since -1 <= sin(u) <= 1
e^x -> +∞
1/∞ = 0 as x -> +∞
But, oddly enough, this only a one-sided asymptote, since
e^-∞ = 0,
e^x-2 -> -2 as x -> -∞
And of course, there is the vertical asymptote at x = ln(2)
http://www.wolframalpha.com/input/?i=sin(3x-7)+%2F+(e%5Ex-2)+for+-5+%3C%3D+x+%3C%3D+5
Let K(x)= (sin(3x-7)) / (e^x-2)
(I'm sorry if it is not clear here, but the denominator in K(x) has the natural number, e, to the power of just x. So "x-2" is not the exponent. "x" is the exponent).
The Question: Does K(x) have a horizontal asymptote?
(I'm pretty sure it doesn't, but I don't know why).
1 answer
2 answers