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Yes. If f has an oblique asymptote (call it y = ax + b), you will have: a = lim x → ± ∞f(x) x. b = lim x → ± ∞f(x) − ax. In your example, lim x → + ∞√4x2 + x + 6 x = 2 and lim x → + ∞√4x2 + x + 6 − 2x = 1 4. The asymptote as x → + ∞ is therefore y = 2x + 1 4. Share. Cite. edited Oct 1, 2012 at 20:31. answered Oct 1 ...
0. When x approaches negative infinity, the original function is approximately f(x) = x − | x | = 2x, so the oblique asymptote is y = 2x. When x approaches positive infinity, f(x) should approach 0, leading to a horizontal asymptote of y = 0. You can check the result by graphing the function. Share.
A student did some exploring, discovered it was wrong, and informed me. Since then I've been curious why you must perform polynomial long division to find the oblique asymptotes. Everything I can find online about it merely states what you have to do, not why you must perform long division.
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How to find the oblique asymptote of this non rational fuction? 0. Oblique Asymptote. 0.
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Now when x = 0 (I do this because since I subsituted the eqn of the asymptote in (1) the resulting equation varies the same way as the asymptote and to find the intercept we just let x = 0) we get y = c thus we have c3 + c = 0 thus c = 0. As a result the asymptote to the curve is y = 2x. Share.
The function f f has an oblique asymptote y = ax + b y = a x + b when x → ∞ x → ∞ iff. limx→∞ f(x) x = a lim x → ∞ f (x) x = a. limx→∞(f(x) − ax) = b lim x → ∞ (f (x) − a x) = b. Similar conditions hold for the case x → −∞ x → − ∞. For f(x) = x arctan(x) f (x) = x arctan (x) we have when x → +∞ x → ...
1. If you've properly split the rational function you have into a "polynomial part" and a "proper rational function part"; that is, p(x) + r(x) s(x) p (x) + r (x) s (x) , where the degree of r(x) r (x) is less than the degree of s(x) s (x), see what happens as x x increases without bound. – J. M. ain't a mathematician.
Those are actually called rational functions. An Oblique asymptote for one of those is the same at $\pm \infty.$ For other functions you can have two distinct oblique asymptotes, $$ \frac{\sqrt{1 + x^6}}{1 + x^2} $$ is roughly $|x|.$ Oh, my original point: you get at most two oblique asymptotes, because you are asking about the graph of a function.