math

y = ( ( ∫ 0 x 2 3 e 1 2 x 2 d x − x 2 3 + 1 ) 1 x 2 ) y=\left( \left( \int_{0}^{\sqrt[3]{x^2}}{e^{\frac{1}{2}x^2}}\mathrm{d}x-x^{\frac{2}{3}}+1 \right)^{\Huge{\frac{1}{x^2}}\normalsize } \right) y=⎝ ⎛​(∫03x2 ​​e21​x2dx−x32​+1)x21​⎠ ⎞​

为了书写和显示方便 , 将复杂表达式拆分称几部分 : { u = 1 x 2 t = ∫ 0 x 2 3 e 1 2 x 2 d x − x 2 3 + 1 y = t u 为了书写和显示方便,将复杂表达式拆分称几部分: \\ \begin{cases} u&=&\frac{1}{x^2} \\ t&=&\small\displaystyle \int_{0}^{\normalsize \sqrt[3]{x^2}} {\normalsize {e^{\frac{1}{2}x^2}}\mathrm{d}x-x^{\frac{2}{3}}+1} \end{cases} \\ y=t^u 为了书写和显示方便,将复杂表达式拆分称几部分:⎩ ⎨ ⎧​ut​==​x21​∫03x2 ​​e21​x2dx−x32​+1​y=tu

y = e ln ⁡ y = e ln ⁡ t u = e u ln ⁡ t 现在容易由 e ln ⁡ t u 看出 , 复合关系 ( 三重复合 ) lim ⁡ x → 0 y = lim ⁡ x → 0 e u ln ⁡ t \\ y=e^{\ln{y}}=e^{\ln{t^u}}=e^{u\ln{t}} \\ 现在容易由e^{\ln{t^u}}看出,复合关系(三重复合) \\ \lim\limits_{x\to{0}}y=\lim\limits_{x\to{0}}e^{u\ln{t}} y=elny=elntu=eulnt现在容易由elntu看出,复合关系(三重复合)x→0lim​y=x→0lim​eulnt

🎈首先判断 lim ⁡ x → 0 t = lim ⁡ x → 0 ( ∫ 0 x 2 3 e 1 2 x 2 d x − x 2 3 + 1 ) = ( ∫ 0 0 e 0 d x ) − 0 + 1 = 1 则 lim ⁡ x → 0 ln ⁡ t = lim ⁡ t → 1 ln ⁡ t = 0 🎈首先判断\lim_{x\to{0}}t=\lim_{x\to{0}}(\small\displaystyle \int_{0}^{\normalsize \sqrt[3]{x^2}} {\normalsize {e^{\frac{1}{2}x^2}}\mathrm{d}x-x^{\frac{2}{3}}+1}) \\=\small\displaystyle (\int_{0}^{0} {e^{0}\mathrm{d}x)-0+1} =1 \\ 则\lim\limits_{x\to{0}}\ln{t}=\lim\limits_{t\to{1}}\ln{t}=0 🎈首先判断x→0lim​t=x→0lim​(∫03x2 ​​e21​x2dx−x32​+1)=(∫00​e0dx)−0+1=1则x→0lim​lnt=t→1lim​lnt=0

再考察 lim ⁡ x → 0 u = 1 x 2 = ∞ 再考察\lim\limits_{x\to 0}u=\frac{1}{x^2}=\infin 再考察x→0lim​u=x21​=∞

lim ⁡ x → 0 u ln ⁡ t 是一个 0 ⋅ ∞ 我们尝试将其转换为 0 0 lim ⁡ x → 0 ln ⁡ t ( 1 u ) = L H o p i t a l 求导 ? 在洛必达求导前 , 先看能不能用等价无穷下给它简化一下 ! ln ⁡ ( x + 1 ) ∼ x 由 t 的表达式 ( ∫ 0 x 2 3 e 1 2 x 2 d x − x 2 3 ) + 1 看出 , 恰好可以 记 v = ∫ 0 x 2 3 e 1 2 x 2 d x − x 2 3 得到 : ln ⁡ t = ln ⁡ ( v + 1 ) ∼ v lim ⁡ x → 0 ln ⁡ t ( 1 u ) = lim ⁡ x → 0 v ( 1 u ) = lim ⁡ x → 0 v ′ 2 x \lim\limits_{x\to{0}} {u\ln{t}}是一个0\cdot \infin \\我们尝试将其转换为\frac{0}{0} \\ \lim\limits_{x\to{0}}\frac{\ln{t}}{(\frac{1}{u})} \xlongequal{LHopital求导?} \\在洛必达求导前,先看能不能用等价无穷下给它简化一下! \\\ln{(x+1)}\sim x 由t的表达式(\small\displaystyle \int_{0}^{\normalsize \sqrt[3]{x^2}} {\normalsize {e^{\frac{1}{2}x^2}}\mathrm{d}x-x^{\frac{2}{3}})+1}看出,恰好可以 \\记v=\small\displaystyle \int_{0}^{\normalsize \sqrt[3]{x^2}} {\normalsize {e^{\frac{1}{2}x^2}}\mathrm{d}x-x^{\frac{2}{3}}} \\得到:\ln{t}=\ln(v+1)\sim v \\\lim\limits_{x\to{0}}\frac{\ln{t}}{(\frac{1}{u})} =\lim\limits_{x\to{0}}\frac{v}{(\frac{1}{u})} =\lim\limits_{x\to{0}}\frac{v'}{2x} x→0lim​ulnt是一个0⋅∞我们尝试将其转换为00​x→0lim​(u1​)lnt​LHopital求导? 在洛必达求导前,先看能不能用等价无穷下给它简化一下!ln(x+1)∼x由t的表达式(∫03x2 ​​e21​x2dx−x32​)+1看出,恰好可以记v=∫03x2 ​​e21​x2dx−x32​得到:lnt=ln(v+1)∼vx→0lim​(u1​)lnt​=x→0lim​(u1​)v​=x→0lim​2xv′​

记 z = ∫ 0 x 2 3 e 1 2 x 2 d x v = z + x 2 3 z ′ = d d x F ( x ) ∣ 0 x 2 3 = d d x ( F ( x 2 3 ) − F ( 0 ) ) = d d x F ( x 2 3 ) = f ( x 2 3 ) ⋅ ( 2 3 x − 1 3 ) = e 1 2 x 4 3 ⋅ ( 2 3 x − 1 3 ) v ′ = z ′ + 2 3 x − 1 3 = e 1 2 x 4 3 ⋅ ( 2 3 x − 1 3 ) + 2 3 x − 1 3 = 2 3 x − 1 3 ( e 1 2 x 4 3 − 1 ) 记z=\small\displaystyle \int_{0}^{\normalsize \sqrt[3]{x^2}} {e^{\frac{1}{2}x^2}}\mathrm{d}x \\ v=z+x^{\frac{2}{3}} \\ z'=\frac{d}{dx}F(x)|_{0}^{x^{\frac{2}{3}}} =\frac{d}{dx}(F(x^{\frac{2}{3}})-F(0)) =\frac{d}{dx}F(x^{\frac{2}{3}}) \\=f(x^{\frac{2}{3}})\cdot(\frac{2}{3}x^{-\frac{1}{3}}) =\huge e^{\large\frac{1}{2}{x^{\frac{4}{3}}}}\normalsize\cdot(\frac{2}{3}x^{-\frac{1}{3}}) \\v'=z'+\frac{2}{3}x^{-\frac{1}{3}} =\huge e^{\large\frac{1}{2}{x^{\frac{4}{3}}}}\normalsize\cdot(\frac{2}{3}x^{-\frac{1}{3}})+\frac{2}{3}x^{-\frac{1}{3}} =\frac{2}{3}x^{-\frac{1}{3}}(\huge e^{\large\frac{1}{2}{x^{\frac{4}{3}}}}\normalsize-1) 记z=∫03x2 ​​e21​x2dxv=z+x32​z′=dxd​F(x)∣0x32​​=dxd​(F(x32​)−F(0))=dxd​F(x32​)=f(x32​)⋅(32​x−31​)=e21​x34​⋅(32​x−31​)v′=z′+32​x−31​=e21​x34​⋅(32​x−31​)+32​x−31​=32​x−31​(e21​x34​−1)

lim ⁡ x → 0 2 3 x − 1 3 ( e 1 2 x 4 3 − 1 ) 2 x = 等价无穷小 e u − 1 ∼ u = lim ⁡ x → 0 1 3 x − 4 3 1 2 x 4 3 = 1 6 \lim_{x\to{0}}\frac{\frac{2}{3}x^{-\frac{1}{3}}(\huge e^{\large\frac{1}{2}{x^{\frac{4}{3}}}}\normalsize-1)}{2x} \xlongequal{等价无穷小e^u-1\sim u} =\lim_{x\to{0}}\frac{1}{3}x^{\frac{-4}{3}}\frac{1}{2}x^{\frac{4}{3}} =\frac{1}{6} x→0lim​2x32​x−31​(e21​x34​−1)​等价无穷小eu−1∼u =x→0lim​31​x3−4​21​x34​=61​