极限
部分表示可能不一定严谨
导数的定义
实数序列的极限
实数序列 \(\{a\}\) 的极限为 \(b\)：
\[\lim_{x\rightarrow\infty}a_x=b\iff \forall \varepsilon>0,\exists N\in N^+,\forall n\ge N(|a_n-b|< \varepsilon)
\]
实数序列 \(\{a\}\) 的极限不为 \(b\)：
\[\lim_{x\rightarrow\infty}a_x\ne b\iff \exists \varepsilon>0,\forall N\in N^+,\exists n\ge N(|a_n-b|> \varepsilon)
\]
实数序列 \(\{a\}\) 不存在极限：
\[\forall b\in R(\lim_{x\rightarrow\infty}a_x\ne b)\iff \forall b\in R,\exists \varepsilon>0,\forall N\in N^+,\exists n\ge N(|a_n-b|> \varepsilon)
\]
复数序列的极限
\[\lim_{n\rightarrow\infty}z_n=x+iy\iff\lim_{n\rightarrow\infty}a_n=x,\lim_{n\rightarrow\infty}b_n=y(z_n =a_n+ib_n,n\in N^+)
\]
逼近 1
从正方向逼近
\[\lim_{x\rightarrow a^+}f(x)=b\iff \forall\varepsilon>0,\exists(\{c\},c_i\ge 0,\lim_{x\rightarrow \infty}c_x=0),\exists N\in N^+,\forall n\ge N,|f(a+c_n)-b|<\varepsilon
\]
\[\lim_{x\rightarrow a^+}f(x)\ne b\iff \exists\varepsilon>0,\forall(\{c\},c_i\ge 0,\lim_{x\rightarrow \infty}c_x=0),\forall N\in N^+,\exists n\ge N,|f(a+c_n)-b|>\varepsilon
\]
从负方向逼近
\[\lim_{x\rightarrow a^-}f(x)=b\iff \forall\varepsilon>0,\exists(\{c\},c_i\ge 0,\lim_{x\rightarrow \infty}c_x=0),\exists N\in N^+,\forall n\ge N,|f(a-c_n)-b|<\varepsilon
\]
\[\lim_{x\rightarrow a^-}f(x)\ne b\iff \exists\varepsilon>0,\forall(\{c\},c_i\ge 0,\lim_{x\rightarrow \infty}c_x=0),\forall N\in N^+,\exists n\ge N,|f(a-c_n)-b|>\varepsilon
\]
逼近 2
\[\lim_{x\rightarrow a^+}f(x)=b\iff \forall\varepsilon>0,\exists \delta>0,\forall0<n<\delta,|f(x+n)-b|<\varepsilon
\]
\[\lim_{x\rightarrow a^+}f(x)\ne b\iff \exists\varepsilon>0,\forall \delta>0,\exists 0<n<\delta,|f(x+n)-b|>\varepsilon
\]
\[\lim_{x\rightarrow a^-}f(x)=b\iff \forall\varepsilon>0,\exists \delta>0,\forall0<n<\delta,|f(x-n)-b|<\varepsilon
\]
\[\lim_{x\rightarrow a^-}f(x)\ne b\iff \exists\varepsilon>0,\forall \delta>0,\exists 0<n<\delta,|f(x-n)-b|>\varepsilon
\]
极限的定义
\[\lim_{x\rightarrow a}f(x)=b\iff \lim_{x\rightarrow a^+}f(x)=b\land \lim_{x\rightarrow a^-}f(x)=b
\]
\[\lim_{x\rightarrow a}f(x)\ne b\iff \lim_{x\rightarrow a^+}f(x)\ne b\lor \lim_{x\rightarrow a^-}f(x)\ne b
\]
不存在极限
\[\forall b\in R\lim_{x\rightarrow a}f(x)\ne b\iff \forall b\in R,\lim_{x\rightarrow a^+}f(x)\ne b\lor \lim_{x\rightarrow a^-}f(x)\ne b
\]
导数的定义
\(x\) 处可导
\[f^{'}(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}h
\]
\(x\) 处不可导（\(IND\) 表示未定义）
\[f^{'}(x)\;IND\iff\forall b\in R\left(\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}h\ne b\right)
\]
洛必达法则
\(\frac00\) 型
若
\[\lim_{x\rightarrow a}f(x)=0
\]
\[\lim_{x\rightarrow a}g(x)=0
\]
且 \(g^{'}(x)\ne 0\)，并 \(\exists \varepsilon >0\)，使得 \((\forall |n|<\varepsilon,n\ne 0)\)，\(f^{'}(x+n)\) 和 \(g^{'}(x+n)\) 存在
\[\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=A
\]
则
\[\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\lim_{x\rightarrow a}\frac{f^{'}(x)}{g^{'}(x)}=A
\]
例如
\[\begin{aligned}
\lim_{x\rightarrow 0}\frac{\sin(x)}x=&\lim_{x\rightarrow 0}\frac{\sin^{'}(x)}{x^{'}}\\
=&\lim_{x\rightarrow 0}\frac{\cos(x)}1\\
=&\frac11\\
=&1\\
\end{aligned}\]
\(\frac\infty\infty\) 型
若
\[\lim_{x\rightarrow a}f(x)=\infty
\]
\[\lim_{x\rightarrow a}g(x)=\infty
\]
且 \(g^{'}(x)\ne 0\)，并 \(\exists \varepsilon >0\)，使得 \((\forall |n|<\varepsilon,n\ne 0)\)，\(f^{'}(x+n)\) 和 \(g^{'}(x+n)\) 存在
\[\lim_{x\rightarrow a^+}\frac{f(x)}{g(x)}=A
\]
则
\[\lim_{x\rightarrow a^+}\frac{f(x)}{g(x)}=\lim_{x\rightarrow a^+}\frac{f^{'}(x)}{g^{'}(x)}=A
\]
\(0\cdot\infty\) 型
将极限为 \(0\) 的项取倒数转移到分母上，变为 \(\frac\infty\infty\) 型
或将极限为 \(\infty\) 的项取倒数转移到分母上，变为 \(\frac00\) 型
例如
\[\begin{aligned}
\lim_{x\rightarrow 0^+}x\ln x=&\lim_{x\rightarrow 0^+}\frac{\ln x}{\frac1x}\\
=&\lim_{x\rightarrow 0^+}\frac{\ln^{'} x}{{\left(\frac1x\right)}^{'}}\\
=&\lim_{x\rightarrow 0^+}\frac{\frac1x}{-\frac1{x^2}}\\
=&\lim_{x\rightarrow 0^+}(-x)\\
=&0\\
\end{aligned}\]
\(\infty-\infty\) 型
变为两个无穷小的倒数，然后通分变为 \(\frac00\) 型
例如
\[\begin{aligned}
\lim_{x\rightarrow 1}\left(\frac1{x-1}-\frac1{\ln x}\right)=&\lim_{x\rightarrow 1}\left(\frac{\ln x-(x-1)}{(x-1)\ln x}\right)\\
=&\lim_{x\rightarrow 1}\left(\frac{(\ln x-(x-1))^{'}}{(x\ln x-\ln x)^{'}}\right)\\
=&\lim_{x\rightarrow 1}\left(\frac{\frac1x-1}{1+\ln x-\frac1x}\right)\\
=&\lim_{x\rightarrow 1}\left(\frac{(\frac1x-1)^{'}}{(1+\ln x-\frac1x)^{'}}\right)\\
=&\lim_{x\rightarrow 1}\left(\frac{-\frac1{x^2}}{\frac1x+\frac1{x^2}}\right)\\
=&\lim_{x\rightarrow 1}\left(-\frac1{x+1}\right)\\
=&-\frac12\\
\end{aligned}\]
\(1^{\infty}\) 型
转化为 \(e^{\ln 1\cdot \infty}=e^{0\cdot\infty}\)，然后用 \(0\cdot\infty\) 的方式计算
例如
\[\Large{\begin{aligned}
\lim_{x\rightarrow\infty}\left(\frac{x^2-1}{x^2+1}\right)^x=&\lim_{x\rightarrow\infty}e^{\left(\ln\left(\dfrac{x^2-1}{x^2+1}\right)\cdot x\right)}\\
=&e^{\left(\lim_{x\rightarrow\infty}\left(\ln\left({\frac{x^2-1}{x^2+1}}\right)\cdot x\right)\right)}\\
=&e^{\left(\lim_{x\rightarrow\infty}\left(\ln\left({1-\frac{2}{x^2+1}}\right)\cdot x\right)\right)}\\
=&e^{\left(\lim_{x\rightarrow\infty}\left(\frac {\large{x}}{{\ln^{-1}\left(1-\frac{2}{x^2+1}\right)}}\right)\right)}\\
=&e^{\left(\lim_{x\rightarrow\infty}\left(\frac {\large{x}^{'}}{{\left(\ln^{-1}\left(1-\frac{2}{x^2+1}\right)\right)^{'}}}\right)\right)}\\
=&e^{\left(\lim_{x\rightarrow\infty}\left(\frac {\large1}{{-\ln^{'}\left(1-\frac{2}{x^2+1}\right)\ln^{-2}\left(1-\frac{2}{x^2+1}\right)}}\right)\right)}\\
=&e^{\left(\lim_{x\rightarrow\infty}\left(\frac {\large1}{{-\frac1{\left(1-\frac{2}{x^2+1}\right)}\frac1{\ln^2\left(1-\frac{2}{x^2+1}\right)}}}\right)\right)}\\
=&e^{\left(\lim_{x\rightarrow\infty}\left({{-{\left(1-\frac{2}{x^2+1}\right)}{\ln^2\left(1-\frac{2}{x^2+1}\right)}}}\right)\right)}\\
=&e^{\left({{-{\left(1-0\right)}{\ln^2\left(1-0\right)}}}\right)}\\
=&e^0\\
=&1\\
\end{aligned}}\]
\(0^0\) 型
类似地化为 \(e^{\ln 0\cdot 0}=e^{0\cdot\infty}\)
例如
\[\large{\begin{aligned}
\lim_{x\rightarrow 0}x^{\frac1{1+\ln x}}=&\lim_{x\rightarrow 0}e^{\frac{\ln x}{1+\ln x}}\\
=&\lim_{x\rightarrow 0}e^{\frac{\ln^{'} x}{(1+\ln x)^{'}}}&\\
=&\lim_{x\rightarrow 0}e^{\frac{\frac1x}{\frac1x}}&\\
=&\lim_{x\rightarrow 0}e^1&\\
=&e&\\
\end{aligned}}\]
\(\infty^0\) 型
转化为 \(e^{\ln\infty\cdot 0}=e^{\infty\cdot 0}\)
例如
\[\large{\begin{aligned}
\lim_{x\rightarrow\infty} (x+1)^{\frac 1x}=&\lim_{x\rightarrow\infty} e^{\left(\ln(x+1)\cdot\frac 1x\right)}\\
=&\lim_{x\rightarrow\infty} e^{\left(\frac {\ln(x+1)}x\right)}\\
=&\lim_{x\rightarrow\infty} e^{\left(\frac {\ln^{'}(x+1)}{x^{'}}\right)}\\
=&\lim_{x\rightarrow\infty} e^{\left(\frac {\frac1{x+1}}{1}\right)}\\
=&\lim_{x\rightarrow\infty} e^{\left(\frac1{x+1}\right)}\\
=&e^0\\
=&1\\
\end{aligned}}\]
浙公网安备 33010602011771号