1
(2010，数三) 设 f ( x ) = ln 10 x , g ( x ) = x 10 , h ( x ) = e x 10 f(x)=\ln^{10}x,\ g(x)=x^{10},\ h(x)=e^{\frac{x}{10}} f ( x ) = ln 10 x , g ( x ) = x 10 , h ( x ) = e 10 x x x x
A. g ( x ) < h ( x ) < f ( x ) g(x)<h(x)<f(x) g ( x ) < h ( x ) < f ( x ) h ( x ) < g ( x ) < f ( x ) h(x)<g(x)<f(x) h ( x ) < g ( x ) < f ( x ) f ( x ) < g ( x ) < h ( x ) f(x)<g(x)<h(x) f ( x ) < g ( x ) < h ( x ) g ( x ) < f ( x ) < h ( x ) g(x)<f(x)<h(x) g ( x ) < f ( x ) < h ( x )
x → + ∞ x\to+\infty x → + ∞ ln β x ≪ x α ( α > 0 ) ≪ a x ( a > 1 ) ≪ x x \ln^\beta x \ll x^\alpha(\alpha>0) \ll a^x(a>1) \ll x^x ln β x ≪ x α ( α > 0 ) ≪ a x ( a > 1 ) ≪ x x n → ∞ n\to\infty n → ∞ ln β n ≪ n α ≪ a n ≪ n ! ≪ n n \ln^\beta n \ll n^\alpha \ll a^n \ll n! \ll n^n ln β n ≪ n α ≪ a n ≪ n ! ≪ n n
答案
按增长阶：对数幂 ≪ \ll ≪ ≪ \ll ≪ ln 10 x ≪ x 10 ≪ e x 10 \ln^{10}x \ll x^{10} \ll e^{\frac{x}{10}} ln 10 x ≪ x 10 ≪ e 10 x f < g < h f<g<h f < g < h 答案：C
2
(2022，数一、二)已知数列{ x n } \{x_n\} { x n } − π 2 ≤ x n ≤ π 2 -\dfrac{\pi}{2}\le x_n \le \dfrac{\pi}{2} − 2 π ≤ x n ≤ 2 π
A. 当lim n → ∞ cos ( sin x n ) \lim\limits_{n\to\infty}\cos(\sin x_n) n → ∞ lim cos ( sin x n ) lim n → ∞ x n \lim\limits_{n\to\infty}x_n n → ∞ lim x n lim n → ∞ sin ( cos x n ) \lim\limits_{n\to\infty}\sin(\cos x_n) n → ∞ lim sin ( cos x n ) lim n → ∞ x n \lim\limits_{n\to\infty}x_n n → ∞ lim x n lim n → ∞ cos ( sin x n ) \lim\limits_{n\to\infty}\cos(\sin x_n) n → ∞ lim cos ( sin x n ) lim n → ∞ sin x n \lim\limits_{n\to\infty}\sin x_n n → ∞ lim sin x n lim n → ∞ x n \lim\limits_{n\to\infty}x_n n → ∞ lim x n lim n → ∞ sin ( cos x n ) \lim\limits_{n\to\infty}\sin(\cos x_n) n → ∞ lim sin ( cos x n ) lim n → ∞ cos x n \lim\limits_{n\to\infty}\cos x_n n → ∞ lim cos x n lim n → ∞ x n \lim\limits_{n\to\infty}x_n n → ∞ lim x n
严格单调复合函数：复合收敛 ⟺ 内层收敛。设 f f f 严格单调且连续 ，复合 y n = f ( x n ) y_n=f(x_n) y n = f ( x n ) { y n } 收敛 ⟺ { x n } 收敛 \{y_n\}\text{收敛} \iff \{x_n\}\text{收敛} { y n } 收敛 ⟺ { x n } 收敛
正向：x n x_n x n ⇒ f ( x n ) \Rightarrow f(x_n) ⇒ f ( x n )
反向：靠反函数来证明：
f f f ⇒ \Rightarrow ⇒ f − 1 f^{-1} f − 1 f ( x n ) f(x_n) f ( x n ) f − 1 f^{-1} f − 1 x n = f − 1 ( f ( x n ) ) x_n = f^{-1}\big(f(x_n)\big) x n = f − 1 ( f ( x n ) ) x n x_n x n
具有对称性的多对一函数（偶函数、周期函数）：复合收敛无法保证内层收敛，总能构造正负震荡特例推翻必然性结论。
对于此题，思路是构造出极限不存在的 x n x_n x n
答案
解：
容易构造出 x n = ( − 1 ) n ⋅ π 4 x_n=(-1)^n\cdot\dfrac{\pi}{4} x n = ( − 1 ) n ⋅ 4 π sin x n = ± 2 2 \sin x_n=\pm\dfrac{\sqrt{2}}{2} sin x n = ± 2 2 cos ( sin x n ) = cos 2 2 \cos(\sin x_n)=\cos\dfrac{\sqrt{2}}{2} cos ( sin x n ) = cos 2 2 lim n → ∞ cos ( sin x n ) \lim\limits_{n\to\infty}\cos(\sin x_n) n → ∞ lim cos ( sin x n ) lim n → ∞ sin ( cos x n ) \lim\limits_{n\to\infty}\sin(\cos x_n) n → ∞ lim sin ( cos x n ) lim n → ∞ sin x n \lim\limits_{n\to\infty}\sin x_n n → ∞ lim sin x n lim n → ∞ x n \lim\limits_{n\to\infty}x_n n → ∞ lim x n u n = cos x n u_n=\cos x_n u n = cos x n u n ∈ [ 0 , 1 ] u_n\in[0,1] u n ∈ [ 0 , 1 ] sin t \sin t sin t [ 0 , 1 ] [0,1] [ 0 , 1 ] lim n → ∞ sin ( u n ) \lim\limits_{n\to\infty}\sin(u_n) n → ∞ lim sin ( u n ) lim n → ∞ u n \lim\limits_{n\to\infty}u_n n → ∞ lim u n D \boldsymbol{D} D
3
(2016，数三)已知函数f ( x ) f(x) f ( x ) lim x → 0 1 + f ( x ) sin 2 x − 1 e 3 x − 1 = 2 \lim\limits_{x \to 0}\dfrac{\sqrt{1+f(x)\sin2x}-1}{e^{3x}-1}=2 x → 0 lim e 3 x − 1 1 + f ( x ) sin 2 x − 1 = 2 lim x → 0 f ( x ) = ‾ \lim\limits_{x \to 0}f(x)=\underline{\quad\quad} x → 0 lim f ( x ) =
常用等价无穷小：
1 + u − 1 ∼ 1 2 u ( u → 0 ) \sqrt{1+u}-1 \sim \dfrac{1}{2}u \quad(u\to0) 1 + u − 1 ∼ 2 1 u ( u → 0 ) e t − 1 ∼ t ( t → 0 ) e^t-1 \sim t \quad(t\to0) e t − 1 ∼ t ( t → 0 ) sin a x ∼ a x \sin ax \sim ax sin a x ∼ a x 答案
当x → 0 x\to0 x → 0 sin 2 x ∼ 2 x \sin2x\sim 2x sin 2 x ∼ 2 x e 3 x − 1 ∼ 3 x e^{3x}-1\sim 3x e 3 x − 1 ∼ 3 x 1 + f ( x ) sin 2 x − 1 ∼ 1 2 f ( x ) sin 2 x ∼ 1 2 f ( x ) ⋅ 2 x = x ⋅ f ( x ) \sqrt{1+f(x)\sin2x}-1 \sim \dfrac{1}{2}f(x)\sin2x \sim \dfrac{1}{2}f(x)\cdot 2x=x\cdot f(x) 1 + f ( x ) sin 2 x − 1 ∼ 2 1 f ( x ) sin 2 x ∼ 2 1 f ( x ) ⋅ 2 x = x ⋅ f ( x )
lim x → 0 x ⋅ f ( x ) 3 x = 2 \lim_{x\to0}\frac{x\cdot f(x)}{3x}=2 x → 0 lim 3 x x ⋅ f ( x ) = 2 约去x ( x ≠ 0 ) x(x\neq0) x ( x = 0 )
lim x → 0 f ( x ) 3 = 2 \lim_{x\to0}\frac{f(x)}{3}=2 x → 0 lim 3 f ( x ) = 2 1 3 lim x → 0 f ( x ) = 2 \frac{1}{3}\lim_{x\to0}f(x)=2 3 1 x → 0 lim f ( x ) = 2 lim x → 0 f ( x ) = 6 \lim_{x\to0}f(x)=6 x → 0 lim f ( x ) = 6 答案：6 \boldsymbol{6} 6
4
(2006，数二) 求极限
lim x → 0 1 x 3 [ ( 2 + cos x 3 ) x − 1 ] \lim_{x \to 0} \frac{1}{x^3}\left[\left(\frac{2+\cos x}{3}\right)^x - 1\right] x → 0 lim x 3 1 [ ( 3 2 + cos x ) x − 1 ] 根据 1 ∞ 1^\infty 1 ∞ α ( x ) → 0 , β ( x ) → ∞ \alpha(x) \to 0,\quad \beta(x)\to \infty α ( x ) → 0 , β ( x ) → ∞ lim ( 1 + α ) β = e lim α β \lim (1+\alpha)^\beta = e^{\,\lim \alpha\beta} lim ( 1 + α ) β = e l i m α β ( 1 + α ( x ) ) β ( x ) − 1 ∼ α ( x ) ⋅ β ( x ) \left(1+\alpha(x)\right)^{\beta(x)}-1 \;\sim\; \alpha(x)\cdot\beta(x) ( 1 + α ( x ) ) β ( x ) − 1 ∼ α ( x ) ⋅ β ( x )
见 极限存在准则 两个重要极限
答案 lim x → 0 1 x 3 [ ( 2 + cos x 3 ) x − 1 ] = lim x → 0 1 x 3 ⋅ x ( cos x − 1 ) 3 = 1 3 lim x → 0 cos x − 1 x 2 = 1 3 ⋅ ( − 1 2 ) = − 1 6 \begin{aligned}
\lim_{x\to0}\frac{1}{x^3}\left[\left(\frac{2+\cos x}{3}\right)^x -1\right]
&=\lim_{x\to0}\frac{1}{x^3}\cdot \frac{x(\cos x-1)}{3}\\
&=\frac13 \lim_{x\to0}\frac{\cos x -1}{x^2}\\
&=\frac13 \cdot \left(-\frac12\right)\\
&=-\frac16
\end{aligned} x → 0 lim x 3 1 [ ( 3 2 + cos x ) x − 1 ] = x → 0 lim x 3 1 ⋅ 3 x ( cos x − 1 ) = 3 1 x → 0 lim x 2 cos x − 1 = 3 1 ⋅ ( − 2 1 ) = − 6 1
题三
(2018，数三)已知实数a , b a,b a , b lim x → + ∞ [ ( a x + b ) e 1 x − x ] = 2 \displaystyle \lim_{x \to +\infty}\left[(ax+b)e^{\frac{1}{x}}-x\right]=2 x → + ∞ lim [ ( a x + b ) e x 1 − x ] = 2 a , b a,b a , b
答案
令t = 1 x t=\dfrac1x t = x 1 x → + ∞ x\to+\infty x → + ∞ t → 0 + t\to0^+ t → 0 +
lim t → 0 + ( a + b t ) e t − 1 t = 2 \lim\limits_{t \to 0^+}\dfrac{(a+bt)e^t-1}{t}=2 t → 0 + lim t ( a + b t ) e t − 1 = 2 分母趋于0，分子极限必为0，代入t = 0 t=0 t = 0 a − 1 = 0 a-1=0 a − 1 = 0 a = 1 a=1 a = 1 a = 1 a=1 a = 1 e t = 1 + t + o ( t ) e^t=1+t+o(t) e t = 1 + t + o ( t )
lim t → 0 + ( 1 + b t ) ( 1 + t + o ( t ) ) − 1 t = lim t → 0 + ( b + 1 ) t + o ( t ) t = b + 1 = 2 \lim_{t \to 0^+}\frac{(1+bt)(1+t+o(t))-1}{t}=\lim_{t \to 0^+}\frac{(b+1)t+o(t)}{t}=b+1=2 t → 0 + lim t ( 1 + b t ) ( 1 + t + o ( t )) − 1 = t → 0 + lim t ( b + 1 ) t + o ( t ) = b + 1 = 2 解得b = 1 b=1 b = 1 a = 1 , b = 1 a=1,b=1 a = 1 , b = 1
题四
求极限 lim n → ∞ ( a 1 n + b 1 n + c 1 n 3 ) n \displaystyle \lim_{n\rightarrow\infty}\left(\frac{a^{\frac{1}{n}}+b^{\frac{1}{n}}+c^{\frac{1}{n}}}{3}\right)^n n → ∞ lim ( 3 a n 1 + b n 1 + c n 1 ) n a > 0 , b > 0 , c > 0 a>0,b>0,c>0 a > 0 , b > 0 , c > 0
答案
解：
令 x = 1 n x=\dfrac1n x = n 1 n → ∞ n\to\infty n → ∞ x → 0 + x\to0^+ x → 0 + lim x → 0 + ( a x + b x + c x 3 ) 1 x \displaystyle\lim_{x\to0^+}\left(\frac{a^x+b^x+c^x}{3}\right)^{\frac1x} x → 0 + lim ( 3 a x + b x + c x ) x 1
记 L = lim x → 0 + ln ( a x + b x + c x 3 ) 1 x = lim x → 0 + 1 x ln a x + b x + c x 3 L= \lim_{x\to0^+}\ln\left(\frac{a^x+b^x+c^x}{3}\right)^{\frac1x}=\lim_{x\to0^+}\frac1x\ln\frac{a^x+b^x+c^x}{3} L = lim x → 0 + ln ( 3 a x + b x + c x ) x 1 = lim x → 0 + x 1 ln 3 a x + b x + c x a x = e x ln a = 1 + x ln a + o ( x ) a^x=e^{x\ln a}=1+x\ln a+o(x) a x = e x l n a = 1 + x ln a + o ( x )
L = lim x → 0 + 1 x ln 3 + x ( ln a + ln b + ln c ) + o ( x ) 3 = lim x → 0 + 1 x ln ( 1 + x ( ln a + ln b + ln c ) 3 + o ( x ) ) = lim x → 0 + 1 x ( x ln a b c 3 + o ( x ) ) = 1 3 ln a b c = ln a b c 3 \begin{aligned}
L &= \lim_{x\to0^+}\frac1x\ln\frac{3+x(\ln a+\ln b+\ln c)+o(x)}{3}\\
&= \lim_{x\to0^+}\frac1x\ln\left(1+\dfrac{x(\ln a+\ln b+\ln c)}{3}+o(x)\right)\\
&=\lim_{x\to0^+}\frac1x\left(\frac{x\ln abc}{3}+o(x)\right)\\
&=\frac13\ln abc\\
&=\ln\sqrt[3]{abc}
\end{aligned} L = x → 0 + lim x 1 ln 3 3 + x ( ln a + ln b + ln c ) + o ( x ) = x → 0 + lim x 1 ln ( 1 + 3 x ( ln a + ln b + ln c ) + o ( x ) ) = x → 0 + lim x 1 ( 3 x ln ab c + o ( x ) ) = 3 1 ln ab c = ln 3 ab c 原极限 = e L = a b c 3 =e^L=\sqrt[3]{abc} = e L = 3 ab c
题 n
(2008，数四)设0 < a < b 0<a<b 0 < a < b lim n → ∞ ( a − n + b − n ) 1 n = ( ) \lim\limits_{n \to \infty}\left(a^{-n}+b^{-n}\right)^{\frac{1}{n}}=(\quad) n → ∞ lim ( a − n + b − n ) n 1 = ( ) a a a a − 1 a^{-1} a − 1 b b b b − 1 b^{-1} b − 1
lim n → ∞ a 1 n + a 2 n + ⋯ + a k n n = max { a 1 , a 2 , … , a k } , a i > 0 \lim\limits_{n\to\infty}\sqrt[n]{a_1^n+a_2^n+\dots+a_k^n} = \max\{a_1,a_2,\dots,a_k\},a_i>0 n → ∞ lim n a 1 n + a 2 n + ⋯ + a k n = max { a 1 , a 2 , … , a k } , a i > 0 底数里最大那个数 。
见 极限存在准则 两个重要极限
答案
由0 < a < b 0<a<b 0 < a < b 0 < a b < 1 0<\dfrac{a}{b}<1 0 < b a < 1
lim n → ∞ ( a − n + b − n ) 1 n = lim n → ∞ ( 1 a n + 1 b n ) 1 n = lim n → ∞ [ 1 b n ( ( b a ) n + 1 ) ] 1 n = lim n → ∞ 1 b ⋅ [ ( b a ) n + 1 ] 1 n = 1 b ⋅ lim n → ∞ ( ( b a ) n + 1 ) 1 n \begin{align*}
\lim_{n \to \infty}\left(a^{-n}+b^{-n}\right)^{\frac{1}{n}}&=\lim_{n \to \infty}\left(\frac{1}{a^n}+\frac{1}{b^n}\right)^{\frac{1}{n}}=\lim_{n \to \infty}\left[\frac{1}{b^n}\left(\left(\frac{b}{a}\right)^n+1\right)\right]^{\frac{1}{n}}\\
&=\lim_{n \to \infty}\frac{1}{b}\cdot\left[\left(\frac{b}{a}\right)^n+1\right]^{\frac{1}{n}}\\
&=\frac{1}{b}\cdot\lim_{n \to \infty}\left(\left(\frac{b}{a}\right)^n+1\right)^{\frac{1}{n}}
\end{align*} n → ∞ lim ( a − n + b − n ) n 1 = n → ∞ lim ( a n 1 + b n 1 ) n 1 = n → ∞ lim [ b n 1 ( ( a b ) n + 1 ) ] n 1 = n → ∞ lim b 1 ⋅ [ ( a b ) n + 1 ] n 1 = b 1 ⋅ n → ∞ lim ( ( a b ) n + 1 ) n 1 因b a > 1 \dfrac{b}{a}>1 a b > 1 lim n → ∞ ( b a ) n = + ∞ \lim\limits_{n \to \infty}\left(\dfrac{b}{a}\right)^n=+\infty n → ∞ lim ( a b ) n = + ∞ lim n → ∞ ( ( b a ) n ) 1 n ≤ lim n → ∞ ( ( b a ) n + 1 ) 1 n ≤ lim n → ∞ ( 2 ⋅ ( b a ) n ) 1 n \lim\limits_{n \to \infty}\left(\left(\dfrac{b}{a}\right)^n\right)^{\frac{1}{n}}\le\lim\limits_{n \to \infty}\left(\left(\dfrac{b}{a}\right)^n+1\right)^{\frac{1}{n}}\le\lim\limits_{n \to \infty}\left(2\cdot\left(\dfrac{b}{a}\right)^n\right)^{\frac{1}{n}} n → ∞ lim ( ( a b ) n ) n 1 ≤ n → ∞ lim ( ( a b ) n + 1 ) n 1 ≤ n → ∞ lim ( 2 ⋅ ( a b ) n ) n 1 b a ≤ lim n → ∞ ( ( b a ) n + 1 ) 1 n ≤ b a ⋅ lim n → ∞ 2 1 n = b a \dfrac{b}{a}\le\lim\limits_{n \to \infty}\left(\left(\dfrac{b}{a}\right)^n+1\right)^{\frac{1}{n}}\le\dfrac{b}{a}\cdot\lim\limits_{n \to \infty}2^{\frac{1}{n}}=\dfrac{b}{a} a b ≤ n → ∞ lim ( ( a b ) n + 1 ) n 1 ≤ a b ⋅ n → ∞ lim 2 n 1 = a b lim n → ∞ ( ( b a ) n + 1 ) 1 n = b a \lim\limits_{n \to \infty}\left(\left(\dfrac{b}{a}\right)^n+1\right)^{\frac{1}{n}}=\dfrac{b}{a} n → ∞ lim ( ( a b ) n + 1 ) n 1 = a b = 1 b ⋅ b a = 1 a = a − 1 =\dfrac{1}{b}\cdot\dfrac{b}{a}=\dfrac{1}{a}=a^{-1} = b 1 ⋅ a b = a 1 = a − 1
题n
求lim x → + ∞ ( x + 1 + x 2 ) 1 x \displaystyle \lim_{x \to +\infty} \big(x+\sqrt{1+x^2}\big)^{\frac{1}{x}} x → + ∞ lim ( x + 1 + x 2 ) x 1
答案
设y = ( x + 1 + x 2 ) 1 x y=\big(x+\sqrt{1+x^2}\big)^{\frac{1}{x}} y = ( x + 1 + x 2 ) x 1 ln y = 1 x ln ( x + 1 + x 2 ) \ln y=\frac{1}{x}\ln\big(x+\sqrt{1+x^2}\big) ln y = x 1 ln ( x + 1 + x 2 ) lim x → + ∞ ln ( x + 1 + x 2 ) x \displaystyle \lim_{x \to +\infty}\frac{\ln\big(x+\sqrt{1+x^2}\big)}{x} x → + ∞ lim x ln ( x + 1 + x 2 ) ∞ ∞ \frac{\infty}{\infty} ∞ ∞ 1 + 2 x 2 1 + x 2 x + 1 + x 2 = 1 + x 2 + x 1 + x 2 ( x + 1 + x 2 ) = 1 1 + x 2 \frac{1+\frac{2x}{2\sqrt{1+x^2}}}{x+\sqrt{1+x^2}}=\frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}\big(x+\sqrt{1+x^2}\big)}=\frac{1}{\sqrt{1+x^2}} x + 1 + x 2 1 + 2 1 + x 2 2 x = 1 + x 2 ( x + 1 + x 2 ) 1 + x 2 + x = 1 + x 2 1 1 1 1 = lim x → + ∞ 1 1 + x 2 = 0 =\displaystyle \lim_{x \to +\infty}\frac{1}{\sqrt{1+x^2}}=0 = x → + ∞ lim 1 + x 2 1 = 0 lim x → + ∞ ln y = 0 \displaystyle \lim_{x \to +\infty}\ln y=0 x → + ∞ lim ln y = 0 lim x → + ∞ y = e 0 = 1 \displaystyle \lim_{x \to +\infty}y=e^0=1 x → + ∞ lim y = e 0 = 1 lim x → + ∞ ( x + 1 + x 2 ) 1 x = 1 \displaystyle \lim_{x \to +\infty} \big(x+\sqrt{1+x^2}\big)^{\frac{1}{x}}=1 x → + ∞ lim ( x + 1 + x 2 ) x 1 = 1
题 n
求 lim n → ∞ 1 + x n + ( x 2 2 ) n n ( x > 0 ) \displaystyle \lim_{n \to \infty} \sqrt[n]{1 + x^n + \left(\frac{x^2}{2}\right)^n}\quad (x>0) n → ∞ lim n 1 + x n + ( 2 x 2 ) n ( x > 0 )
答案
记 A = max { 1 , x , x 2 2 } A=\max\left\{1,x,\dfrac{x^2}{2}\right\} A = max { 1 , x , 2 x 2 }
A n n < 1 + x n + ( x 2 2 ) n n < 3 A n n = A ⋅ 3 n \sqrt[n]{A^n} < \sqrt[n]{1+x^n+\left(\frac{x^2}{2}\right)^n} < \sqrt[n]{3A^n}=A\cdot\sqrt[n]{3} n A n < n 1 + x n + ( 2 x 2 ) n < n 3 A n = A ⋅ n 3 因 lim n → ∞ 3 n = 1 \lim\limits_{n\to\infty}\sqrt[n]{3}=1 n → ∞ lim n 3 = 1 = A = max { 1 , x , x 2 2 } =A=\max\left\{1,x,\dfrac{x^2}{2}\right\} = A = max { 1 , x , 2 x 2 }
当 0 < x < 1 0<x<1 0 < x < 1 max { 1 , x , x 2 2 } = 1 \max\left\{1,x,\dfrac{x^2}{2}\right\}=1 max { 1 , x , 2 x 2 } = 1 1 1 1
当 1 ≤ x < 2 1\le x<2 1 ≤ x < 2 max { 1 , x , x 2 2 } = x \max\left\{1,x,\dfrac{x^2}{2}\right\}=x max { 1 , x , 2 x 2 } = x x x x
当 x = 2 x=2 x = 2 max { 1 , 2 , 4 2 } = 2 \max\left\{1,2,\dfrac{4}{2}\right\}=2 max { 1 , 2 , 2 4 } = 2 2 2 2
当 x > 2 x>2 x > 2 max { 1 , x , x 2 2 } = x 2 2 \max\left\{1,x,\dfrac{x^2}{2}\right\}=\dfrac{x^2}{2} max { 1 , x , 2 x 2 } = 2 x 2 x 2 2 \dfrac{x^2}{2} 2 x 2
lim n → ∞ 1 + x n + ( x 2 2 ) n n = { 1 , 0 < x < 1 , x , 1 ≤ x ≤ 2 , x 2 2 , x > 2. \lim_{n \to \infty} \sqrt[n]{1 + x^n + \left(\frac{x^2}{2}\right)^n}=
\begin{cases}
1, & 0<x<1, \\
x, & 1\le x\le 2, \\
\dfrac{x^2}{2}, & x>2.
\end{cases} n → ∞ lim n 1 + x n + ( 2 x 2 ) n = ⎩ ⎨ ⎧ 1 , x , 2 x 2 , 0 < x < 1 , 1 ≤ x ≤ 2 , x > 2. 题目
设 x 1 > 0 x_1>0 x 1 > 0 x n + 1 = 1 2 ( x n + 1 x n ) , n = 1 , 2 , ⋯ x_{n+1}=\dfrac12\left(x_n+\dfrac{1}{x_n}\right),\ n=1,2,\cdots x n + 1 = 2 1 ( x n + x n 1 ) , n = 1 , 2 , ⋯ lim n → ∞ x n \displaystyle\lim_{n\to\infty}x_n n → ∞ lim x n
由 x 1 > 0 x_1>0 x 1 > 0 x n > 0 x_n>0 x n > 0 x n + 1 = 1 2 ( x n + 1 x n ) ≥ x n ⋅ 1 x n = 1 x_{n+1}=\dfrac12\left(x_n+\dfrac1{x_n}\right)\ge\sqrt{x_n\cdot\dfrac1{x_n}}=1 x n + 1 = 2 1 ( x n + x n 1 ) ≥ x n ⋅ x n 1 = 1 1 1 1 x n + 1 − x n = 1 2 ( x n + 1 x n ) − x n = 1 − x n 2 2 x n ≤ 0 x_{n+1}-x_n=\dfrac12\left(x_n+\dfrac1{x_n}\right)-x_n=\dfrac{1-x_n^2}{2x_n}\le0 x n + 1 − x n = 2 1 ( x n + x n 1 ) − x n = 2 x n 1 − x n 2 ≤ 0 { x n } \{x_n\} { x n } lim n → ∞ x n \lim\limits_{n\to\infty}x_n n → ∞ lim x n A A A A = 1 2 ( A + 1 A ) A=\dfrac12\left(A+\dfrac1A\right) A = 2 1 ( A + A 1 ) 2 A 2 = A 2 + 1 2A^2=A^2+1 2 A 2 = A 2 + 1 A 2 = 1 A^2=1 A 2 = 1 x n ≥ 1 x_n\ge1 x n ≥ 1 A = 1 A=1 A = 1 lim n → ∞ x n = 1 \displaystyle\lim_{n\to\infty}x_n=1 n → ∞ lim x n = 1
如何让豆包生成考研数学题目答案
写出完整题目和标准答案，注意，标准答案是不可能像豆包那样拆解各种标题的，就给我一段完整的求解而已，不要啰嗦
