Oblique AsymptotesDate: 04/03/2002 at 23:47:45 From: Jo Subject: Oblique limits Hi, I'm really stuck on how to calculate oblique limits. Vertical and horizontal limits are fine, but I don't even know where to start with oblique ones. For example this question. x^2 / (x-3) It seems simple but there is an oblique limit somewhere that I can't find. How do I find it? What method do I use? Thanks for your time. Jo Date: 04/04/2002 at 12:56:03 From: Doctor Peterson Subject: Re: Oblique limits Hi, Jo. I'm assuming that by "limit" you mean what I call an "asymptote." Here is how I would do it: First, I notice that the denominator has a lower degree than the numerator, and I divide both by x so that the highest power of x in the former is 0: x^2 x f(x) = ----- = ------- x - 3 1 - 3/x When x gets large, 3/x gets very small, so this will get close to x/1 = x. The asymptote therefore is parallel to y = x. To find the actual asymptote, look at the limit of f(x) - x: x^2 x^2 - x(x - 3) 3x 3 lim (----- - x) = lim -------------- = lim ----- = lim ------- x->oo x - 3 x->oo x - 3 x->oo x - 3 x->oo 1 - 3/x This limit is 3; so for large x, f(x) is close to x + 3, which is the asymptote. Rather than this two-step method (finding the slope and then the offset, or y-intercept), you can just use polynomial division: f(x) = x^2/(x-3) = x + 3 + 9/(x-3) The last term becomes negligible for large x, so the asymptote is x + 3 again. Here are some other versions of these two methods: Asymptotes http://mathforum.org/dr.math/problems/moshfeghian4.13.96.html Nonvertical Asymptotes http://mathforum.org/dr.math/problems/amrita.10.16.01.html - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 04/04/2002 at 22:00:16 From: jo Subject: oblique limits Question submitted via WWW: thanks very much for your help. i haven't found another place more useful for help with my questions. its very much appreciated. thanks. |
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