A continuous distribution's density often has an infinite domain, and its total area still has to equal exactly 1. That means "what happens to f(x) as x grows without bound" is a question this course returns to constantly. It starts here, with the exponential and normal distributions.
Picture zooming a graph's camera out further and further to the right. If the curve settles onto a flat, level line no matter how far you zoom, that flat line's height is the limit at infinity. If the curve keeps climbing, or diving, as the camera pulls back forever, no settled height exists. In that case the limit is not a finite number.
Limits at infinity are values L that f(x) approaches as x grows without bound, written \displaystyle\lim_{x\to\infty} f(x) = L. In that case L is called a horizontal asymptote of f: the flat height the graph settles onto far to the right. For a rational function, the outcome depends on comparing the degrees of numerator and denominator. If the denominator's degree is larger, the limit is 0, since the bottom grows faster and shrinks the fraction toward nothing. If the degrees match, the limit is the ratio of leading coefficients. If the numerator's degree is larger, the limit does not exist as a finite number, because the function grows without bound instead.
Drag the coefficients, watch the asymptote move
Drag a_0/a_1 (numerator) and b_0/b_1 (denominator) — the horizontal asymptote's height is always the ratio of the two x-coefficients.
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Worked example
Degree of denominator wins
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1
Evaluate \displaystyle\lim_{x\to\infty} \frac{3x+1}{x^2-4}.
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2
The denominator's degree (2) exceeds the numerator's degree (1).
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3\boxed{\lim_{x\to\infty}\frac{3x+1}{x^2-4} = 0}
Worked example
Matching degrees
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1
Evaluate \displaystyle\lim_{x\to\infty} \frac{4x^2+x}{2x^2-5}.
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2
The degrees match (both 2), so the limit is the ratio of leading coefficients, \frac{4}{2}.
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3\boxed{\lim_{x\to\infty}\frac{4x^2+x}{2x^2-5} = 2}
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Where this goes next
Chapter 8 (Exponential and normal distributions)
An exponential density f(x) = \lambda e^{-\lambda x} has to satisfy \lim_{x\to\infty} f(x) = 0 for its total area to come out finite at all. That is the same horizontal-asymptote reasoning practiced here, just with an exponential curve standing in for a rational one.
Check your understanding (counts toward your progress)
Evaluate \displaystyle\lim_{x\to\infty}\dfrac{x^2-1}{5x^2-3}.
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Why
Same-degree top and bottom — the limit is the ratio of leading coefficients, \dfrac{1}{5} \approx 0.2.
Correct answer
Also accepted:
Why
Evaluate \displaystyle\lim_{x\to\infty}\dfrac{7x^2+1}{9x^2-3}.
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Why
Same-degree top and bottom — the limit is the ratio of leading coefficients, \dfrac{7}{9} \approx 0.7778.
Correct answer
Also accepted:
Why
What is \displaystyle\lim_{x\to\infty} \frac{5x^2+1}{2x^2-3}?
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0 is the answer when the DENOMINATOR has the higher degree — here the degrees match.
The limit doesn't grow without bound here — the degrees match, so it settles on a finite ratio.
The ratio isn't 1 — it's the ratio of the LEADING coefficients, 5 and 2, not just any coefficients.
Why
The degrees match (both 2), so the limit is the ratio of leading coefficients: 5/2.
Compare the degrees of numerator and denominator first — that decides which case you're in.
Determine \displaystyle\lim_{x\to\infty}\frac{x^3+1}{x^2-4} by comparing the degrees of numerator and denominator.
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0 is the answer when the DENOMINATOR has the higher degree — here the NUMERATOR's degree is higher.
The ratio of leading coefficients (1) only gives the limit when the degrees are EQUAL — here they're not.
1/4 doesn't match either case here — recheck which polynomial has the higher degree.
Why
The numerator's degree (3) is higher than the denominator's (2), so the fraction grows without bound: the limit is ∞.
Compare the DEGREES of numerator and denominator first — the higher one wins the growth race.
Practice: Reading degree comparisons for limits at infinity (5 problems · doesn’t affect your progress)
A rational function's degree in the numerator equals its degree in the denominator. What determines its limit as x\to\infty?
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Constant terms become negligible for large x — only the highest-power (leading) terms matter.
The NUMBER of terms doesn't matter at all — only the degree (highest power) and leading coefficient do.
Equal degrees is exactly the case where a finite limit DOES exist — it's the ratio of leading coefficients.
Why
When the degrees match, every lower-power term becomes negligible for large x, leaving just the ratio of the leading coefficients.
The ratio of the two leading coefficients
Evaluate \displaystyle\lim_{x\to\infty}\frac{4x+7}{x^2-1}.
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Why
The denominator's degree (2) exceeds the numerator's degree (1), so the limit is 0.
Correct answer
Also accepted:
Why
Find \displaystyle\lim_{x\to\infty}\frac{9x^3-2x}{3x^3+x^2}.
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Why
Both degrees are 3, so the limit is the ratio of leading coefficients: 9/3 = 3.
Correct answer
Also accepted:
Why
Only the highest-power terms in numerator and denominator matter for the limit.
Compute \displaystyle\lim_{x\to\infty}\frac{x^4}{x^2+1}.
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0 happens when the DENOMINATOR has the higher degree — here it's the numerator that's higher.
1 would apply only if the degrees matched — they don't here.
There's no ratio of leading coefficients to take when the degrees aren't equal — that shortcut only works in the equal-degree case.
Why
The numerator's degree (4) exceeds the denominator's (2), so the fraction grows without bound as x→∞: the limit is ∞.
Compare the degrees first: 4 versus 2.
∞
A candidate density function is f(x)=\dfrac{5}{x^2+1}. Does \displaystyle\lim_{x\to\infty} f(x)=0, as required for its total area to possibly be finite?
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The constant 5 is only the NUMERATOR — what matters for the limit is comparing the numerator's degree (0, since it's just a constant) to the denominator's degree (2).
This function does have a well-defined limit — it's a ratio of polynomials with the denominator's degree higher, the exact case that gives limit 0.
Whether the constant is prime is irrelevant — this works for ANY constant, since only the DEGREES of numerator and denominator decide the limit.
Why
Treat the numerator 5 as a degree-0 polynomial. The denominator's degree (2) is higher, so by the same rule used throughout this section, the limit is 0 — satisfying the requirement for the density's tail to vanish.
Treat the numerator 5 as a degree-0 polynomial, then compare degrees exactly as usual.
Yes — the denominator's degree (2) exceeds the numerator's degree (0), so the ratio shrinks to 0