Exercises — Epsilon-delta definition of a limit — formal proofs

$\forall \epsilon >0\exists \delta >0\text{such that}0<|x-a|<\delta ⟹|f(x)-L|<\epsilon .$ $0<$ wala part $x=a$ ko khud exclude karta hai: limit $a$ ki taraf approach describe karti hai, aur isse koi fark nahi padta ki $f$, $a$ par defined bhi hai ya nahi.

Adversary pehle $\epsilon$ choose karta hai; tum baad mein $\delta$ se reply karte ho. Tumhara $\delta$, $\epsilon$ par depend kar sakta hai (yahi toh poora point hai — ek chhota ring ek chhote wobble ki demand karta hai). Tumhara $\delta$, $x$ par depend nahi kar sakta, kyunki $x$ tumhare $\delta$ commit karne ke baad choose hota hai.

Output distance hai $|f(x)-L|=|7-7|=0$, jo har $\epsilon >0$ ke liye $<\epsilon$ hai, $x$ chahe kuch bhi ho. Proof. $\epsilon >0$ lelo. $\delta =1$ choose karo (koi bhi positive number kaam karega). Agar $0<|x-5|<\delta$, toh $|7-7|=0<\epsilon$. $■$

Scratchwork. Output distance: $|(3x-5)-7|=|3x-12|=3|x-4|.$ Factor hai constant $3$. $3|x-4|<\epsilon$ force karne ke liye $|x-4|<\epsilon /3$ chahiye. $\delta =\epsilon /3$ lo. Proof. $\epsilon >0$ lelo, $\delta =\epsilon /3$ choose karo. Agar $0<|x-4|<\delta$ toh $|(3x-5)-7|=3|x-4|<3\delta =3\cdot \frac{\epsilon }{3}=\epsilon .■$

Scratchwork. $|(-4x+2)-6|=|-4x-4|=4|x+1|=4|x-(-1)|$. Absolute value minus sign ko absorb kar leta hai; factor hai $4$. $\delta =\epsilon /4$ lo. Proof. $\epsilon >0$ lelo, $\delta =\epsilon /4$ choose karo. Agar $0<|x+1|<\delta$ toh $|(-4x+2)-6|=4|x+1|<4\delta =\epsilon .■$

$\left|\frac{x}{2}-1\right|=\frac{1}{2}|x-2|$. Factor hai $\frac{1}{2}$. $\frac{1}{2}|x-2|<\epsilon$ banane ke liye $|x-2|<2\epsilon$ chahiye. $\delta =2\epsilon$ lo. Proof. $\epsilon >0$ lelo, $\delta =2\epsilon$ choose karo. Agar $0<|x-2|<\delta$ toh $\left|\frac{x}{2}-1\right|=\frac{1}{2}|x-2|<\frac{1}{2}\delta =\epsilon .$ $■$ Dhyan do ki factor 1 se kam tha, toh $\delta$, $\epsilon$ se bada aaya — ek gentle function wide window tolerate karta hai.

Scratchwork. $|x^2-9|=|x-3||x+3|$. Factor $|x+3|$ constant nahi hai — pehle ise bound karo. Preliminary $|x-3|<1$ impose karo, toh $2<x<4$, hence $5<x+3<7$ aur $|x+3|<7$. Tab $|x^2-9|<7|x-3|.$ $7|x-3|<\epsilon$ chahiye, matlab $|x-3|<\epsilon /7$. Dono ko $\boxed{\delta =\min (1,\epsilon /7)}$ se enforce karo. Proof. $\epsilon >0$ lelo, $\delta =\min (1,\epsilon /7)$. Agar $0<|x-3|<\delta$: kyunki $\delta \le 1$, $|x+3|<7$; kyunki $\delta \le \epsilon /7$, $|x^2-9|=|x-3||x+3|<7|x-3|<7\cdot \frac{\epsilon }{7}=\epsilon .■$

Scratchwork. $x^2+x-2=(x-1)(x+2)$, toh $|f(x)-L|=|x-1||x+2|$. Preliminary $|x-1|<1$ deta hai $0<x<2$, toh $2<x+2<4$ aur $|x+2|<4$. Tab $|f(x)-2|<4|x-1|$. $|x-1|<\epsilon /4$ chahiye. $\delta =\min (1,\epsilon /4)$ lo. Proof. $\epsilon >0$ lelo, $\delta =\min (1,\epsilon /4)$. Agar $0<|x-1|<\delta$: $|x+2|<4$ aur $|x^2+x-2|=|x-1||x+2|<4|x-1|<4\cdot \frac{\epsilon }{4}=\epsilon .■$

Scratchwork. $\left|\frac{1}{x}-\frac{1}{5}\right|=\left|\frac{5-x}{5x}\right|=\frac{|x-5|}{5|x|}.$ Dangerous piece hai $\frac{1}{|x|}$: agar $x$ zero tak pahunch jaye toh ye blow up ho jaata. Toh $x$ ko zero se door rakhna hoga. Preliminary $|x-5|<1$ deta hai $4<x<6$. Kyunki is interval par $x>4>0$ hai, $x$ positive hai, toh $|x|=x>4$; isliye $\frac{1}{|x|}<\frac{1}{4}$. Tab $\left|\frac{1}{x}-\frac{1}{5}\right|=\frac{|x-5|}{5|x|}<\frac{|x-5|}{5\cdot 4}=\frac{|x-5|}{20}.$ $\frac{|x-5|}{20}<\epsilon$ chahiye, matlab $|x-5|<20\epsilon$. $\delta =\min (1,20\epsilon )$ lo. Proof. $\epsilon >0$ lelo, $\delta =\min (1,20\epsilon )$. Agar $0<|x-5|<\delta$: toh $4<x<6$, toh $x>0$ aur $|x|=x>4$; isliye $\left|\frac{1}{x}-\frac{1}{5}\right|=\frac{|x-5|}{5|x|}<\frac{|x-5|}{20}<\frac{20\epsilon }{20}=\epsilon .■$

Key fact: har $\theta$ ke liye $\left|\sin \theta \right|\le 1$, toh jab bhi $x\ne 0$ ho $\left|\sin \frac{1}{x}\right|\le 1$. Isliye $\left|x\sin \frac{1}{x}-0\right|=|x|\left|\sin \frac{1}{x}\right|\le |x|=|x-0|.$ Wild oscillating factor constant $1$ se bounded hai — koi bhi preliminary restriction nahi chahiye. $\delta =\epsilon$ lo. Proof. $\epsilon >0$ lelo, $\delta =\epsilon$ choose karo. Agar $0<|x-0|<\delta$ toh $\left|x\sin \frac{1}{x}\right|\le |x|<\delta =\epsilon .■$ Ye $\epsilon$–$\delta$ proof ke andar rehne wala squeeze idea hai; dekho Limit Laws (sum, product, quotient).

Yahan $f(x)=1$ for $x>0$ aur $f(x)=-1$ for $x<0$. Negation: dikhao ki koi $\epsilon >0$ har $\delta$ aur har candidate $L$ ko defeat karta hai. $\epsilon =1$ lo. Koi bhi candidate $L$ aur koi bhi $\delta >0$ fix karo. Window ke andar do explicit points construct karo: $x_{+}=\delta /2>0$ aur $x_{-}=-\delta /2<0$ lo. Dono $0<|x|<\delta$ satisfy karte hain, aur $f(x_{+})=1$, $f(x_{-})=-1$. Agar dono $L$ ke $1$ ke andar hote, toh triangle inequality se $2=|1-(-1)|=|(1-L)+(L-(-1))|\le |1-L|+|L-(-1)|<1+1=2,$ ek strict contradiction ($2<2$ false hai). Toh $x_{+},x_{-}$ mein se kam se kam ek $L$ se $\ge 1$ door hai. Kyunki hum ne har $\delta >0$ aur har $L$ ke liye aise points produce kiye, limit exist nahi karta. $■$ $\epsilon =1$ kyun? Do output values $2$ apart hain; half-width $1$ (full width $2$) ka ek band dono ko contain nahi kar sakta.

$x>0$ ke liye, output distance hai $|\sqrt{x}-0|=\sqrt{x}$. Hum chahte hain $\sqrt{x}<\epsilon$, jo (squaring, valid kyunki dono sides $\ge 0$) $x<{\epsilon }^2$ hai. $\delta ={\epsilon }^2$ lo. Proof. $\epsilon >0$ lelo, $\delta ={\epsilon }^2$ choose karo. Agar $0<x<\delta ={\epsilon }^2$ toh $\sqrt{x}<\sqrt{{\epsilon }^2}=\epsilon$. $■$ Dhyan do ki non-linear link hai: output ko ek factor se shrink karne ke liye, input window ko uske square se shrink karna padta hai.

Idea: hume output distance $|(f+g)-(L+M)|$ control karni hai. Ise triangle inequality se split karo: $|(f(x)+g(x))-(L+M)|\le |f(x)-L|+|g(x)-M|.$ Har piece ko $\epsilon /2$ se neeche force kiya ja sakta hai, toh sum $\epsilon$ se neeche rahega. $\epsilon$ ko aadha karna hi trick hai. Proof. $\epsilon >0$ lelo. Kyunki $\lim f=L$, ek ${\delta }_1>0$ hai jahan $0<|x-a|<{\delta }_1\Rightarrow |f(x)-L|<\epsilon /2$. Kyunki $\lim g=M$, ek ${\delta }_2>0$ hai jahan $0<|x-a|<{\delta }_2\Rightarrow |g(x)-M|<\epsilon /2$. $\delta =\min ({\delta }_1,{\delta }_2)$ choose karo. Agar $0<|x-a|<\delta$, toh dono conditions hold karti hain, toh $|(f+g)-(L+M)|\le |f-L|+|g-M|<\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon .■$ Ye exactly Limit Laws (sum, product, quotient) mein sum rule hai.

Scratchwork. Factor: $x^3-8=(x-2)(x^2+2x+4)$, toh $|x^3-8|=|x-2||x^2+2x+4|$. Preliminary $|x-2|<1$ deta hai $1<x<3$. Is interval par $x^2+2x+4$ increasing aur positive hai; $x=3$ par ye $9+6+4=19$ equals karta hai, toh $|x^2+2x+4|<19$. Tab $|x^3-8|<19|x-2|$. $|x-2|<\epsilon /19$ chahiye. $\delta =\min (1,\epsilon /19)$ lo. Proof. $\epsilon >0$ lelo, $\delta =\min (1,\epsilon /19)$. Agar $0<|x-2|<\delta$: kyunki $\delta \le 1$, $|x^2+2x+4|<19$; kyunki $\delta \le \epsilon /19$, $|x^3-8|=|x-2||x^2+2x+4|<19|x-2|<19\cdot \frac{\epsilon }{19}=\epsilon .■$

Pehle simplify karo ($h\ne 0$ ke liye): $\frac{(3+h{)}^2-9}{h}=\frac{9+6h+h^2-9}{h}=\frac{6h+h^2}{h}=6+h.$ Toh output distance hai $|(6+h)-6|=|h|=|h-0|$. Factor hai constant $1$. $\delta =\epsilon$ lo. Proof. $\epsilon >0$ lelo, $\delta =\epsilon$ choose karo. Agar $0<|h-0|<\delta$, toh quotient $6+h$ equals karta hai (defined kyunki $h\ne 0$), aur $\left|\frac{(3+h{)}^2-9}{h}-6\right|=|h|<\delta =\epsilon .■$ Hence $f^{\prime }(3)=6$, jo $x=3$ par $\frac{d}{dx}{x}^2=2x$ se match karta hai.