$f(x)$ 一阶连续可导 $\quad\Rightarrow\quad$ $\lim\limits_{x \to 0} f’(x_0 + x) = f’(x_0)$

$$ \begin{aligned} \lim\limits_{x\to0}\dfrac{\displaystyle\int_0^{x^2}f(t)dt}{x^2\displaystyle\int_0^xf(t)dt} &\xlongequal{L’} \lim\limits_{x\to0}\dfrac{2x f(x^2)}{2x\displaystyle\int_0^xf(t)dt + x^2f(x)} \\ &= \lim\limits_{x\to0}\dfrac{2f(x^2)}{2\displaystyle\int_0^xf(t)dt + xf(x)} \\ &\xlongequal{L’} \lim\limits_{x\to0}\dfrac{4xf’(x^2)}{3f(x) + xf’(x)} \\ &= \lim\limits_{x\to0}\dfrac{4f’(x^2)}{3 \cdot \dfrac{f(x)}{x} + f’(x)} \\ &= \frac{4f’(0)}{3 \cdot \lim\limits_{x\to0} \dfrac{f(x)}{x} + f’(0)} \\ &= \frac{4f’(0)}{3f’(0) + f’(0)} \\ &= 1 \\ \end{aligned} \\ $$

$$ \left(x = 1 \\ y = 1 \right) $$

$$ f(x)=\left(\begin{aligned} x &= \cos(t) \ y &= \sin(t) \ z &= \frac xy \end{aligned} \right.$$

常用极限结论：$\lim\limits_{n\to\infty} x^n = \begin{cases} 0 & ,|x| < 1 \\ \infty & ,|x| > 1 \\ 1 & ,x = 1 \\ \not\exists &,x=-1 \end{cases}$

反推：$AP = C$，则 $\begin{pmatrix}A&C\\0&B\end{pmatrix} \begin{pmatrix}E_n&-P\\0&E_n\end{pmatrix}=\begin{pmatrix}A&0\\0&B\end{pmatrix} \Rightarrow r(A) + r(B)$
正推不成立，反例读者自构不难

$y = \begin{cases} 1, &x = 1 \\ x^2 + 2, &x = 2 \end{cases}$

$\begin{pmatrix} A & B & E \\ C & D &2X \end{pmatrix}$