Question bank — Infinite limits and limits at infinity — vertical - horizontal asymptotes

False. Undefined hone ka matlab ek removable hole bhi ho sakta hai (jaise $x=3$ par $\frac{x^2-9}{x-3}$). Vertical asymptote ke liye ek one-sided limit ka $\pm \infty$ hona zaroori hai, sirf "yahan koi value nahi" kaafi nahi hai.

False. ${\lim }_{x\to 0}\frac{1}{x}$ exist nahi karta (sides aapas mein disagree karti hain $+\infty$ vs $-\infty$), phir bhi $x=0$ ek vertical asymptote hai — ek buri side kaafi hai.

False. $+\infty$ koi number nahi hai; yeh shorthand hai "har bound se zyada badhta hai" ke liye. Finite definition ke hisaab se limit exist nahi karti — hum sirf describe karte hain ki yeh kaise fail hoti hai.

True. Asymptote sirf long-run ($x\to \pm \infty$) behaviour ko govern karta hai. $\frac{\sin x}{x}$ jaisi curve $y=0$ ko infinitely baar cross karti hai phir bhi uski HA $y=0$ hai.

False. $x=a$ par function undefined hai (ya explode karta hai), isliye graph kabhi line $x=a$ ko touch ya cross nahi karta.

False. Iski do ho sakti hain — ek $x\to +\infty$ se, ek $x\to -\infty$ se — jaise $\frac{\sqrt{4x^2+1}}{x}\to \pm 2$.

False. Vertical asymptotes denominator ke un zeros se aati hain jo cancel nahi hote; ek polynomial mein finitely many roots hote hain, isliye finitely many asymptotes hoti hain.

False. Iski koi horizontal asymptote nahi hoti, lekin ek slant (oblique) asymptote ho sakti hai — dekho Slant (oblique) asymptotes via polynomial division.

False. Sirf $p>0$ ke liye. Agar $p<0$ toh yeh $x^{|p|}\to \infty$ hai; agar $p=0$ toh yeh constant $1$ hai.

Error: $\sqrt{x^2}=|x|$ hota hai, aur $x\to -\infty$ ke liye, $|x|=-x$ hota hai. Sahi value $-2$ hai; student ne sign flip ko ignore kiya.

Error: numerator bhi $0$ hai (form $\frac{0}{0}$). Cancel karne par $x+3\to 6$ milta hai, isliye yeh ek removable hole hai $(3,6)$ par, asymptote nahi. Dekho Continuity and removable discontinuities.

Error: dono sides disagree karti hain — right se $+\infty$, left se $-\infty$ — isliye ek single (two-sided) $\infty$ galat hai. Sahi yeh hai: limit exist nahi karti (haalaanki $x=0$ phir bhi ek VA hai). Dekho One-sided limits.

Error: $\frac{\infty }{\infty }$ indeterminate hai, $1$ nahi. Top aur bottom ko $x^2$ se divide karo taaki $\frac{3}{5}$ mile. (Ya L'Hôpital's rule for indeterminate forms use karo.)

Error: $(x-3{)}^2\to 0^{+}$ dono sides se hota hai (square kabhi negative nahi hota), aur numerator $\to 4>0$ hai, isliye dono sides $+\infty$ deti hain. Limit $+\infty$ hai ($x=3$ par VA hai).

Error: leading-coefficient rule ke liye equal degrees chahiye. Yahaan top degree $3>2$ hai, isliye limit $+\infty$ hai aur koi horizontal asymptote nahi hai.

Error: sirf ek side ka infinite hona kaafi hai, aur dono sides ke alag signs ho sakte hain (jaise $\frac{1}{x}$: right se $+\infty$, left se $-\infty$).

Yeh denominator ko nonzero leading constant $b_n$ ki taraf le jaata hai, isliye fraction ek clean "constant over constant" ban jaata hai; baaki har term ek constant times $1/x^k\to 0$ ban jaati hai.

Ek fixed nonzero top aur ek shrinking bottom ka matlab hai "ek fixed amount mein kitne tiny pieces fit hote hain" — yeh count bina kisi bound ke badhta hai; sign top aur bottom ke signs se $a$ ke paas fix hota hai.

Denominator ek side se positive values ke through $0$ approach kar sakta hai aur doosri side se negative values ke through, jo poore fraction ka sign flip kar deta hai — isliye dono directions alag-alag explode ho sakti hain.

Kyunki infinity par limit ka ek precise meaning hota hai: kisi bhi $\epsilon >0$ ke liye hum $N={\epsilon }^{-1/p}$ exhibit karte hain taaki $x>N$ hone par $1/x^p<\epsilon$ ho jaaye. Rigour wahi hai jo Limits — formal epsilon-delta definition demand karta hai.

Equal degrees ke saath leading terms dono $x^n$ se divide karne ke baad survive karte hain, aur ratio $a_m/b_n$ bachta hai; ek chhote top ke saath, har surviving term mein $x$ ki negative power hoti hai jo vanish ho jaati hai, aur $0$ milta hai.

Common factor cancel karne ke baad limit ek finite number hoti hai; agar hum us single point ko fill kar dein toh graph perfectly smooth hoga, isliye height kabhi explode nahi hoti.

Square root symbol nonnegative root return karta hai; agar $x$ negative hai, toh $x$ khud negative hai aur woh output nahi ho sakta, isliye nonnegative result guarantee karne ke liye $|x|$ zaroori hai.

Haan, agar doosri side ${\lim }_{x\to a^{-}}f=\pm \infty$ ho. Ek infinite one-sided limit kaafi hai line $x=a$ ko vertical asymptote banane ke liye.

Yeh $\pi /2$ hai; left side ${\lim }_{x\to -\infty }\arctan x=-\pi /2$ hai. Toh $\arctan$ ki do horizontal asymptotes hain, $y=\pm \pi /2$.

$e^{-x}=1/e^x\to 0$, isliye right par HA $y=0$ hai. (Left par, $e^{-x}\to +\infty$ — wahaan koi HA nahi aur koi finite value nahi.)

Nahi. $\frac{0}{0}$ indeterminate hai: cancel karne ke baad yeh ek finite number de sakta hai (hole), ya ek leftover factor phir bhi $\pm \infty$ tak blow up ho sakta hai (asymptote), jaise $\frac{x-3}{(x-3{)}^2}=\frac{1}{x-3}$.

Sirf ek line, $y=L$; ek hi value share karne wali do limits ek single asymptote describe karti hain jise dono ends se approach kiya jaata hai.

Nahi. Ek non-constant polynomial $x\to \pm \infty$ ke saath $\pm \infty$ tak bhaag jaata hai, isliye yeh kabhi kisi finite $L$ par settle nahi hota — koi horizontal asymptote nahi (aur koi vertical bhi nahi).

Horizontal asymptote $y=5$ (trivially, dono taraf ki limit $5$ hai), aur koi vertical asymptote nahi kyunki yeh kabhi blow up nahi karti.