If there is, I would use the easy rule of making a ratio out of the coefficients of the highest degree terms containing x. There has to be trick as solving it to find whether there is such an asymptote or not, although I have not been able to find one through google.
f(x) = (3x^2-48)/(x^2-9x+20)
... would have one at 3. (ratio of leading terms with x is 3/1)
An answer on how to do this in a most simple and non-mathy as possible way gets the full 5 stars.
f(x) = (3x^2-48)/(x^2-9x+20)
... would have one at 3. (ratio of leading terms with x is 3/1)
An answer on how to do this in a most simple and non-mathy as possible way gets the full 5 stars.
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If the degree of the numerator is greater than the degree of the denominator, then there is no horizontal asymptote. If the degree of the denominator is greater, then the horizontal asymptote is y = 0. And, like you've stated, if the numerator and denominator are equal in degree, then the horizontal asymptote is equal to the ratio of the leading coefficient of the numerator to the leading coefficient of the denominator.
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but you just did it. limit of ratio of leading terms, 3x²/x², as x → ∞ is 3/1 = 3
if bottom exponent is bigger, the limit becomes n/∞ = 0
if top is bigger, no horizontal asymptote at all.
if bottom exponent is bigger, the limit becomes n/∞ = 0
if top is bigger, no horizontal asymptote at all.