
Kenneth S. answered 10/08/17
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vertical asymptotes: if i set denominator to 0, then sinx=0 and x= 0 or npi?
CORRECT...in fact, x = n(pi) is sufficient because n can be an integer, including zero.
As x→ ±infinity, y also approaches infinity (wildly) and perhaps you should use a graphing calculator (not in Degree Mode) to show this function's behavior. It's not H.A. behavior because that would result in y → a constant, as x increases without bound.
In the vicinity of zero, y → 1 because the limit of x/sinx is 1, a well known result from early Calculus.

Kenneth S.
as x approaches infinity, sin x varies from -1 to 1 and thus there is no limit (a specific real number L).
I repeat this sentence from earlier answer: It's not H.A. behavior because that would result in y → a constant, as x increases without bound.
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