Infinite Limits and Vertical Asymptotes
When f(x) grows without bound near a point, the limit is infinite (\(\pm\infty\)) — and that behavior pins down a vertical asymptote. The sign of the blow-up can differ on each side, so one-sided limits do the real work.
By the end you'll be able to decide whether f blows up to \(+\infty\) or \(-\infty\) on each side of a point using a quick sign check, and use infinite limits to locate vertical asymptotes.
Predict: as x approaches 2 from the right vs the left, does f blow up to \(+\infty\) or \(-\infty\) on each side? Drag x toward the dashed asymptote line to check the sign.
This is \(f(x) = \dfrac{1}{x-2}\). The slider skips right over x = 2 — f isn't defined there — and the readout below tracks the sign of \(x-2\), which decides everything. Then switch to \(g(x) = \dfrac{x+1}{(x-3)^2}\) and watch the squared denominator send both sides up to \(+\infty\). Hover or focus the curve for exact values.
x = 3.00 · x − 2 = +1.00 (positive) · f(x) = 1.00 → heading to +∞ on the right of the asymptote x = 2
A limit is infinite when f(x) grows without bound near a — which locates a vertical asymptote at x = a — and one-sided limits describe the sign of the blow-up on each side.
Divide a fixed nonzero number by something tiny and the result is enormous: \(1/0.001 = 1000\), \(1/0.000001 = 1{,}000{,}000\). That's the whole mechanism. When substitution gives (nonzero)/0, the numerator holds steady while the denominator shrinks to nothing, so the quotient's magnitude explodes — and the sign of that tiny denominator decides whether you rocket up to \(+\infty\) or plunge to \(-\infty\). Test one value just left and one just right of a, and you know both sides.
We write \(\lim_{x\to a} f(x) = +\infty\) (or \(-\infty\)) when f(x) increases (or decreases) without bound as x approaches a. This describes behavior — it is not a claim that the limit exists as a number, because ∞ is not a value f gets close to. The one-sided limits \(x \to a^-\) and \(x \to a^+\) can carry different signs; when they disagree, the two-sided limit does not exist, but each one-sided statement still precisely describes the graph. Whenever a limit is \(+\infty\) or \(-\infty\) at x = a, the line x = a is a vertical asymptote of the graph.
In the plot above, the curve hugs the dashed asymptote line and shoots off the top or bottom of the frame — the graph never touches the line, it just runs alongside it ever more steeply. With \(1/(x-2)\) the two arms point opposite ways: up on the right, down on the left. Switch to \((x+1)/(x-3)^2\) and both arms point up, because the squared denominator is positive on both sides. Odd vs even power of the vanishing factor is exactly the difference you see.
One caution from the previous lesson: not every 0 in a denominator means an infinite limit. When substitution gives the indeterminate form 0/0, factoring can cancel the offending factor and leave a perfectly finite limit. Here the situation is different: \(1/(x-2)\) at x = 2 gives 1/0 — nonzero over zero — and no factoring can rescue it, because nothing in the numerator cancels the vanishing denominator. Substitute first and read the form: 0/0 means "do more algebra"; (nonzero)/0 means "infinite limit — now find the sign on each side."
Find the vertical-asymptote behavior of \(f(x) = \dfrac{x+1}{(x-3)^2}\). At x = 3 the numerator \(\to 4\) (nonzero) and the denominator \((x-3)^2 \to 0^+\) from both sides — a square is never negative. So positive-over-small-positive gives \[\lim_{x\to 3^-} f(x) = +\infty \quad\text{and}\quad \lim_{x\to 3^+} f(x) = +\infty.\] The graph rises to \(+\infty\) on both sides of the vertical asymptote x = 3.
Analyze \(h(x) = \dfrac{3}{x-5}\) near x = 5. Substitution gives 3/0 — nonzero over zero — so the limit is infinite and x = 5 is a vertical asymptote. From the right (try x = 5.001): \(x - 5\) is a small positive, so \(h(x) \to +\infty\). From the left (try x = 4.999): \(x - 5\) is a small ____, so \(h(x) \to\) ____
Reveal the answer
From the left, \(x - 5\) is a small negative, so \(3/(x-5)\) is a huge negative: \(\lim_{x\to 5^-} h(x) = -\infty\). The two sides disagree, so the two-sided limit does not exist — but each one-sided statement exactly describes the graph: up on the right of x = 5, down on the left. Same shape you saw for \(1/(x-2)\) in the explorer above.
More info — a quick sign-check recipe (and two pitfalls)
Recipe: once substitution gives (nonzero)/0, pick a test value just left of a (like 1.999 for a = 2) and just right (2.001), and track only the signs of the numerator and denominator — positive over small-positive blows up to \(+\infty\), positive over small-negative to \(-\infty\), and so on. Two pitfalls from this lesson's source material: don't call an infinite limit "existent" — writing "\(= \infty\)" describes behavior, while a finite limit is a genuinely different statement; and don't confuse a vertical asymptote (an infinite limit at a finite a) with a horizontal asymptote (a finite limit as \(x \to \pm\infty\) — that's the next lesson). For more sign-analysis practice, work the Paul's Online Notes link in Dive deeper below.
Check your understanding
For \(f(x) = \dfrac{1}{x-2}\), what are \(\lim_{x\to 2^+} f(x)\) and \(\lim_{x\to 2^-} f(x)\)?
For \(k(x) = \dfrac{2-x}{(x+1)^2}\), what happens on BOTH sides of \(x = -1\)?
Compare \(\lim_{x\to 3} \dfrac{x^2-9}{x-3}\) with \(\lim_{x\to 3} \dfrac{x+9}{x-3}\). Which statement is correct?
Recap
- \(\lim_{x\to a} f(x) = \pm\infty\) means f grows (or falls) without bound near a — a description of behavior, not a number the limit equals.
- Substitution giving (nonzero)/0 signals an infinite limit — and then x = a is a vertical asymptote. (0/0 is different: it's indeterminate and may hide a finite limit.)
- Find the sign of the blow-up side by side: test a value just left and just right of a and track the sign of the shrinking denominator.
- An even power like \((x-a)^2\) in the denominator is positive on both sides, so both one-sided limits share one sign; an odd power splits them, and then the two-sided limit does not exist.
Dive deeper
- Paul's Online Notes — Infinite Limits Analyze sign of the blow-up on each side and identify vertical asymptotes
Sources
- Infinite Limits and Vertical Asymptotes