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).
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
To determine whether K(x) has a horizontal asymptote, we need to analyze the behavior of the function as x approaches positive infinity or negative infinity.
First, let's focus on what happens as x approaches positive infinity. We want to find the limit of K(x) as x approaches positive infinity.
K(x) = (sin(3x-7)) / (e^x-2)
As x gets larger, the term e^x in the denominator grows exponentially, meaning it will dominate the entire expression. This is because the exponential function grows much faster than the sine function.
So, the numerator (sin(3x-7)) doesn't have a significant effect on the overall behavior of the function as x approaches positive infinity. On the other hand, the denominator (e^x-2) becomes very large as x gets larger.
Since the denominator becomes very large, while the numerator does not, the limit of K(x) as x approaches positive infinity will be 0.
Now let's consider what happens as x approaches negative infinity. Again, we want to find the limit of K(x) as x approaches negative infinity.
As x becomes more negative, the denominator e^x will approach 0 (since e^x approaches 0 as x approaches negative infinity). In this case, the numerator also approaches 0.
So, as x approaches negative infinity, both the numerator and denominator approach 0. This means that the limit of K(x) as x approaches negative infinity is also 0.
Since the limit of K(x) is 0 as x approaches both positive and negative infinity, we can conclude that K(x) has a horizontal asymptote at y = 0 (x-axis).