极限知识
- limx→0sinxx=1,limx→0(1+x)1x=e\lim\limits_{x\rightarrow 0} \frac{\sin x}{x}=1,\lim\limits_{x\rightarrow 0}(1+x)^{\frac{1}{x}}=ex→0limxsinx=1,x→0lim(1+x)x1=e
- x→0x\rightarrow 0x→0时
(1) x∼sinx∼tanx∼arcsinx∼arctanx∼ln(1+x)∼ex−1x\sim \sin x \sim \tan x \sim \arcsin x \sim \arctan x \sim ln(1+x) \sim e^x-1x∼sinx∼tanx∼arcsinx∼arctanx∼ln(1+x)∼ex−1
(2)1−cosx∼12x21−cosax∼a2x21-\cos x \sim \frac12x^2 \quad 1-cos^a x \sim \frac a2 x^21−cosx∼21x21−cosax∼2ax2
(3)1+x−1∼12x(1+x)a−1∼ax\sqrt{1+x}-1\sim \frac12x \quad (1+x)^a-1 \sim ax1+x−1∼21x(1+x)a−1∼ax
夹逼定理
若在x0x_0x0的邻域内,恒有g(x)≤f(x)≤h(x)g(x)\leq f(x)\leq h(x)g(x)≤f(x)≤h(x),且limx→x0g(x)=limx→x0h(x)=A\lim\limits_{x\rightarrow x_0} g(x)=\lim\limits_{x\rightarrow x_0}h(x)=Ax→x0limg(x)=x→x0limh(x)=A,则limx→x0f(x)=A\lim\limits_{x\rightarrow x_0}f(x)=Ax→x0limf(x)=A
例1:
求limx→∞(1n2+1+2n2+2+⋯+nn2+n)\lim\limits_{x\rightarrow \infty}(\dfrac{1}{n^2+1}+\dfrac{2}{n^2+2}+\dots+\dfrac{n}{n^2+n})x→∞lim(n2+11+n2+22+⋯+n2+nn)
解:∵limx→∞(1n2+n+2n2+n+⋯+nn2+n)\because \lim\limits_{x\rightarrow \infty}(\dfrac{1}{n^2+n}+\dfrac{2}{n^2+n}+\dots+\dfrac{n}{n^2+n})∵x→∞lim(n2+n1+n2+n2+⋯+n2+nn)
≤limx→∞(1n2+1+2n2+2+⋯+nn2+n)\qquad \leq \lim\limits_{x\rightarrow \infty}(\dfrac{1}{n^2+1}+\dfrac{2}{n^2+2}+\dots+\dfrac{n}{n^2+n})≤x→∞lim(n2+11+n2+22+⋯+n2+nn)
≤limx→∞(1n2+1+2n2+1+⋯+nn2+1)\qquad \leq\lim\limits_{x\rightarrow \infty}(\dfrac{1}{n^2+1}+\dfrac{2}{n^2+1}+\dots+\dfrac{n}{n^2+1})≤x→∞lim(n2+11+n2+12+⋯+n2+1n)
左侧:limx→∞(1n2+n+2n2+n+⋯+nn2+n)=limx→∞n2+n2n2+n=12\lim\limits_{x\rightarrow \infty}(\dfrac{1}{n^2+n}+\dfrac{2}{n^2+n}+\dots+\dfrac{n}{n^2+n})=\lim\limits_{x\rightarrow \infty}\dfrac{\frac{n^2+n}{2}}{n^2+n}=\dfrac12x→∞lim(n2+n1+n2+n2+⋯+n2+nn)=x→∞limn2+n2n2+n=21
右侧:limx→∞(1n2+1+2n2+1+⋯+nn2+1)=limx→∞n2+n2n2+1=12\lim\limits_{x\rightarrow \infty}(\dfrac{1}{n^2+1}+\dfrac{2}{n^2+1}+\dots+\dfrac{n}{n^2+1})=\lim\limits_{x\rightarrow \infty}\dfrac{\frac{n^2+n}{2}}{n^2+1}=\dfrac12x→∞lim(n2+11+n2+12+⋯+n2+1n)=x→∞limn2+12n2+n=21
∴limx→∞(1n2+1+2n2+2+⋯+nn2+n)=12\therefore\lim\limits_{x\rightarrow \infty}(\dfrac{1}{n^2+1}+\dfrac{2}{n^2+2}+\dots+\dfrac{n}{n^2+n})=\dfrac12∴x→∞lim(n2+11+n2+22+⋯+n2+nn)=21
例2:
求limx→∞(1n2+1+1n2+2+⋯+1n2+n)\lim\limits_{x\rightarrow \infty}(\dfrac{1}{\sqrt{n^2+1}}+\dfrac{1}{\sqrt{n^2+2}}+\dots+\dfrac{1}{\sqrt{n^2+n}})x→∞lim(n2+11+n2+21+⋯+n2+n1)
解:∵nn2+n≤(1n2+1+1n2+2+⋯+1n2+n)≤nn2+1\because\dfrac{n}{\sqrt{n^2+n}}\leq (\dfrac{1}{\sqrt{n^2+1}}+\dfrac{1}{\sqrt{n^2+2}}+\dots+\dfrac{1}{\sqrt{n^2+n}})\leq \dfrac{n}{\sqrt{n^2+1}}∵n2+nn≤(n2+11+n2+21+⋯+n2+n1)≤n2+1n
左侧:limx→∞nn2+n=limx→∞11+1n=1\lim\limits_{x\rightarrow \infty}\dfrac{n}{\sqrt{n^2+n}}=\lim\limits_{x\rightarrow \infty}\dfrac{1}{\sqrt{1+\dfrac{1}{n}}}=1x→∞limn2+nn=x→∞lim1+n11=1
右侧:limx→∞nn2+1=limx→∞11+1n2=1\lim\limits_{x\rightarrow \infty}\dfrac{n}{\sqrt{n^2+1}}=\lim\limits_{x\rightarrow \infty}\dfrac{1}{\sqrt{1+\dfrac{1}{n^2}}}=1x→∞limn2+1n=x→∞lim1+n211=1
∴limx→∞(1n2+1+1n2+2+⋯+1n2+n)=1\therefore \lim\limits_{x\rightarrow \infty}(\dfrac{1}{\sqrt{n^2+1}}+\dfrac{1}{\sqrt{n^2+2}}+\dots+\dfrac{1}{\sqrt{n^2+n}})=1∴x→∞lim(n2+11+n2+21+⋯+n2+n1)=1