For what values of x, if any, does #f(x) = 1/((x-2)sin(pi+(8pi)/x) # have vertical asymptotes?

Answer 1

This function has a vertical asymptote for #x=2# and for every other #x# of the form #8/k#, for #k \in \mathbb{Z}#

A rational function has a vertical asymptote if its denominator equals zero. In your case, the denominator is

(x-2)\sin(pi+(8pi)/x)

As every product, it equals zero if and only if at least one of its factors equals zero.

It is easy to find out that the first factor equals zero if #x=2#.
As for the second, we can first of all note that #sin(\pi+z)=-\sin(z)#

So, we need to find out when

-\sin((8pi)/x)=0 \iff \sin((8pi)/x)=0

This function has an infinite number of solutions, given by

(8pi)/x=k\pi \iff 8/x=k \iff x=8/k

So, for #x=...-8/4,-8/3,-8/2,-8,8,8/2,8/3,8/4...# the denominator equals zero, and the function has a vertical asymptote.
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Answer 2

The function f(x) has vertical asymptotes at x = 2 and x = 0.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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