Infinite limits and limits at infinity — vertical - horizontal asymptotes

1. Infinite limits (vertical asymptotes)

WHAT happens & WHY: This typically occurs when you divide a nonzero number by something heading to $0$ . As the denominator $\to 0$ , the fraction's size $\to\infty$ . The sign of the result depends on the signs of numerator and denominator near $a$ — that's why we check left and right separately.

2. Limits at infinity (horizontal asymptotes)

Rational functions — derive the rule

Take $f(x)=\dfrac{a_mx^m+\dots}{b_nx^n+\dots}$ . HOW we evaluate $x\to\infty$ : divide top and bottom by the highest power in the denominator, $x^n$ , so each term becomes a constant times $1/x^k$ which we can send to $0$ .

$\frac{a_mx^m+\dots+a_0}{b_nx^n+\dots+b_0}=\frac{a_mx^{m-n}+\dots}{b_n+\dots}.$

If $m<n$ : numerator has only negative powers $\to0$ . Limit $=0$ , asymptote $y=0$ .
If $m=n$ : leading terms survive. Limit $=\dfrac{a_m}{b_n}$ , asymptote $y=a_m/b_n$ .
If $m>n$ : numerator grows without bound. Limit $=\pm\infty$ , no horizontal asymptote (maybe a slant one).

Figure — Infinite limits and limits at infinity — vertical - horizontal asymptotes

3. Forecast-then-Verify drill

Predict before reading the answer:

$\lim_{x\to\infty}\dfrac{2x^3+1}{x^2}$ ? → degree $3>2$ , $+\infty$ , no HA.
$\lim_{x\to 2^-}\dfrac{1}{x-2}$ ? → $x-2\to0^-$ , so $\dfrac{1}{0^-}=-\infty$ .
$\lim_{x\to\infty} e^{-x}$ ? → $e^{-x}=1/e^x\to0$ , HA $y=0$ .
$\lim_{x\to\infty}\arctan x$ ? → $\pi/2$ , HA $y=\pi/2$ (and $-\pi/2$ at $-\infty$ ).

Recall Feynman: explain it to a 12-year-old

Imagine a road (the graph). A vertical wall ( $x=a$ ) is a place where the road suddenly rockets straight up to the sky or plunges down — you can get super close to that spot left/right but the road's height goes crazy. A horizontal ceiling ( $y=L$ ) is when you walk forever to the right and the road flattens out, hugging a flat line but never quite touching it. Vertical wall = $x$ frozen, height explodes. Horizontal ceiling = $x$ runs away, height calms down.

Active recall #flashcards/maths

Definition of

\lim_{x\to a}f(x)=+\infty

in words

For any

M

, there's

\delta

0<|x-a|<\delta

forces

f(x)>M

When is

x=a

a vertical asymptote?

When a one-sided limit at

a

+\infty

-\infty

Denominator

\to0

AND numerator

\to0

— asymptote or not?

Possibly a removable hole; factor/cancel first.

\lim_{x\to\infty}1/x^p

for

p>0

0

HA of a rational function when deg(top)=deg(bottom)?

y=

ratio of leading coefficients.

HA when deg(top)>deg(bottom)?

None (limit is

\pm\infty

\sqrt{x^2}

equals?

|x|

; equals

-x

when

x<0

\lim_{x\to-\infty}\arctan x

-\pi/2

\lim_{x\to 0^-}1/x

-\infty

Max number of horizontal asymptotes a function can have?

Two (one at

+\infty

, one at

-\infty

Connections

Limits — formal epsilon-delta definition
One-sided limits
Continuity and removable discontinuities
Curve sketching using first and second derivatives
Slant (oblique) asymptotes via polynomial division
L'Hôpital's rule for indeterminate forms

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, do alag-alag "blow up" situations hote hain limits mein. Pehla: infinite limit. Yahan $x$ kisi finite number $a$ ke paas jaata hai, lekin function ki value $f(x)$ upar ya neeche $\infty$ ki taraf bhaag jaati hai. Jaise $1/x$ mein jab $x$ chhota positive hota hai to value rocket ki tarah upar, aur chhota negative hone par neeche. Aisi jagah par graph mein ek vertical asymptote banta hai — ek seedhi khadi deewar $x=a$ .

Dusra: limit at infinity. Ab $x$ khud $\infty$ ya $-\infty$ ki taraf bhaag raha hai, aur dekhna hai $f(x)$ kis finite number $L$ par settle hota hai. Yeh $L$ ek horizontal asymptote $y=L$ deta hai — ek flat ceiling jise graph chhune ki koshish karta hai par chhuta nahin. Rational functions ke liye trick simple hai: top aur bottom dono ko denominator ki sabse badi power se divide karo, phir $1/x^p \to 0$ laga do. Degree equal ho to leading coefficients ka ratio, top chhota ho to $0$ , top bada ho to koi horizontal asymptote nahin.

Do bade dhyaan dene wale points: (1) Sirf denominator zero hone se vertical asymptote nahin banta — agar numerator bhi zero ho raha hai ( $\frac00$ form), to factor cancel ho sakta hai aur sirf ek hole banega. (2) $\sqrt{x^2}=|x|$ , aur jab $x\to-\infty$ ho to $|x|=-x$ — yahan sign flip karna mat bhoolna, warna answer ka chinh galat aa jaayega.

Yeh cheez kyun important hai? Kyunki asymptotes graph ka skeleton hote hain. Bina ek bhi point plot kiye, tum keh sakte ho ki curve kahan deewar se takraayega aur kahan flat ho jaayega — yeh curve sketching ka 80/20 hai.

Infinite limits and limits at infinity — vertical - horizontal asymptotes

1. Infinite limits (vertical asymptotes)

2. Limits at infinity (horizontal asymptotes)

Rational functions — derive the rule

3. Forecast-then-Verify drill

Active recall #flashcards/maths

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Calculus I — Limits & Derivatives

Connections