Дифференциал. Правило Бернулли-Лопиталя. 21.11.2025

Дифференциал функции

$$\begin{array}{c}\begin{array}{c}dy=f^{\prime }(x)\Delta x\\ dy=f^{\prime }(x)dx\end{array}-\text{Дифференциал функуции, линейная часть её приращения.}\\ \\ dy=f^{\prime }(x)dx=>f^{\prime }(x)=\frac{dy}{dx}\\ \\ \Delta x-\text{Мало }=>\Delta y\approx dy\\ \Delta y=f(x+\Delta x)-f(x)\approx f^{\prime }(x)\Delta x=dy\\ f(x+\Delta x)\approx f^{\prime }(x)\Delta x+f(x)-\text{Первое приближение}\end{array}$$

$y=\cos xx=\frac{\pi }{6}\Delta x=\frac{\pi }{36}\Delta y-?dy-?$

$$\begin{array}{c}\Delta y=\cos (x+\Delta x)-\cos x\\ \Delta y=\cos (\frac{\pi }{6}+\frac{\pi }{36})-\cos \frac{\pi }{6}\\ \Delta y=\cos \frac{7\pi }{36}-\cos \frac{\pi }{6}\approx 0.046\\ dy=-\sin x\Delta x\\ dy=-\sin \frac{\pi }{6}\cdot \frac{\pi }{36}\approx -0.043\end{array}$$

$\arcsin 0.05\approx ?$

$$\begin{array}{c}y=\arcsin xx=0\Delta x=0.05\\ y^{\prime }=\sqrt{\frac{1}{1-x^2}}\\ \arcsin (x+\Delta x)\approx \frac{1}{\sqrt{1-x^2}}\Delta x+\arcsin x\approx 0.05\end{array}$$

$\arctan 1.04\approx ?$

$$\begin{array}{c}y=\arctan xx=1\Delta x=0.04\\ f(x+\Delta x)\approx f^{\prime }(x)\cdot \Delta x+f(x)\\ f(1+0.04)\approx \frac{1}{1+1}\cdot 0.04+\arctan 1\approx 0.02+\frac{\pi }{4}\end{array}$$

$\ln 1,2\approx ?$

$$\begin{array}{c}x=1\Delta x=0,2f(x)=\ln x\\ f(x+\Delta x)\approx f^{\prime }(x)\cdot \Delta x+f(x)\\ \ln (1+0,2)\approx \frac{1}{1}\cdot 0,2+\ln (1)\\ \ln (1,2)\approx 0,2\end{array}$$

$\sqrt{1,2}\approx ?$

$$\begin{array}{c}f(x+\Delta x)\approx f^{\prime }(x)\cdot \Delta x+f(x)\\ \sqrt{\underset{x}{1}+\underset{\Delta x}{0,2}}\approx \frac{1}{2}\cdot 1^{-\frac{1}{2}}\cdot 0,2+1\approx 1,1\\ \sqrt{1,21-0,01}\approx \frac{1}{2}\cdot 1,21^{-\frac{1}{2}}\cdot (-0,01)+\sqrt{1,21}=\\ =-\frac{1}{2}\cdot \frac{1}{\sqrt{1,21}}\cdot \frac{1}{100}+\sqrt{1,21}=-\frac{1}{2}\cdot \frac{1}{1,1}\cdot \frac{1}{100}+1,1=\\ =-\frac{1}{220}+1,1\end{array}$$

Правило Бернулли-Лопиталя

$$\begin{array}{c}1)f(x)\text{ и }\phi (x)\in D=>\forall x:a-h<x<a+h,{\phi }^{\prime }(x)\ne 0\\ 2)f(x)\text{ и }\phi (x)-\text{б.б. (или б.м.)}\\ \text{Тогда, если }\exists \underset{x\to a}{\lim }\frac{f^{\prime }(x)}{{\phi }^{\prime }(x)}=[\frac{0}{0}]\text{ Или }=[\frac{\infty }{\infty }]\\ =>\underset{x\to a}{\lim }\frac{f(x)}{\phi (x)}=\underset{x\to a}{\lim }\frac{f^{\prime }(x)}{{\phi }^{\prime }(x)}\end{array}$$

${\lim }_{x\to 0}\frac{x\cos x-\sin x}{x^3}$

$$\begin{array}{c}\underset{x\to 0}{\lim }\frac{x\cos x-\sin x}{x^3}=\left[\frac{0}{0}\right]\stackrel{\text{b-l}}{=}\\ =\underset{x\to 0}{\lim }\frac{1\cdot \cos x+x\cdot (-\sin x)-\cos x}{3x^2}=\underset{x\to 0}{\lim }\frac{-x\sin x}{3x^2}=\\ =\underset{x\to 0}{\lim }\frac{-\sin x}{3x}=-\frac{1}{3}\end{array}$$

${\lim }_{x\to 0}\frac{e^{3x}-3x-1}{{\sin }^{2}5x}$

$$\begin{array}{c}\underset{x\to 0}{\lim }\frac{e^{3x}-3x-1}{{\sin }^{2}5x}=[\frac{0}{0}]=\underset{x\to 0}{\lim }\frac{3e^{3x}-3}{5\sin 10x}=[\frac{0}{0}]=\underset{x\to 0}{\lim }\frac{9e^{3x}}{50\cos 10x}=[\frac{9}{50}]=0.18\end{array}$$

${\lim }_{x\to +\infty }\frac{\ln x}{\sqrt[3]{x}}$

$$\underset{x\to +\infty }{\lim }\frac{\ln x}{\sqrt[3]{x}}=[\frac{+\infty }{+\infty }]=\underset{x\to +\infty }{\lim }-\frac{1}{3\sqrt[3]{x^7}}=[-\frac{1}{\infty }]=0$$

${\lim }_{x\to 1}(1-x)\tan \frac{\pi x}{2}$

$$\begin{array}{c}\underset{x\to 1}{\lim }(1-x)\tan \frac{\pi x}{2}=[0\cdot \infty ]=\underset{x\to 1}{\lim }\frac{\tan \frac{\pi x}{2}}{\frac{1}{1-x}}=[\frac{\infty }{\infty }]=\underset{x\to 1}{\lim }\frac{\frac{\pi }{2}(1-x{)}^2}{{\cos }^2\frac{\pi x}{2}}=[\frac{0}{0}]=\\ =\underset{x\to 1}{\lim }\frac{1-x}{\cos \frac{\pi x}{2}}=[\frac{0}{0}]=\underset{x\to 1}{\lim }\frac{-1}{-\sin (\frac{\pi x}{2})\frac{\pi }{2}}=\frac{\pi }{2}\end{array}$$

${\lim }_{x\to 1}(\ln x\cdot \ln (x-1))$

$$\begin{array}{c}\underset{x\to 1+0}{\lim }(\ln x\cdot \ln (x-1))=[0\cdot (-\infty )]{=}^{б.л.}\underset{x\to 1+0}{\lim }(\frac{\ln (x-1)}{\frac{1}{\ln x}})=[\frac{-\infty }{0}]\\ =-\underset{x\to 1+0}{\lim }(\frac{x\ln x^2}{1-x})=[-\frac{0}{0}]{=}^{б.л.}-\underset{x\to 1+0}{\lim }(\frac{\frac{1}{x}2\ln x}{1})=0\end{array}$$

${\lim }_{x\to \frac{\pi }{2}}(\frac{x}{\cot x}-\frac{\pi }{2\cos x})$

$$\underset{x\to \frac{\pi }{2}}{\lim }(\frac{x}{\cot x}-\frac{\pi }{2\cos x})=[\infty -\infty ]=\underset{x\to \frac{\pi }{2}}{\lim }\frac{2x\sin x-\pi }{2\cos x}\stackrel{L^{\prime }H}{=}\underset{x\to \frac{\pi }{2}}{\lim }\frac{2\sin x+2x\cos x}{-2\sin x}=-1$$

${\lim }_{0+0}{x}^{\frac{3}{4+\ln x}}$

$$\begin{array}{c}L=\underset{x\to 0+}{\lim }{x}^{\frac{3}{4+\ln x}}\\ \ln L=\underset{x\to 0+}{\lim }\ln \left(x^{\frac{3}{4+\ln x}}\right)=\underset{x\to 0+}{\lim }\frac{3\ln x}{4+\ln x}\\ \underset{x\to 0+}{\lim }\frac{3\ln x}{4+\ln x}=\left[\frac{\infty }{\infty }\right]=\underset{x\to 0+}{\lim }\frac{(3\ln x{)}^{\prime }}{(4+\ln x{)}^{\prime }}=\\ =\underset{x\to 0+}{\lim }\frac{\frac{3}{x}}{\frac{1}{x}}=\underset{x\to 0+}{\lim }3=3\end{array}$$