Exercises — L'Hôpital's rule — proof using linear approximation, 0 - 0, ∞ - ∞, other indeterminate forms

Type check. Top $\to \sin 0=0$, bottom $\to 0$. Form hai $\frac{0}{0}$ ✔ — L'Hôpital allowed hai. Top aur bottom ko alag-alag differentiate karo (quotient rule nahi): $\frac{(\sin x{)}^{\prime }}{(x{)}^{\prime }}=\frac{\cos x}{1}.$ Limit lo: $\cos 0=1$. $\boxed{\underset{x\to 0}{\lim }\frac{\sin x}{x}=1.}$

Type check. $\tan 0=0$, bottom $\to 0$: form $\frac{0}{0}$ ✔. Yaad karo $\frac{d}{dx}\tan x={\sec }^{2}x=\frac{1}{{\cos }^{2}x}$. $\frac{(\tan x{)}^{\prime }}{(x{)}^{\prime }}=\frac{{\sec }^{2}x}{1}\stackrel{x\to 0}{\to }\frac{1}{{\cos }^{2}0}=1.$ $\boxed{\underset{x\to 0}{\lim }\frac{\tan x}{x}=1.}$

Pehle type check karo! Top $\to \cos 0=1$, bottom $\to 0+1=1$. Yeh hai $\frac{1}{1}$ — indeterminate nahi. L'Hôpital allowed nahi hai; answer sirf direct substitution se milega: $\boxed{\underset{x\to 0}{\lim }\frac{\cos x}{x+1}=\frac{1}{1}=1.}$ Agar tumne blindly differentiate kiya hota, toh milta $\frac{-\sin x}{1}\to 0$ — jo galat answer hai. Form gate optional nahi hai.

Type check. $e^0-1=0$, $\sin 0=0$: form $\frac{0}{0}$ ✔. Differentiate karo: top $(e^{2x}-1{)}^{\prime }=2e^{2x}$ (chain rule), bottom $(\sin 3x{)}^{\prime }=3\cos 3x$. $\frac{2e^{2x}}{3\cos 3x}\stackrel{x\to 0}{\to }\frac{2\cdot 1}{3\cdot 1}=\frac{2}{3}.$ $\boxed{\underset{x\to 0}{\lim }\frac{e^{2x}-1}{\sin 3x}=\frac{2}{3}.}$

Type check. $0-\sin 0=0$, $0^3=0$: $\frac{0}{0}$ ✔. Pehla application: $\frac{1-\cos x}{3x^2}$. Abhi bhi $\frac{0}{0}$ (top $\to 1-1=0$). Doosra application: $\frac{\sin x}{6x}$. Abhi bhi $\frac{0}{0}$. Teesra application: $\frac{\cos x}{6}\to \frac{1}{6}$. $\boxed{\underset{x\to 0}{\lim }\frac{x-\sin x}{x^3}=\frac{1}{6}.}$ Dobara kyun repeat karein? Har naya fraction phir se legitimate $\frac{0}{0}$ hai, toh rule re-apply hota hai. Hum tab rukते hain jab top ya bottom dono vanish karna band kar dein.

Type check. Jab $x\to 0^{+}$: $x^2\to 0$ aur $\ln x\to -\infty$, toh form hai $0\cdot (-\infty )$ — indeterminate hai lekin fraction nahi, toh L'Hôpital abhi isko touch nahi kar sakta. Fraction mein convert karo. $-\infty$ wale part ko top par rakho: $x^2\ln x=\frac{\ln x}{1/x^2},\text{form }\frac{-\infty }{\infty }✔.$ Differentiate karo: top $(\ln x{)}^{\prime }=\frac{1}{x}$, bottom $(x^{-2}{)}^{\prime }=-2x^{-3}$. $\frac{1/x}{-2/x^3}=\frac{x^3}{-2x}=-\frac{x^2}{2}\stackrel{x\to 0^{+}}{\to }0.$ $\boxed{\underset{x\to 0^{+}}{\lim }{x}^2\ln x=0.}$ $\ln x$ ko top par kyun rakha? Iska derivative $1/x$ simple hai; alternative split $\frac{x^2}{1/\ln x}$ ka derivative ugly hota. Woh split chuno jo simplify kare.

Type check. Jab $x\to 0$, $\frac{1}{x}\to \infty$ aur $\frac{1}{\sin x}\to \infty$: form $\infty -\infty$. Indeterminate hai — do infinities kuch bhi cancel ho sakti hain. Common denominator par combine karo taaki $\frac{0}{0}$ expose ho: $\frac{1}{x}-\frac{1}{\sin x}=\frac{\sin x-x}{x\sin x},\text{form }\frac{0}{0}✔.$ Pehla application: top $(\sin x-x{)}^{\prime }=\cos x-1$; bottom $(x\sin x{)}^{\prime }=\sin x+x\cos x$. $\frac{\cos x-1}{\sin x+x\cos x},\text{abhi bhi }\frac{0}{0}.$ Doosra application: top $(\cos x-1{)}^{\prime }=-\sin x$; bottom $(\sin x+x\cos x{)}^{\prime }=\cos x+\cos x-x\sin x=2\cos x-x\sin x$. $\frac{-\sin x}{2\cos x-x\sin x}\stackrel{x\to 0}{\to }\frac{0}{2-0}=0.$ $\boxed{\underset{x\to 0}{\lim }\left(\frac{1}{x}-\frac{1}{\sin x}\right)=0.}$

Type check. $x^3\to \infty$, $e^x\to \infty$: form $\frac{\infty }{\infty }$ ✔. Teen baar apply karo (polynomial degree har baar ek kam hoti hai; $e^x$ unchanged rehta hai): $\frac{x^3}{e^x}\to \frac{3x^2}{e^x}\to \frac{6x}{e^x}\to \frac{6}{e^x}\stackrel{x\to \infty }{\to }0.$ $\boxed{\underset{x\to \infty }{\lim }\frac{x^3}{e^x}=0.}$ Interpretation. Exponential kisi bhi fixed power ko beat karta hai — dekho Exponential & Logarithm Growth Rates. Neeche ki picture dikhati hai $e^x$ ko $x^3$ ko overtake karte hue, phir kabhi peeche nahi dekhta.

Type check. Jab $x\to (\pi /2{)}^{-}$, $\cos x\to 0^{+}$, toh $\sec x=\frac{1}{\cos x}\to +\infty$ aur $\tan x=\frac{\sin x}{\cos x}\to +\infty$: form $\infty -\infty$. Combine karo: $\sec x-\tan x=\frac{1}{\cos x}-\frac{\sin x}{\cos x}=\frac{1-\sin x}{\cos x},\text{form }\frac{0}{0}✔.$ Ek baar apply karo: top $(1-\sin x{)}^{\prime }=-\cos x$; bottom $(\cos x{)}^{\prime }=-\sin x$. $\frac{-\cos x}{-\sin x}=\frac{\cos x}{\sin x}\stackrel{x\to \pi /2}{\to }\frac{0}{1}=0.$ $\boxed{\underset{x\to (\pi /2{)}^{-}}{\lim }(\sec x-\tan x)=0.}$

Type check. Base $\to 1$, exponent $\to \infty$: form $1^{\infty }$ — indeterminate hai, aur exponents differentiation ke liye invisible hote hain jab tak hum unhe neeche na laayen. Log lo. Maano $y={\left(1+\frac{3}{x}\right)}^x$, toh $\ln y=x\ln \left(1+\frac{3}{x}\right),\text{form }\infty \cdot 0.$ $\to 0$ wale factor ko top par $1/x$ ke upar rakhke fraction mein convert karo: $\ln y=\frac{\ln (1+3/x)}{1/x},\text{form }\frac{0}{0}✔.$ Differentiate karo. Top: $\frac{d}{dx}\ln (1+3/x)=\frac{-3/x^2}{1+3/x}$. Bottom: $\frac{d}{dx}(1/x)=-1/x^2$. $\frac{-3/x^2}{1+3/x}\cdot \frac{1}{-1/x^2}=\frac{3}{1+3/x}\stackrel{x\to \infty }{\to }\frac{3}{1}=3.$ Toh $\ln y\to 3$, isliye $y\to e^3$. $\boxed{\underset{x\to \infty }{\lim }{\left(1+\frac{3}{x}\right)}^x=e^3.}$

Type check. Base $x\to 0^{+}$, exponent $x\to 0$: form $0^0$ — indeterminate. Log lo. Maano $y=x^x$, toh $\ln y=x\ln x$, form $0\cdot (-\infty )$. Convert karo: $\ln y=\frac{\ln x}{1/x}$, form $\frac{-\infty }{\infty }$ ✔. Differentiate karo: $\frac{1/x}{-1/x^2}=-x\stackrel{x\to 0^{+}}{\to }0$. Toh $\ln y\to 0$, isliye $y\to e^0=1$. $\boxed{\underset{x\to 0^{+}}{\lim }{x}^x=1.}$

Type check. Base $\to \infty$, exponent $\to 0$: form ${\infty }^0$ — indeterminate. Log lo. $y=x^{1/x}\Rightarrow \ln y=\frac{\ln x}{x}$, form $\frac{\infty }{\infty }$ ✔. Differentiate karo: $\frac{1/x}{1}=\frac{1}{x}\stackrel{x\to \infty }{\to }0$. Toh $\ln y\to 0$, isliye $y\to e^0=1$. $\boxed{\underset{x\to \infty }{\lim }{x}^{1/x}=1.}$

Type check. Top $\to \infty$, bottom $\to \infty$: form $\frac{\infty }{\infty }$. Lekin dekho L'Hôpital kya karta hai: top ${}^{\prime }=e^x$; bottom ${}^{\prime }=\frac{2e^{2x}}{2\sqrt{e^{2x}+1}}=\frac{e^{2x}}{\sqrt{e^{2x}+1}}$. Ratio $=\frac{e^x\sqrt{e^{2x}+1}}{e^{2x}}=\frac{\sqrt{e^{2x}+1}}{e^x}$ — fraction flip ho gaya aur hamesha ke liye cycle karta hai. RHS limit kabhi resolve nahi hoti, toh rule kuch nahi deta. Algebra use karo. Root mein se $e^{2x}$ factor karo: $\frac{e^x}{\sqrt{e^{2x}+1}}=\frac{e^x}{e^x\sqrt{1+e^{-2x}}}=\frac{1}{\sqrt{1+e^{-2x}}}\stackrel{x\to \infty }{\to }\frac{1}{\sqrt{1+0}}=1.$ $\boxed{\underset{x\to \infty }{\lim }\frac{e^x}{\sqrt{e^{2x}+1}}=1.}$ Lesson: L'Hôpital require karta hai ki RHS limit exist kare. Jab yeh cycle kare, toh woh hypothesis fail hoti hai — algebra ki taraf retreat karo.

Direct (sahi) evaluation. $\frac{x+\sin x}{x}=1+\frac{\sin x}{x}.$ Kyunki $-1\le \sin x\le 1$, $\left|\frac{\sin x}{x}\right|\le \frac{1}{x}\to 0$ (squeeze). Isliye $\boxed{\underset{x\to \infty }{\lim }\frac{x+\sin x}{x}=1.}$ L'Hôpital kya deta. Top ${}^{\prime }=1+\cos x$, bottom ${}^{\prime }=1$, ratio $=1+\cos x$, jo $[0,2]$ mein oscillate karta hai aur jiska koi limit nahi. Koi contradiction nahi. L'Hôpital ka theorem kehta hai: agar $\lim f^{\prime }/g^{\prime }$ exist kare, toh woh $\lim f/g$ ke barabar hai. Yeh kuch nahi kehta jab $\lim f^{\prime }/g^{\prime }$ exist na kare. Implication ek hi direction mein chalta hai — non-existent $f^{\prime }/g^{\prime }$ limit $f/g$ ko condemn nahi karta. Yahan $f/g$ ka bilkul theek limit hai, $1$.

$\sin x$ ki Taylor expansion $0$ ke paas hai $\sin x=x-\frac{x^3}{6}+\frac{x^5}{120}-\cdots .$ Toh $x-\sin x=\frac{x^3}{6}-\frac{x^5}{120}+\cdots$, aur $\frac{x-\sin x}{x^3}=\frac{1}{6}-\frac{x^2}{120}+\cdots \stackrel{x\to 0}{\to }\frac{1}{6}.$ $\boxed{\underset{x\to 0}{\lim }\frac{x-\sin x}{x^3}=\frac{1}{6}.}$ Kyun agree karte hain. L'Hôpital $k$ baar ek $\frac{0}{0}$ par apply karne se $k$-th order terms ka ratio isolate hota hai; yeh exactly leading nonzero Taylor coefficients ka ratio hai. Yahan numerator ka pehla surviving term hai $\frac{x^3}{6}$ aur denominator ka hai $x^3$, jo deta hai $\frac{1/6}{1}=\frac{1}{6}$ — wahi $\frac{1}{6}$ jo humne teen derivative passes se grind kiya. Yeh L'Hôpital hai orbit se dekha hua: leading Taylor terms compare karo.

Determinate form ($\frac{\text{finite}}{\text{nonzero}}$)? ::: Bas substitute karo — L'Hôpital forbidden hai. Product $0\cdot \infty$? ::: Fraction $f/(1/g)$ ki tarah rewrite karo, woh split chuno jo simplify kare. Difference $\infty -\infty$? ::: $\frac{0}{0}$ expose karne ke liye common denominator lo. Power $0^0,{\infty }^0,1^{\infty }$? ::: $\ln$ lo, solve karo, phir result ko exponentiate karo. Rule cycle kare ya RHS oscillate kare? ::: Ise chodo — algebra, squeeze, ya Taylor series use karo.