Math · Calculus I · Worked example
Find a limit at infinity of a rational function
Find the limit of (3x² − 5x + 1)/(2x² + 7) as x → ∞, and name the horizontal asymptote.
Divide by the highest power
The highest power in the denominator is x². Divide every term of the numerator and the denominator by it; for x ≠ 0 the value does not change.
Let x grow
5/x, 1/x² and 7/x² all tend to 0, so the numerator tends to 3 and the denominator to 2.
Name the asymptote
The limit is 3/2 = 1.5, so y = 3/2 is a horizontal asymptote. The same steps give 3/2 as x → −∞. The limit is the ratio of the leading coefficients because the degrees are equal.
Compare other degrees
For (4x + 1)/(x² − 3) the numerator has the lower degree, and dividing by x² sends everything above to 0: the limit is 0. For 5x³/(x² + 1) the numerator has the higher degree, and the function grows without bound.
Result
The limit is 3/2, and y = 3/2 is a horizontal asymptote at both ends.
Your turn
Find the limit of (6x³ − x)/(1 − 2x³) as x → ∞.
Show the answer and explanation
−3.
Divide by x³: (6 − 1/x²)/(1/x³ − 2) → 6/(−2) = −3. The degrees are equal, so the limit is the ratio of the leading coefficients.
Keep exploring
Limits & one-sided behavior opens with this function as x → ∞ and verifies 3/2. Enter −∞ as the approach point to check the left end.
Return to the concept →Sources and scope
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Try in the workspace
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Check the limit in Limits & one-sided behavior Graph the function and its asymptote Open worked example on a board End behavior in Math ReferenceYour existing work stays on this device. Examples open as editable copies.