前置知识
习题
求 lim x → 0 + e sin 2 x − cos ( 2 x ) − 2 x x 2 \lim\limits_{x\rightarrow0^+}\dfrac{e^{\sin^2x}-\cos(2\sqrt x)-2x}{x^2} x→0+limx2esin2x−cos(2x)−2x
解：
sin
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\qquad \sin x=x+o(x^2)
sinx=x+o(x2)
e u 2 = 1 + u 2 + o ( u 2 ) ( u → 0 ) \qquad e^{u^2}=1+u^2+o(u^2) \qquad (u\rightarrow0) eu2=1+u2+o(u2)(u→0)
\qquad 当 x → 0 x\rightarrow0 x→0时， u = sin x ∼ x u=\sin x\sim x u=sinx∼x，所以
e sin 2 x = 1 + [ x + o ( x 2 ) ] 2 + o ( x 2 ) = 1 + x 2 + o ( x 2 ) \qquad e^{\sin^2x}=1+[x+o(x^2)]^2+o(x^2)=1+x^2+o(x^2) esin2x=1+[x+o(x2)]2+o(x2)=1+x2+o(x2)
\qquad 因为
cos ( 2 x ) = 1 − 1 2 ! ( 2 x ) 2 + 1 4 ! ( 2 x ) 4 + o ( x 2 ) = 1 − 2 x + 2 3 x 2 + o ( x 2 ) \qquad \cos (2\sqrt x)=1-\dfrac{1}{2!}(2\sqrt x)^2+\dfrac{1}{4!}(2\sqrt x)^4+o(x^2)=1-2x+\dfrac 23x^2+o(x^2) cos(2x)=1−2!1(2x)2+4!1(2x)4+o(x2)=1−2x+32x2+o(x2)
\qquad 所以
\qquad 原式 = lim x → 0 + [ 1 + x 2 + o ( x 2 ) ] − [ 1 − 2 x + 2 3 x 2 + o ( x 2 ) ] − 2 x x 2 = lim x → 0 + 1 3 x 2 + o ( x 2 ) x 2 = 1 3 =\lim\limits_{x\rightarrow0^+}\dfrac{[1+x^2+o(x^2)]-[1-2x+\frac 23x^2+o(x^2)]-2x}{x^2}=\lim\limits_{x\rightarrow0^+}\dfrac{\frac 13x^2+o(x^2)}{x^2}=\dfrac 13 =x→0+limx2[1+x2+o(x2)]−[1−2x+32x2+o(x2)]−2x=x→0+limx231x2+o(x2)=31
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