Toán Lớp 12: \lim _{x\to 0}\left(\frac{\sqrt[5]{1+10x}-\sqrt[3]{1+3x}}{\arcsin \left(3x+x^2\right)-\tan \left(2x+x^3\right)}\right)
toán cao cấp giúp mình với ạ:<
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Toán Lớp 12: \lim _{x\to 0}\left(\frac{\sqrt[5]{1+10x}-\sqrt[3]{1+3x}}{\arcsin \left(3x+x^2\right)-\tan \left(2x+x^3\right)}\right)
toán cao cấp giúp mình với ạ:<
Comments ( 1 )
\quad\lim\limits_{x\to 0}\dfrac{\sqrt[5]{1+10x}-\sqrt[3]{1+3x}}{\arcsin \left(3x+x^2\right)-\tan \left(2x+x^3\right)}\\
\text{Áp dụng vô cùng bé tương đương khi $x\to 0$ ta được:}\\
\sqrt[5]{1 + 10x} = (1 + 10x)^{\tfrac15} \sim \dfrac15\cdot 10x = 2x\\
\sqrt[3]{1 + 3x} = (1 + 3x)^{\tfrac13} \sim \dfrac13\cdot 3x = x\\
\arcsin(3x + x^2) \sim 3x + x^2\\
\tan(2x + x^3)\sim 2x + x^3\\
x^2\sim 0\\
x^3 \sim 0\\
\text{Khi đó:}\\
\quad\lim\limits_{x\to 0}\dfrac{\sqrt[5]{1+10x}-\sqrt[3]{1+3x}}{\arcsin \left(3x+x^2\right)-\tan \left(2x+x^3\right)}\\
= \lim\limits_{x\to 0}\dfrac{2x – x}{3x -2x}\\
= \lim\limits_{x\to 0}\dfrac{x}{x}\\
= 1
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