4.1.4 · D2Calculus I — Limits & Derivatives
Visual walkthrough — Infinite limits and limits at infinity — vertical - horizontal asymptotes
Step 1 — What "sending to infinity" even looks like
WHAT. Before any algebra, let us watch the function. Pick bigger and bigger: and plot the height .
WHY. A limit at infinity is a question about a trend, not a single value. We never reach infinity; we ask "where is the height heading?" You cannot answer that without first seeing the trend with your own eyes — so we sample it.
PICTURE. In the figure the pale-yellow dots march right. Their heights start scattered () but then hug a flat blue line. That blue line is the horizontal asymptote we are chasing.
Step 2 — Why raw numbers can't finish the job
WHAT. Look at the numerator and denominator separately as grows.
WHY. Naïvely you'd plug in : top , bottom . That gives — a symbol that means "I don't know yet", an indeterminate form. Both parts explode, and a race between two runaways has no obvious winner. We need a fairer way to compare their speeds.
PICTURE. Two chalk curves both rocket upward off the top of the board — the pink (top) and blue (bottom). You can't read a ratio off two things running to the ceiling. That is the whole problem, drawn.
Step 3 — The key move: divide top and bottom by the biggest denominator power
WHAT. The largest power downstairs is . Divide every term, top and bottom, by :
=\frac{\dfrac{3x^2}{x^2}-\dfrac{x}{x^2}}{\dfrac{5x^2}{x^2}+\dfrac{7}{x^2}} =\frac{\;3-\dfrac{1}{x}\;}{\;5+\dfrac{7}{x^2}\;}.$$ Term by term: - $\dfrac{3x^2}{x^2}=3$ — the top's leading term becomes a plain constant. - $\dfrac{x}{x^2}=\dfrac1x$ — a shrinking piece. - $\dfrac{5x^2}{x^2}=5$ — the bottom's leading term becomes a plain constant. - $\dfrac{7}{x^2}$ — another shrinking piece. **WHY.** Dividing top and bottom by the *same* thing $x^2$ doesn't change the fraction's value (it's multiplying by $\tfrac{x^2}{x^2}=1$). But it **converts every growing term into either a constant or a fraction of the form $\tfrac{1}{x^p}$** — and those tiny fractions we already know how to kill. We are turning an unreadable race into a tidy sum of things that either *stand still* or *fade to zero*. **PICTURE.** The figure shows the four terms as separate chalk bars. The two constant bars ($3$ and $5$) stay tall; the two "$1/x^p$" bars visibly shrink as $x$ grows — annotated with down-arrows "$\to 0$". ![[deepdives/dd-maths-4.1.04-d2-s03.png]] --- ## Step 4 — Kill the shrinking terms with the master fact **WHAT.** Send $x\to\infty$. Every $\tfrac{1}{x^p}$ with $p>0$ dies: $$\frac1x\to0,\qquad \frac{7}{x^2}\to0.$$ **WHY.** This is the one engine the parent note built from the $\varepsilon$–$N$ definition: for any tininess $\varepsilon$ you demand, once $x>\varepsilon^{-1/p}$ the term is smaller than $\varepsilon$. It **must** reach zero. We use it because it is the *only* tool that converts "$x$ gets huge" into an exact number we can write down. **PICTURE.** The two shrinking chalk bars from Step 3 fade to nothing (dotted stubs on the board), while the constants $3$ and $5$ remain solid. ![[deepdives/dd-maths-4.1.04-d2-s04.png]] > [!recall]- Why $1/x^p\to 0$ in one breath > Want $\tfrac{1}{x^p}<\varepsilon$? ::: Solve for $x$: need $x>\varepsilon^{-1/p}$; past that point the term stays smaller than any $\varepsilon$ you name, which is exactly what "$\to 0$" means. --- ## Step 5 — Read off the limit and draw the ceiling **WHAT.** With the shrinking terms gone: $$\lim_{x\to\infty}\frac{3-\frac1x}{5+\frac{7}{x^2}}=\frac{3-0}{5+0}=\frac{3}{5}.$$ Each surviving symbol: $3$ = numerator's leading coefficient, $5$ = denominator's leading coefficient. The limit is their **ratio**. **WHY.** Only the highest-power terms had the muscle to survive division by $x^2$; everything else was outrun and vanished. So the *leading coefficients alone* dictate the destination — this is the degree-equal rule ("$m=n\Rightarrow y=a_m/b_n$") *derived*, not memorised. **PICTURE.** The blue horizontal line $y=\tfrac35$ is drawn, and the curve is shown creeping up to hug it without touching — a ceiling. ![[deepdives/dd-maths-4.1.04-d2-s05.png]] > [!formula] Degree-equal shortcut, now earned > If $\deg(\text{top})=\deg(\text{bottom})$, then $\displaystyle\lim_{x\to\pm\infty}\frac{a_mx^m+\cdots}{b_nx^n+\cdots}=\frac{a_m}{b_n}$. > Here $\dfrac35=0.6$ — check against Step 1's dots, which flattened right around $0.6$. ✓ --- ## Step 6 — The other two degree cases, side by side **WHAT.** Same divide-by-$x^n$ trick, but change which side wins. - **Top smaller** ($m<n$), e.g. $\dfrac{x}{5x^2+7}=\dfrac{1/x}{5+7/x^2}\to\dfrac{0}{5}=0$: ceiling at $y=0$. - **Top bigger** ($m>n$), e.g. $\dfrac{2x^3+1}{x^2}=2x+\dfrac1{x^2}\to+\infty$: **no** ceiling; it flies off (possibly a [[Slant (oblique) asymptotes via polynomial division|slant asymptote]]). **WHY.** The division always leaves the winner as a constant and the loser as a $1/x^p$. If the *top* is the loser it dies to $0$; if the top is the winner an honest positive power of $x$ survives and grows forever. Covering all three cases means the reader never meets a rational function they can't classify. **PICTURE.** Three mini-panels: top-heavy curve blasting up (no ceiling), bottom-heavy curve flattening to $y=0$, and equal-degree curve flattening to $y=\tfrac35$. ![[deepdives/dd-maths-4.1.04-d2-s06.png]] > [!mnemonic] BOTU (from the parent) > **B**ottom bigger → $0$. **O**dds even, i.e. equal degrees → ratio of leading coefficients. **T**op bigger → b**U**st, no horizontal asymptote. --- ## Step 7 — The sign trap when $x\to-\infty$ (roots) **WHAT.** March **left** instead of right, and add a square root: $$\lim_{x\to-\infty}\frac{\sqrt{4x^2+1}}{x}.$$ Factor $x^2$ out of the root — but carefully: $\sqrt{x^2}=|x|$, and for $x<0$, $|x|=-x$. $$\sqrt{4x^2+1}=|x|\sqrt{4+\tfrac1{x^2}}=-x\sqrt{4+\tfrac1{x^2}} \;\Rightarrow\;\frac{-x\sqrt{4+1/x^2}}{x}=-\sqrt{4+\tfrac1{x^2}}\to-2.$$ **WHY.** A square root is never negative, but $x$ *is* negative on the left. If you blindly wrote $\sqrt{x^2}=x$ you would carry a wrong sign all the way to a wrong ceiling. Pulling the root out as $-x$ keeps the arithmetic honest. This is why a curve can own **two different** horizontal ceilings — $+2$ on the right, $-2$ on the left. **PICTURE.** The board shows both branches: the right branch flattening to $y=+2$ (blue), the left branch flattening to $y=-2$ (pink), with the sign flip circled. ![[deepdives/dd-maths-4.1.04-d2-s07.png]] > [!mistake] Writing $\sqrt{x^2}=x$ as $x\to-\infty$ > **Why it feels right:** it's true for the positive $x$ we usually meet. > **The fix:** $\sqrt{x^2}=|x|$, and $|x|=-x$ when $x<0$. Forgetting this flips the sign of your whole answer. --- ## Step 8 — The *vertical* wall: the same idea, denominator hitting zero **WHAT.** Switch questions: keep $x$ **finite** but push the denominator to $0$ with a nonzero top, e.g. $\dfrac{x+1}{(x-3)^2}$ near $x=3$. **WHY.** A limit at infinity asked "what happens when $x$ runs away?" A vertical asymptote asks the mirror question: "what happens when the *height* runs away while $x$ stays put?" A nonzero number over something shrinking to $0$ has size exploding to $\infty$; the **sign** is fixed by the signs of top and bottom on each side (that's why [[One-sided limits]] matter). Here $(x-3)^2\ge0$, so the bottom is $0^+$ from both sides and the top is $+4$: $+\infty$ both ways. **PICTURE.** A vertical dashed wall at $x=3$; the curve rockets to $+\infty$ on both sides, contrasted with a small inset of $1/x$ where the two sides *disagree* ($+\infty$ vs $-\infty$). ![[deepdives/dd-maths-4.1.04-d2-s08.png]] > [!mistake] "Bottom $=0$ always means a vertical wall" > **The fix:** if the top also $\to0$ you have $\tfrac00$ — factor and cancel first; you may get a [[Continuity and removable discontinuities|removable hole]], not a wall. Example: $\dfrac{x^2-9}{x-3}=x+3\to6$, a hole at $(3,6)$. --- ## The one-picture summary ![[deepdives/dd-maths-4.1.04-d2-s09.png]] The final board puts both stories together: the **horizontal ceiling** $y=\tfrac35$ that $\tfrac{3x^2-x}{5x^2+7}$ hugs as $x$ runs away, and a **vertical wall** where a denominator hits zero — the two ways a graph's skeleton is fixed. > [!recall]- Feynman retelling — the whole walkthrough in plain words > We wanted to know where the curve's height goes when $x$ walks off to the right forever. Plugging in "infinity" just gave $\tfrac{\infty}{\infty}$ — a shrug. So we played fair: we divided every piece, top and bottom, by the biggest power of $x$ downstairs. That turned every term into either a *number that stands still* (the leading coefficients $3$ and $5$) or a *fraction that fades to nothing* ($\tfrac1x,\tfrac{7}{x^2}$). Once the fading pieces died, only $\tfrac{3}{5}$ was left — the ceiling. Same recipe classifies every rational function: bottom wins ⇒ ceiling at $0$, top wins ⇒ no ceiling, tie ⇒ ratio of the leaders. Going *left* to $-\infty$ we had to remember a square root is never negative, so $\sqrt{x^2}=-x$ there, which can flip the ceiling's sign. And the mirror-image question — height exploding while $x$ stays put — gives vertical walls, provided the top isn't also zero (else it's just a fixable hole). ## Connections - [[Infinite limits and limits at infinity — vertical - horizontal asymptotes]] - [[One-sided limits]] - [[Continuity and removable discontinuities]] - [[Slant (oblique) asymptotes via polynomial division]] - [[L'Hôpital's rule for indeterminate forms]] - [[Limits — formal epsilon-delta definition]] - [[Curve sketching using first and second derivatives]]