Exercises — One-sided limits — left-hand, right-hand

Tool 1 — Direction padho. $x\to a^{+}$ ka matlab hai $x$ $a$ ki taraf slide karta hai aur $a$ se bada rehta hai (right se). $x\to a^{-}$ ka matlab hai $x$ chhota rehta hai (left se). Chhota sa sign ek direction hai, kabhi "minus number" nahi. Tool 2 — Function ka sahi piece chuno. Piecewise rule ke liye, arrow ki side batati hai kaun sa formula active hai. $a$ par right se aa rahe ho? Woh rule use karo jo $x>a$ ke liye hold karta hai. Tool 3 — Handshake. Two-sided limit ${\lim }_{x\to a}f(x)$ tab hi exist karta hai jab dono one-sided limits exist karein AUR equal hon. ${\lim }_{x\to a}f(x)=L⟺{\lim }_{x\to a^{-}}f(x)={\lim }_{x\to a^{+}}f(x)=L$

Arrow kya kehta hai: $5^{-}$ = $5$ ke neeche se approach karo (values $5$ se kam). $5^{+}$ = upar se approach karo (values $5$ se zyaada).

• $x\to 5^{-}$: jaise $4.9,4.99,4.999$ — sab $5$ ki taraf creep kar rahe hain upar se.
• $x\to 5^{+}$: jaise $5.1,5.01,5.001$ — sab $5$ ki taraf creep kar rahe hain neeche se. Koi bhi negative nahi hai. Minus sign describe karta hai $5$ ki kaun si side, $x$ ki sign nahi. Saatey ke saatey numbers positive hain.

Handshake apply karo. Dono sides exist karti hain aur dono $4$ ke barabar hain, toh woh handshake karte hain: ${\lim }_{x\to 2}f(x)=4.$ $f(2)=9$ ko kyun ignore kiya jaata hai: ek limit describe karta hai ki $f$ $2$ ke paas kidhar ja raha hai, kabhi nahi woh single value jo $2$ par baith rahi hai. Point $(2,9)$ ek akela dot hai jo curve ke upar float kar raha hai; approach phir bhi height $4$ par aim karti hai.

• $x<a$ (neeche se) $\Rightarrow$ minus sign: ${\lim }_{x\to a^{-}}$.
• $x>a$ (upar se) $\Rightarrow$ plus sign: ${\lim }_{x\to a^{+}}$. Parent note se mnemonic: PLUS upper side se push karta hai, MINUS lower side se.

Left ($x<1$): active rule hai $2x+1$. Target $1$ substitute karo: ${\lim }_{x\to 1^{-}}(2x+1)=2(1)+1=3.$ Right ($x>1$): active rule hai $5x-2$: ${\lim }_{x\to 1^{+}}(5x-2)=5(1)-2=3.$ Handshake: dono $3$ ke barabar hain, isliye ${\lim }_{x\to 1}f(x)=3$. Value $f(1)=10$ koi role nahi khaelti.

Right ($x>0$): yahaan $|x|=x$ hai (yahi positive ke liye absolute value ki definition hai), toh $f(x)=\frac{x}{x}=1$. Isliye ${\lim }_{x\to 0^{+}}f=1$. Left ($x<0$): yahaan $|x|=-x$ hai (negative ko positive mein flip karta hai), toh $f(x)=\frac{-x}{x}=-1$. Isliye ${\lim }_{x\to 0^{-}}f=-1$. Handshake fail: $1\ne -1$, isliye ${\lim }_{x\to 0}f(x)$ exist nahi karta.

Figure mein kya dekhna hai: black curve sab positive $x$ ke liye height $+1$ par flat baithta hai aur sab negative $x$ ke liye height $-1$ par flat. $x=0$ par do open circles dikhate hain ki koi bhi height $0$ par reach nahi hoti; unke beech red dotted segment jump hai size $2$ ka — visual proof ki dono sides handshake nahi kar saktin.

Right ($x$ just above $4$, jaise $4.001$): greatest integer $\le 4.001$ hai $4$. Toh ${\lim }_{x\to 4^{+}}\lfloor x\rfloor =4$. Left ($x$ just below $4$, jaise $3.999$): greatest integer $\le 3.999$ hai $3$. Toh ${\lim }_{x\to 4^{-}}\lfloor x\rfloor =3$. Kyunki $3\ne 4$, $x=4$ par koi two-sided limit nahi. Yeh mismatch har integer par hota hai.

Figure mein kya dekhna hai: graph flat black steps ki ek staircase hai, har ek ek unit wide. Har step left par ek solid dot pe khatam hoti hai (value included) aur right par open circle (value excluded). $x=4$ par red dotted segment dikhata hai ki neeche wala step height $3$ par baitha hai jabki agla step height $4$ tak jump karta hai — left limit lower step padhti hai, right limit upper wali.

$u=x-3$ let karo, toh $h=\frac{1}{u}$ aur $x\to 3$ ban jaata hai $u\to 0$. Right ($x>3\Rightarrow u>0$, tiny positive): chhota positive denominator ek huge positive output deta hai: ${\lim }_{x\to 3^{+}}h=+\infty$. Left ($x<3\Rightarrow u<0$, tiny negative): chhota negative denominator ek huge negative output deta hai: ${\lim }_{x\to 3^{-}}h=-\infty$. Verdict: sides disagree karti hain aur koi bhi finite nahi, isliye two-sided limit exist nahi karti. $x=3$ par ek vertical asymptote hai.

Figure mein kya dekhna hai: red dashed vertical line asymptote $x=3$ mark karti hai. Black curve follow karo: line ke bilkul right mein woh $+\infty$ ki taraf upar rocket karta hai; bilkul left mein woh $-\infty$ ki taraf neeche plunge karta hai. Dono arms opposite directions mein escape karte hain — yahi opposition exactly kyun hai ki sides disagree karti hain.

Pehle domain: $\sqrt{x-2}$ ko $x-2\ge 0$ chahiye, yaani $x\ge 2$. $x<2$ ke liye koi real domain nahi hai. Left ($x<2$): impossible — approach karne ke liye koi points nahi hain. ${\lim }_{x\to 2^{-}}f$ exist nahi karta (domain nahi). Right ($x>2$): allowed, aur jaise $x\to 2^{+}$, $x-2\to 0^{+}$, toh $\sqrt{x-2}\to 0$. Isliye ${\lim }_{x\to 2^{+}}f=0$. Two-sided: fail, kyunki ek required side (left) ka koi domain nahi. Domain edge par, sirf ek side exist karti hai.

Pehle naive substitution (aur kyun fail hoti hai): $x=3$ ko $\frac{x^2-9}{x-3}$ mein daalna $\frac{0}{0}$ deta hai — ek indeterminate form, toh akeli substitution yahaan bekar hai. Humein expression ko reshape karna hoga. $x\ne 3$ ke liye numerator factor karo: $x^2-9=(x-3)(x+3)$, toh $\frac{x^2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3(x\ne 3).$ Cancel kyun legal hai: $x=3$ ke paas lekin $3$ par nahi, factor $x-3$ nonzero hai, isliye us se divide karna theek hai. Right: ${\lim }_{x\to 3^{+}}(x+3)=6$. Left: ${\lim }_{x\to 3^{-}}(x+3)=6$. Handshake: dono $=6$, toh ${\lim }_{x\to 3}g(x)=6$. Akela value $g(3)=5$ irrelevant hai — yeh ek removable discontinuity hai.

Left ($x\le 2$, use $x^2+k$): ${\lim }_{x\to 2^{-}}(x^2+k)=4+k$. Right ($x>2$, use $4x-3$): ${\lim }_{x\to 2^{+}}(4x-3)=8-3=5$. Handshake force karo: left $=$ right set karo: $4+k=5⟹k=1.$ $k=1$ ke saath dono sides $5$ ke barabar hain, toh ${\lim }_{x\to 2}f(x)=5$ exist karta hai.

Right ($x>1$, use $x^2+2$): ${\lim }_{x\to 1^{+}}(x^2+2)=1+2=3$. Left ($x<1$, use $ax+b$): ${\lim }_{x\to 1^{-}}(ax+b)=a+b$. Condition 1 — handshake: dono sides target $3$ se match karni chahiye: $a+b=3.$ Yeh do unknowns ke saath ek equation hai, toh infinitely many $(a,b)$ kaam karte hain — jaise $a=1,b=2$ ya $a=0,b=3$. Continuity already kyun satisfied hai: right limit $3=f(1)$ hai, toh jab ek baar $a+b=3$ ho jaaye toh full limit $3=f(1)$ hai. Continuity ko limit $=$ value chahiye, aur dono $3$ hain. Answer: koi bhi $(a,b)$ jisme $a+b=3$ ho; ek clean choice hai $a=1,b=2$.

Humein har side par alag constants aur $0$ par ek akela value chahiye: $f(x)=\{\begin{array}{ll}-3& x<0\\ 100& x=0\\ +3& x>0\end{array}$ Check: left branch constant $-3\Rightarrow {\lim }_{x\to 0^{-}}=-3$; right branch constant $+3\Rightarrow {\lim }_{x\to 0^{+}}=+3$; aur $f(0)=100$ by construction. Sides disagree karti hain ($-3\ne 3$), toh yeh ek genuine jump hai aur two-sided limit exist nahi karti — bilkul jaisa design kiya tha.

Setup: kisi bhi real $t$ ke liye, floor yeh obey karta hai $t-1<\lfloor t\rfloor \le t$. $t=\frac{1}{x}$ rakho: $\frac{1}{x}-1<\lfloor \frac{1}{x}\rfloor \le \frac{1}{x}.$ $x$ se multiply karo. Kyunki $x\to 0^{+}$ mein $x>0$ hai, toh inequalities apni direction maintain karti hain: $x\left(\frac{1}{x}-1\right)<x\lfloor \frac{1}{x}\rfloor \le x\cdot \frac{1}{x},$ $1-x<x\lfloor \frac{1}{x}\rfloor \le 1.$ Squeeze: jaise $x\to 0^{+}$, left bound $1-x\to 1$ aur right bound constant $1$ hai. Dono beech wale ko $1$ par squeeze karte hain: ${\lim }_{x\to 0^{+}}x\lfloor \frac{1}{x}\rfloor =1.$ One-sided kyun matter karta hai: humein inequality ko flip kiye bina multiply karne ke liye $x>0$ chahiye tha; left side par sign flip bounds ko badal deta hai.

Case A: $r(x)=\frac{x+1}{(x-2{)}^2}$. $x=2$ ke paas numerator $x+1\to 3>0$. Denominator $(x-2{)}^2$ ek square hai, hamesha $>0$, $0$ tak shrink karta hua. Toh positive-over-tiny-positive dono sides se $+\infty$ deta hai: ${\lim }_{x\to 2^{-}}r=+\infty ,{\lim }_{x\to 2^{+}}r=+\infty .$ Dono sides behaviour mein agree karti hain ($+\infty$), phir bhi two-sided limit exist nahi karta (infinite koi number nahi) — halaanki hum kehte hain woh $+\infty$ ki taraf diverge karta hai. Case B: $\frac{x+1}{x-2}$. Ab denominator sign badalta hai. $x\to 2^{+}$ ke liye, $x-2\to 0^{+}$: $\frac{3}{0^{+}}=+\infty$. $x\to 2^{-}$ ke liye, $x-2\to 0^{-}$: $\frac{3}{0^{-}}=-\infty$. Opposite infinities. Lesson: denominator mein even power sign ko dono sides par same rakhti hai; odd power use flip karti hai. Dekho Vertical asymptotes.

Left ($x<0$, use $\frac{\sin x}{x}$): yeh expression kisi bhi side se $1$ ki taraf tend karta hai, toh ${\lim }_{x\to 0^{-}}f=1$. Right ($x>0$, use $\frac{1-\cos x}{x}$): yeh standard limit $0$ hai, toh ${\lim }_{x\to 0^{+}}f=0$. Handshake fail: $1\ne 0$, toh two-sided limit exist nahi karti — aur isliye $c$ ki koi bhi value $f$ ko $0$ par continuous nahi bana sakti. Continuity ke liye pehle two-sided limit ka exist karna zaroori hai; yahaan woh kabhi nahi karta. Yeh ek jump discontinuity hai height $c$ par ek floating point ke saath.