using the definition, we need to compute the limit of (sin x)/x as x goes to infinity (and minus infinity). think about what happens as you plug in very large numbers. the range of sin(x) is between -1 and 1, while the x in the denominator can be as large as you like. the numerator is bounded, while the denominator is not. that implies that the limit will be 0. so y = 0 is the horizontal asymptote (and there is only one).

hope this helps!

hi russell!

while the arctan function is the inverse of the tan function, y = arctan(x) does not have an infinite number of asymptotes. the domain of arctan(x) is all values of x (from negative infinity to positive infinity). on the other hand, the range of arctan(x) is limited -pi/2 < arctan(x) < pi/2. the two limits, -pi/2 and pi/2 are the two asymptotes of the function.

the range reflects the "principle value" of the arctan relation. if we did not limit the range, then, as you have guessed, arctan(x) would appear as a "flipped" version of tan(x). however, the issue here is that arctan(x) would then have multiple values of y for a given value of x. why does this pose a problem? well, the definition of a function requires that each input (value of x) has a single output (value of y). in order for this to be true for arctan(x), the range of the function must be limited to the range i mention above. and this range results in two asymptotes for arctan(x): -pi/2 and pi/2 🙂

i hope this helps! if you have further questions, feel free to leave another comment below my response. 🙂

ok, so the tan(x) function has an infinite number of vertical asymptotes.
and arctan(x) is “just” the tan(x) function flipped onto the y-axis, no?

for example 2 = arctan(tan(2)) … no?