Back in the discontinuity zoo, one creature ran off to \(\pm\infty\) — the infinite discontinuity. Here’s its mechanism: when the denominator approaches \(0\) but the numerator doesn’t, the fraction grows without bound. We write \(\lim=\infty\), but careful — that’s shorthand for “grows forever,” not a real number. The limit technically does not exist. Watch a function approach its vertical asymptote:
What you’re looking at: \(f(x)=\frac{1}{x-2}\). As \(x\) nears the dashed asymptote at \(2\), the output rockets toward \(\pm\infty\) — up from the right, down from the left.
A vertical asymptote lives where the denominator is zero but the numerator isn’t. The one-sided limits run to \(+\infty\) or \(-\infty\) — so the two-sided limit does not exist.
The other kind of infinite limit asks: where does \(f\) settle as \(x\) marches off to \(+\infty\) or \(-\infty\)? If it levels toward a fixed height \(L\), that height is a horizontal asymptote. For rational functions, the answer is decided by a race between the degrees of the top and bottom:
What you’re looking at: a rational function whose top and bottom degrees you control. The dashed teal line is the horizontal asymptote, if one exists. Watch it move as the race changes.
The race rule, made precise. To find \(\lim_{x\to\pm\infty}\) of a rational function, compare the degrees — or divide every term by the highest power of \(x\) in the denominator and watch the small terms vanish. Step through all three cases:
A function can cross its horizontal asymptote — the HA only describes the end behavior, far out where \(x\) is huge. Near the middle, anything goes.
Here’s the reward for all that work on continuity. The Intermediate Value Theorem says something that feels obvious once you see it, but is surprisingly powerful: if \(f\) is continuous on \([a,b]\), and \(N\) is any value between \(f(a)\) and \(f(b)\), then \(f\) must equal \(N\) somewhere in between. A continuous curve can’t skip a value — no gaps, remember? Watch a root get cornered:
What you’re looking at: \(f(x)=x^3+x-1\), continuous everywhere. \(f(0)=-1\) (below zero) and \(f(1)=1\) (above zero). Since it’s continuous, it must cross zero in between — the IVT guarantees a root.
Continuity is essential. The function \(\frac1x\) jumps from \(-\infty\) to \(+\infty\) across \(0\) without ever being zero — because it’s discontinuous there. No continuity, no guarantee.
The exam tests one move above all: a sign change proves a root. If a continuous function is negative at one point and positive at another, it crossed zero somewhere between. For each case, decide whether the IVT guarantees what’s claimed:
The IVT proves a root exists — it never tells you where, or how many. And it needs continuity: drop that and all bets are off.
“\(\lim=\infty\)” means grows without bound — the limit doesn’t actually exist. It’s a description of how it fails.
End behavior of a rational function is decided by degrees: bottom bigger gives \(y=0\), equal gives the lead ratio, top bigger gives no horizontal asymptote.
A horizontal asymptote describes the far ends only. The graph may cross it freely in the middle — it’s a destination, not a wall.
The whole guarantee rests on no gaps. A single discontinuity lets the function skip the value — the theorem collapses.
Five problems — vertical and horizontal asymptotes, and the IVT.
Eight fast checks across the whole lesson.
An infinite limit means the limit does not exist — \(\infty\) is shorthand for unbounded growth, not a number it reaches. Say “DNE (infinite).”
Vertical: where the function blows up (denominator zero). Horizontal: where it settles as \(x\to\pm\infty\) (the degree race). Different questions entirely.
It can — freely, in the middle. The horizontal asymptote only governs the far ends. Only vertical asymptotes are truly forbidden territory.
The IVT requires \(f\) continuous on the whole closed interval. Skip that check and you can “prove” false things — like \(\frac1x\) hitting 0.
It only guarantees a root exists — never where it is or how many there are. It’s an existence theorem, not a locator.
The target value has to lie between \(f(a)\) and \(f(b)\). If it’s outside that range, the IVT promises nothing.