate limit of $\sin(x\pi)$ : Now we will calculate the limit of $\sin(x\pi)$ as $x$ approaches $-5$ . The limit of a sine function as the input approaches a certain value is simply the sine of that value. Therefore, we need to calculate $\sin(-5\pi)$ .
Calculate $\sin(-5\pi)$ : Calculating $\sin(-5\pi)$ , we know that the sine function has a period of $2\pi$ , which means $\sin(-5\pi)$ is the same as $\sin(-5\pi + 2\pi n)$ for any integer $n$ . Choosing $n=2$ , we get $\sin(-5\pi + 4\pi) = \sin(-\pi)$ , which is the same as $\sin(\pi)$ .
Sine of $\pi$ is $0$ : The sine of $\pi$ is $0$ . Therefore, $\sin(-5\pi) = 0$ .
Limit of $f(x)$ as $x$ approaches $-5$ is $0$ : Since $\sin(-5\pi) = 0$ , the limit of $f(x)$ as $x$ approaches $-5$ is $0$ .