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Calculus learning

Limits at Infinity: Leading Terms, Signs and Horizontal Asymptotes

Limits at infinity describe what happens as the variable grows without bound. Polynomial and rational functions can be handled with leading terms, without choosing an arbitrary large test number. Signs still matter, especially for odd powers and negative infinity.

Direct answer

For a rational function, compare numerator and denominator degrees; equal degrees give the ratio of leading coefficients.

End behavior at +∞ and −∞ can differ. These are limits, not evaluations at a numeric value called infinity.

Limits at Infinity worked method

For (3x²+1)/(2x²−5), divide numerator and denominator by x². The vanishing terms leave 3/2 at either infinite end.

If the denominator degree is larger, dividing by its leading power leaves zero in the numerator, so the rational limit is zero.

If the numerator degree is larger, compare the excess power and the ratio of leading coefficients to decide growth and sign.

For x³/(x²+1), the leading behavior is x. The positive-infinity limit is +∞ and the negative-infinity limit is −∞.

Radicals require extra care: √(x²)=|x|, not x. This is why √(x²+1)/x has different signs at the two ends. The preset remains in the exact rational-rule scope.

1
lim x→±∞ (3x²+1)/(2x²−5)=3/2
2
lim x→±∞ x/(x²+1)=0
3
lim x→−∞ x³/(x²+1)=−∞

Worked examples

lim x→∞ (5x+2)/(2x−1)
52\frac{5}{2}

Equal degrees.

lim x→∞ x/(x²+1)
00

Denominator degree is larger.

lim x→−∞ x³/(x²+1)
−∞-\infty

Odd excess power changes sign.

Common mistakes

  • • Substituting ∞ as though it were a real number.
  • • Cancelling leading coefficients while ignoring degree.
  • • Replacing √(x²) with x for negative x.

Try the supported calculator preset

The preset opens a supported expression with its variable and conditions. Teaching examples elsewhere in this guide are reference explanations, not claims that the engine supports every method.

This example uses the engine’s exact polynomial/rational end-behavior analysis, not large-number substitution.

極限:3 x2+12 x2−5\frac{3\,x^{2} + 1}{2\,x^{2} - 5}
Open the calculator without a preset: 極限

References

Frequently asked questions

Is an infinite limit a horizontal asymptote?+

No. A finite end limit L gives a horizontal asymptote y=L.

Can the two ends have different limits?+

Yes. Check +∞ and −∞ separately.

What input specifies infinity?+

The calculator accepts infinity/inf in the approach field; the preset preserves the approach and direction.

Conclusion

Limits at Infinity starts with the stated domain and conditions. Recheck those before carrying a worked example into a different problem.

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