For what values of x, if any, does $f (x) = \frac{1}{(x - 2) sin x}$ $f(x) = 1/((x-2)sinx)$ have vertical asymptotes?

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For what values of x, if any, does $f (x) = \frac{1}{(x - 2) sin x}$ $f(x) = 1/((x-2)sinx)$ have vertical asymptotes?

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Answer 1 · 2016-03-13T05:14:40

$x = 2$ $x=2$ , $x = 2 π k$ $x=2pik$ , and $x = π + 2 π k$ $x=pi+2pik$ where $k \in Z$ $k in Z$

Explanation:

Vertical asymptotes occur whenever the denominator equals 0. To find them, we simply set the denominator to 0 and solve, like thus:
$(x - 2) sin x = 0$ $(x-2)sinx=0$
$x - 2 = 0$ $x-2=0$ and $sin x = 0$ $sinx=0$
$x = 2$ $x=2$ and $x = 2 π k, π + 2 π k$ $x=2pik, pi+2pik$ where $k \in Z$ $k in Z$ ( $k$ $k$ is an integer)

We have infinitely many vertical asymptotes, at $x = 2$ $x=2$ , $x = 2 π k$ $x=2pik$ , and $x = π + 2 π k$ $x=pi + 2pik$ . We can see this on the graph of the function below:
enter image source here

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For what values of x, if any, does $f (x) = \frac{1}{(x - 2) sin x}$ $f(x) = 1/((x-2)sinx)$ have vertical asymptotes?

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Explanation:

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For what values of x, if any, does $f (x) = \frac{1}{(x - 2) sin x}$ $f(x) = 1/((x-2)sinx)$ have vertical asymptotes?