## Monday, October 02, 2006

### 2.6 Limits at Infinity; Horiz. Asymptotes

This section deals with infinite limits and horizontal asmyptotes. There are two main ways to find a horizontal asymptote.
The first way, which I prefer, is graphically. Plug in the equation to your grapher and visually see where the asymptote is. If it is not obvious, look at the table of values and find the y value that is not quite reached in the graph as x approaches infinity.
The second way is to algebraically find the asymptote(s). To do this with a rational function, you must divide the top leading term by the bottom leading term to find the asymptote- for example: (x^3 +10)/(x^2 -3) the asymptote would be y=x. --[because x cubed divided by x squared equals x....... in the grand scheme of things, that + 10 and -3 will not make a big difference on a graph dealing with large numbers being cubed and squared, so we ignore those].
If there is a leading coefficient and the same degree on top and bottom, you do the same thing, just with the coefficients ... for example: as x approaches infinity, (4x^2 + 7x + 3)/ (2x^2 - x+4) ....4/2= 2 so the asymptote would be [horizontal] at y=2.
An extremely effective way to know which way the graph is heading [in terms of infinity and negative infinity] is to make a 2x2 chart:[the left term is as x approaches neg. inf., the right= inf]

even += up/up
even -= down/down
odd+= down/up
odd-= up/down

so for a problem Mr. French might throw at you on a quiz/test:
find the horizontal asymptote as x approaches +infinity: (sqrt(2x^2 +5))/(3x-5)

to solve this, you must first realize the +5 and -5 are meaningless so scratch them out.
the sqrt of 2x^2= rad2 x
rad2 x divided by 3x = (rad2) / 3 ... [x's cancel]
so the asymptote occurs at y=(rad2)/ 3 when x approaches positive infinity