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

WHAT: vertical asymptotes tab aate hain jab denominator $0$ ho jaata hai aur top nonzero rehta hai. $x-4=0\Rightarrow x=4$ set karo. Top $5\ne 0$ hai wahan. Answer: vertical asymptote $x=4$. (Dono sides par sign check karo taaki shape pata chale: $x\to 4^{+}$ par, $x-4\to 0^{+}$, toh $5/0^{+}=+\infty$; $x\to 4^{-}$ par, $5/0^{-}=-\infty$.)

WHY this rule: top degree $=1$, bottom degree $=1$ — equal degrees. Engine 2 ke hisaab se limit leading coefficients ka ratio hai. ${\lim }_{x\to \pm \infty }\frac{7x}{2x+9}=\lim \frac{7}{2+9/x}=\frac{7}{2}.$ Answer: horizontal asymptote $y=\frac{7}{2}$.

Top degree $3$, bottom degree $2$: top bada hai. Numerator tez grow karta hai, toh limit $\pm \infty$ hai — koi leveling off nahi. Answer: Nahi horizontal asymptote.

WHY divide by $x^2$: denominator mein highest power $x^2$ hai; divide karne se constants aur woh terms expose hoti hain jo mar jaati hain. $={\lim }_{x\to \infty }\frac{4-\frac{3}{x}+\frac{1}{x^2}}{2+\frac{1}{x}-\frac{5}{x^2}}=\frac{4-0+0}{2+0-0}=2.$ Answer: $2$ (horizontal asymptote $y=2$).

$x^2$ se divide karo (bottom ki highest power): $={\lim }_{x\to \infty }\frac{\frac{6}{x}+\frac{2}{x^2}}{1+\frac{1}{x^2}}=\frac{0+0}{1+0}=0.$ Top degree $<$ bottom degree, toh yeh $0$ hona hi tha. Answer: $0$ (horizontal asymptote $y=0$).

WHAT: top $\to 5+2=7$ (nonzero, positive). Bottom $(x-5{)}^2\to 0$, aur ek square hamesha $\ge 0$ hota hai, toh yeh dono sides se $0^{+}$ approach karta hai. Positive $÷0^{+}=+\infty$ left aur right dono se. Answer: $+\infty$. Vertical asymptote $x=5$; curve dono sides se upar rocket karta hai.

Top $\to 3-1=2$ (positive, nonzero). Bottom $x-3\to 0$; uska sign flip hota hai $3$ ke across.

• Right, $x\to 3^{+}$: $x-3\to 0^{+}$. Positive $÷0^{+}=+\infty$.
• Left, $x\to 3^{-}$: $x-3\to 0^{-}$. Positive $÷0^{-}=-\infty$. Dono sides disagree karte hain, toh two-sided ${\lim }_{x\to 3}$ exist nahi karta — lekin ek single infinite one-sided limit kaafi hai. Answer: $x=3$ hai ek vertical asymptote; right par $+\infty$, left par $-\infty$.

WHY be careful: $\sqrt{9x^2+4}$ mein $\sqrt{x^2}=|x|$ hai, aur $x\to -\infty$ ke liye $|x|=-x$ hota hai. Root ke andar se $x^2$ factor karo: $\sqrt{9x^2+4}=|x|\sqrt{9+\frac{4}{x^2}}=-x\sqrt{9+\frac{4}{x^2}}(x<0).$ Top aur bottom ko $x$ se divide karo: $\frac{-x\sqrt{9+4/x^2}}{2x+1}=\frac{-\sqrt{9+4/x^2}}{2+\frac{1}{x}}\to \frac{-\sqrt{9}}{2}=\frac{-3}{2}.$ Answer: $-\frac{3}{2}$ (ek left-side horizontal asymptote). (Right par, $|x|=+x$ hota hai toh $+\frac{3}{2}$ milta hai — do alag asymptotes.)

Bottom factor karo: $x^2-x-6=(x-3)(x+2)$. Zeros $x=3$ aur $x=-2$ par.

• $x=-2$ par: top $x+2$ bhi $0$ hai — ek $\frac{0}{0}$ form. Cancel karo: $\frac{x+2}{(x-3)(x+2)}=\frac{1}{x-3}(x\ne -2).$ Toh $x=-2$ ek removable hole hai, wall nahi. Uski height: $\frac{1}{-2-3}=-\frac{1}{5}$, hole at $(-2,-\frac{1}{5})$.
• $x=3$ par: reduced form $\frac{1}{x-3}$ mein nonzero top hai, bottom $\to 0$ → genuine vertical asymptote. Answer: vertical asymptote sirf $x=3$ par; removable hole at $\left(-2,-\frac{1}{5}\right)$.

Pehle sab kuch factor karo: top $=x(2x+3)$, bottom $=(x-2)(x+2)$. Koi common factor nahi → koi holes nahi. Vertical: bottom ke zeros $x=2$ aur $x=-2$ par, tops wahan nonzero hain → walls at $x=\pm 2$. Sign check (numerator $x(2x+3)$, roots $0$ aur $-\frac{3}{2}$ par):

• $x=2$ ke paas: top $=2(7)=14>0$. Right ($x\to 2^{+}$): $(x-2)\to 0^{+},(x+2)>0$ → bottom $\to 0^{+}$ → $+\infty$. Left: bottom $\to 0^{-}$ → $-\infty$.
• $x=-2$ ke paas: top $=(-2)(-1)=2>0$. Right ($x\to -2^{+}$): $(x-2)<0,(x+2)\to 0^{+}$ → bottom $\to 0^{-}$ → $-\infty$. Left: bottom $\to 0^{+}$ → $+\infty$. Horizontal: equal degrees $2=2$ → leading coefficients ka ratio $=\frac{2}{1}=2$. Toh $y=2$ dono sides par. Answer: vertical $x=2$ aur $x=-2$; horizontal $y=2$; koi holes nahi. Skeleton neeche figure mein dekho.

WHY no HA: top degree $2>$ bottom degree $1$, toh ${\lim }_{x\to \pm \infty }f=\pm \infty$ — koi leveling line nahi. Slant asymptote (dekho Slant (oblique) asymptotes via polynomial division): divide karo. $\frac{x^2+1}{x-1}=x+1+\frac{2}{x-1}.$ Jab $x\to \pm \infty$, remainder $\frac{2}{x-1}\to 0$, toh $f(x)$ line $y=x+1$ se chipka rehta hai. Answer: koi horizontal asymptote nahi; slant asymptote $y=x+1$; aur ek vertical asymptote bhi $x=1$ par.

Jab $x\to +\infty$: $e^x\to \infty$. Top aur bottom ko $e^x$ se divide karo: $\frac{1-e^{-x}}{1+e^{-x}}\to \frac{1-0}{1+0}=1.$ Jab $x\to -\infty$: $e^x\to 0$ directly, toh $\frac{e^x-1}{e^x+1}\to \frac{0-1}{0+1}=-1.$ Answer: do horizontal asymptotes, $y=1$ (right) aur $y=-1$ (left).

WHY the trick: yeh ek $\infty -\infty$ indeterminate form hai; conjugate se multiply karo taaki yeh ek aisi fraction mein convert ho jise hum divide kar sakein. $\sqrt{x^2+6x}-x=\frac{(x^2+6x)-x^2}{\sqrt{x^2+6x}+x}=\frac{6x}{\sqrt{x^2+6x}+x}.$ $x\to \infty$ ke liye, $\sqrt{x^2+6x}=x\sqrt{1+6/x}$ (yahan $x>0$ toh $|x|=x$). Top aur bottom ko $x$ se divide karo: $\frac{6}{\sqrt{1+6/x}+1}\to \frac{6}{\sqrt{1}+1}=\frac{6}{2}=3.$ Answer: $3$. (Equivalently curve $y=\sqrt{x^2+6x}$ slant line $y=x+3$ se chipka rehta hai.)

Form check: jab $x\to \infty$, $\ln x\to \infty$ aur $x\to \infty$ → $\frac{\infty }{\infty }$, indeterminate. L'Hôpital (top aur bottom ko alag alag differentiate karo): ${\lim }_{x\to \infty }\frac{\ln x}{x}={\lim }_{x\to \infty }\frac{1/x}{1}={\lim }_{x\to \infty }\frac{1}{x}=0.$ Growth reasoning: $\ln x$ kisi bhi positive power of $x$ se dheemar grow karta hai, toh denominator jeet jaata hai. Answer: $0$ (is function ke liye horizontal asymptote $y=0$).

WHAT each condition forces:

• HA $y=3$ equal degrees ke saath → leading-coefficient ratio $=3$; sabse simple hai degree 1 over degree 1: $f(x)=\frac{3x+b}{x-1}$ (leading ratio $3/1=3$ ✓).
• Vertical asymptote at $x=1$ → denominator $x-1$ wahan nonzero top ke saath: top $=3(1)+b=3+b$, chahiye $\ne 0$.
• Origin se guzarti hai → $f(0)=0$: $\frac{3(0)+b}{0-1}=-b=0\Rightarrow b=0$. Answer: $f(x)=\frac{3x}{x-1}$. Check: $f(0)=0$ ✓; $x=1$ par top $=3\ne 0$ → wall ✓; ${\lim }_{x\to \pm \infty }=3$ ✓.

Vertical wall ::: nonzero over something $\to 0$; sign left/right se; check karo top bhi $0$ toh nahi (hole). Horizontal ceiling ::: denominator ki top power se divide karo; $1/x^k$ khatam karo; equal degrees → coefficient ratio. Top degree one higher ::: koi HA nahi, lekin polynomial division se ek slant asymptote milta hai. Indeterminate $\frac{0}{0},\frac{\infty }{\infty },\infty -\infty$ ::: reshape karo (factor, conjugate) ya L'Hôpital.