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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.
Not necessary to perform long division as it is not clear why it should give slant asymptote any way. Better to go like this below, If y= mx+c is asymptote then it must be true that lim x tends to infinity of f(x)-(mx+c) is zero.
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.
Types of Asymptotes. There are three types of asymptotes that a rational function could have: horizontal, vertical, or slant (oblique). Figure 3 is the graph of {eq}\frac {4x^2-6} {x^2+8} {/eq ...
To find the oblique asymptote, you must use polynomial long division, and then analyze the function as it approaches infinity. Taking the limit first, like HallsofIvy did, is wrong because 11/x and 1/x approach infinity at different rates, and therefore add to the numerator and denominator in slightly different ways.
Is there a theorem that can tell us the conditions for which function will have an oblique asymptote? $\endgroup$ – Deadpool 36701 Commented Jun 25, 2020 at 18:44
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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.
0. In order to find an oblique asymptote, we need to find some function ϕ(x) = kx + n ϕ (x) = k x + n or in other words, to find k k and n n. Finding n n is pretty straightforward: limx→∞(f(x) − ϕ(x)) limx→∞(f(x) − kx − n) limx→∞(f(x) − kx) = 0 = 0 = n lim x → ∞ (f (x) − ϕ (x)) = 0 lim x → ∞ (f (x) − k x − ...