determine all asymtote(s) of f(x)= 1/(x(x-1)^2) using limits
Oh my, look around x = 0 and x = 1 and x = infinity and x = -infinity
http://www.wolframalpha.com/input/?i=plot+1%2F(x(x-1))
and y=0
Oh, yes :)O
Oh, but I said where x = +/- infinity
yes, you did. My bad.
To determine the asymptotes of a function using limits, we need to analyze the behavior of the function as it approaches certain values. In the case of rational functions like f(x) = 1/(x(x-1)^2), there are three types of asymptotes to consider: vertical, horizontal, and slant (or oblique) asymptotes.
1. Vertical Asymptotes:
Vertical asymptotes occur when the function approaches infinity or negative infinity as x approaches a certain value. To find the vertical asymptotes of a function, we need to identify any values of x where the denominator becomes zero but the numerator does not. In this case, we have two potential vertical asymptotes: x = 0 and x = 1, as the denominator becomes zero at these points.
2. Horizontal Asymptotes:
Horizontal asymptotes occur when the function approaches a constant value as x approaches positive or negative infinity. To find the horizontal asymptotes, we evaluate the limit of the function as x approaches infinity and negative infinity.
As x approaches infinity:
lim (x→∞) [1/(x(x-1)^2)] = 0
Since the limit approaches zero, the horizontal asymptote is y = 0.
As x approaches negative infinity:
lim (x→-∞) [1/(x(x-1)^2)] = 0
Again, the limit approaches zero, which means the horizontal asymptote is y = 0.
Therefore, f(x) has a horizontal asymptote at y = 0.
3. Slant Asymptotes:
For rational functions, slant asymptotes only occur when the degree of the numerator is exactly one greater than the degree of the denominator. In this case, the numerator has a degree of 0, while the denominator has a degree of 3. Therefore, f(x) does not have a slant asymptote.
To summarize:
- Vertical asymptotes: x = 0 and x = 1
- Horizontal asymptote: y = 0
- No slant asymptote.
These are the asymptotes of the given function f(x) = 1/(x(x-1)^2).