Exercises — Continuity — definition, types of discontinuity (removable, jump, infinite)

Dono one-sided limits equal aur finite hain ($=3$), toh two-sided limit exist karta hai. Sirf ek cheez galat hai — missing value. Condition 2 ✓ (limit exists), condition 1 ✗ (value undefined) → removable. Patch: $f(2)=3$ define karo.

Kam se kam ek side infinite hai (yahan dono), toh limit finite number ke roop mein exist nahi karta. → Infinite discontinuity ($x=5$ par ek vertical asymptote). Dekho Vertical asymptotes and rational functions.

Dono one-sided limits finite hain par unequal ($1\ne 4$) → jump discontinuity. Jump size $=|L^{+}-L^{-}|=|4-1|=3$. ($f(0)$ ki value yahan type ke liye irrelevant hai — koi bhi single value do alag heights ke beech ka gap band nahi kar sakti.)

Step 1: $f(3)=\frac{9-9}{0}=\frac{0}{0}$ — undefined. Step 2: Numerator factor karo: $x^2-9=(x-3)(x+3)$. $x\ne 3$ ke liye, $\frac{(x-3)(x+3)}{x-3}=x+3.$ Factor kyun kiya? Limit point $x=3$ ko ignore karti hai; common factor cancel karne se woh height reveal hoti hai jise graph approach karta hai. Toh ${\lim }_{x\to 3}f(x)=3+3=6$. Verdict: limit exist karta hai ($=6$), value undefined → removable. Patch: $f(3)=6$.

Step 1: $f(4)=\frac{6}{0}$ — undefined (numerator $\ne 0$, denominator $=0$: cancellation possible nahi). Step 2: $4$ se thoda neeche, $x-4$ ek tiny negative hai, numerator $\approx 6>0$, toh $\frac{6}{{\text{tiny}}^{-}}\to -\infty$. Thoda upar, $x-4$ ek tiny positive hai, toh $\to +\infty$. $L^{-}=-\infty ,L^{+}=+\infty .$ Verdict: infinite discontinuity ($x=4$ par vertical asymptote).

Step 1: $x=1$ condition $x\ge 1$ satisfy karta hai, toh doosra piece use karo: $f(1)=1^2+3=4$. (Defined.) Step 2: $L^{-}={\lim }_{x\to 1^{-}}(2x+1)=2(1)+1=3,L^{+}={\lim }_{x\to 1^{+}}(x^2+3)=4.$ Verdict: dono finite, $3\ne 4$ → jump, size $|4-3|=1$.

Step 1: $g(0)=2$ (defined). Step 2: Standard limit $\underset{u\to 0}{\lim }\frac{\sin u}{u}=1$ use karo. Yeh tool kyun? Direct substitution $\frac{0}{0}$ deta hai; standard limit exactly $\frac{\sin (\cdot )}{\cdot }$ resolve karne ke liye bana machine hai. Rewrite karo taaki argument denominator se match kare: $\frac{\sin 3x}{x}=3\cdot \frac{\sin 3x}{3x}\underset{x\to 0}{\to }3\cdot 1=3.$ Verdict: limit $=3$ exist karta hai aur finite hai, lekin $g(0)=2\ne 3$ → removable ($g(0)=3$ redefine karo). Removable kyun, jump nahi? Left aur right dono same value $3$ approach karte hain; sirf plotted dot misplaced hai.

Denominator factor hota hai: $x^2-4=(x-2)(x+2)$, $x=2$ aur $x=-2$ par zero. Dono suspect points hain. $f(x)=\frac{x-2}{(x-2)(x+2)}=\frac{1}{x+2}(x\ne 2).$ $x=2$ par: $(x-2)$ cancel ho jaata hai, toh ${\lim }_{x\to 2}f(x)=\frac{1}{2+2}=\frac{1}{4}$ (finite), lekin $f(2)=\frac{0}{0}$ undefined → removable, patch $f(2)=\frac{1}{4}$. $x=-2$ par: factor $(x+2)$ cancel nahi hota. ${\lim }_{x\to -2^{-}}\frac{1}{x+2}=-\infty$, ${\lim }_{x\to -2^{+}}\frac{1}{x+2}=+\infty$ → infinite ($x=-2$ par asymptote). Summary: $x=2$ par removable, $x=-2$ par infinite.

$x=2$ par continuity ka matlab hai $L^{-}=L^{+}=f(2)$.

• $L^{-}={\lim }_{x\to 2^{-}}(3x-1)=5$.
• $f(2)=L^{+}=k(2)+4=2k+4$ (doosra piece value bhi deta hai kyunki woh $x\ge 2$ hai). Barabar set karo: $2k+4=5\Rightarrow 2k=1\Rightarrow \boxed{k=\frac{1}{2}}$. Check: $k=\frac{1}{2}$ ke saath, $L^{+}=f(2)=5=L^{-}$. Teeno conditions hold hoti hain — pen kabhi nahi uthta.

Sirf do seams $x=1$ aur $x=3$ break kar sakti hain; har middle/edge piece ek line hai (apne aap mein continuous). $x=1$ par: left value $=1+1=2$; right approach $=a(1)+b=a+b$. Continuity ⇒ a + b = 2. \tag{i} $x=3$ par: left approach $=a(3)+b=3a+b$; right value $=2(3)-3=3$. Continuity ⇒ 3a + b = 3. \tag{ii} (i) ko (ii) se subtract karo: $2a=1\Rightarrow a=\frac{1}{2}$. Phir $b=2-\frac{1}{2}=\frac{3}{2}$. $\boxed{a=\frac{1}{2},b=\frac{3}{2}.}$ Check: $x=1$ par, $\frac{1}{2}+\frac{3}{2}=2$ ✓; $x=3$ par, $\frac{3}{2}+\frac{3}{2}=3$ ✓.

Absolute value $x-3$ ke sign ke hisaab se split hoti hai:

• $x>3$ ke liye: $|x-3|=x-3$, toh $f(x)=\frac{x-3}{x-3}=+1$.
• $x<3$ ke liye: $|x-3|=-(x-3)$, toh $f(x)=\frac{-(x-3)}{x-3}=-1$. Isliye $L^{+}=+1$, $L^{-}=-1$, dono finite par unequal. Aur $f(3)=\frac{0}{0}$ undefined hai. Verdict: jump, size $|1-(-1)|=2$. Removable kyun nahi, jabki $\frac{0}{0}$ hai? $\frac{0}{0}$ form yeh guarantee nahi karta ki limit exist karta hai — yahan dono sides alag constants par settle hoti hain, toh koi single patch kaam nahi karta.

$p$ ek polynomial hai, isliye closed interval $[0,1]$ par continuous hai — including donoon endpoints, jinhe ham relevant one-sided limit se check karte hain (kahin koi holes, jumps, ya poles nahi, toh yeh automatic hai). Endpoints evaluate karo: $p(0)=0-0+1=1>0,p(1)=1-4+1=-2<0.$ Sign change hai: $p(0)>0$ aur $p(1)<0$. Intermediate Value Theorem ke zariye, ek continuous function jo values $+1$ aur $-2$ leta hai, use beech mein har value leni padti hai — including $0$ — kahin $(0,1)$ mein. Isliye $p$ ka wahan kam se kam ek root hai. $■$ Jis property par rely kiya: closed interval par continuity — pen $x$-axis ko bina uthe cross karta hai, toh woh zero skip nahi kar sakta.

Step 1: $x=1$ condition $x\le 1$ satisfy karta hai, toh $f(1)=1^2=1$ (defined). Step 2:

• $L^{-}={\lim }_{x\to 1^{-}}{x}^2=1$.
• $L^{+}$: $x>1$ ke liye, $\frac{x^2-1}{x-1}=\frac{(x-1)(x+1)}{x-1}=x+1\to 2$. Toh $L^{-}=1$, $L^{+}=2$ — dono finite, unequal. Verdict: jump, size $|2-1|=1$. Subtlety: right piece apne aap mein removable lagta hai (ek $\frac{0}{0}$ hole), lekin left piece ke saath jodne par dono sides disagree karti hain → combined function ka $x=1$ par ek jump hai, aur $f(1)=1$ poora gap fix nahi kar sakta. Yaad karo Differentiability implies continuity: kyunki $f$ yahan continuous bhi nahi hai, toh woh $x=1$ par definitely differentiable bhi nahi ho sakta.

Kisi formula se type seedha kaise padhein? ::: Zero denominator with ek cancelling numerator → removable; non-cancelling → infinite; different-height pieces → jump. Kisi bhi $\frac{0}{0}$ par pehla sawaal kya hai? ::: Kya dono one-sided limits agree karte hain aur finite rehte hain? Tabhi removable. Ek unknown constant ke liye continuity kya maangti hai? ::: Har seam par $L^{-}=L^{+}=f(a)$ — har boundary ke liye ek equation. IVT ko root guarantee karne ke liye kya chahiye? ::: Closed interval par continuity, plus endpoints par sign change.