1.夹逼准则
定义:设{ a n a_n an}, { b n b_n bn}, { c n c_n cn}为实数列, a n ≤ b n ≤ c n a_n≤b_n≤c_n an≤bn≤cn, 且
lim n → ∞ a n = lim n → ∞ c n = l \lim_{n \to \infty} a_n= \lim_{n \to \infty} c_n= l n→∞liman=n→∞limcn=l, 则 lim n → ∞ b n = l \lim_{n \to \infty} b_n=l n→∞limbn=l
例1: lim n → ∞ ( 1 n 2 + 1 + 1 n 2 + 1 + . . . . . . + 1 n 2 + n ) \lim_{n \to \infty} (\frac{1}{\sqrt{n^2+1}} +\frac{1}{\sqrt{n^2+1}}+......+\frac{1}{\sqrt{n^2+n}}) n→∞lim(n2+11+n2+11+......+n2+n1)
解:令 b n = 1 n 2 + 1 + 1 n 2 + 1 + . . . . . . + 1 n 2 + n b_n=\frac{1}{\sqrt{n^2+1}} +\frac{1}{\sqrt{n^2+1}}+......+\frac{1}{\sqrt{n^2+n}} bn=n2+11+n2+11+......+n2+n1
由 5 8 < 5 7 < 5 6 \frac{5}{8}<\frac{5}{7}<\frac{5}{6} 85<75<65,
可得出思路:
令 a n = 1 n 2 + n + 1 n 2 + n + . . . . . . + 1 n 2 + n a^n=\frac{1}{\sqrt{n^2+n}} +\frac{1}{\sqrt{n^2+n}}+......+\frac{1}{\sqrt{n^2+n}} an=n2+n1+n2+n1+......+n2+n1
c n = 1 n 2 + 1 + 1 n 2 + 1 + . . . . . . + 1 n 2 + 1 c^n=\frac{1}{\sqrt{n^2+1}} +\frac{1}{\sqrt{n^2+1}}+......+\frac{1}{\sqrt{n^2+1}} cn=n2+11+n2+11+......+n2+11
则: lim n → ∞ a n = lim n → ∞ n n 2 + n = lim n → ∞ n n 2 = 1 \lim_{n \to \infty} a_n= \lim_{n \to \infty} \frac{n}{\sqrt{n^2+n}} = \lim_{n \to \infty} \frac{n}{\sqrt{n^2}} =1 n→∞liman=n→∞limn2+nn=n→∞limn2n=1
lim n → ∞ c n = lim n → ∞ n n 2 + 1 = lim n → ∞ n n 2 = 1 \lim_{n \to \infty} c_n= \lim_{n \to \infty} \frac{n}{\sqrt{n^2+1}} = \lim_{n \to \infty} \frac{n}{\sqrt{n^2}} =1 n→∞limcn=n→∞limn2+1n=n→∞limn2n=1
∵ lim n → ∞ a n = lim n → ∞ c n = 1 \lim_{n \to \infty} a_n= \lim_{n \to \infty} c_n= 1 n→∞liman=n→∞limcn=1 ,且 a n ≤ b n ≤ c n a_n≤b_n≤c_n an≤bn≤cn,所以
lim n → ∞ b n = 1 \lim_{n \to \infty} b_n= 1 n→∞limbn=1
2.两个重要极限
2.1 0 0 型 ⟹ lim x → 0 s i n x x = 1 \frac{0}{0} 型\Longrightarrow \lim_{x \to 0} \frac{sinx}{x} = 1 00型⟹x→0limxsinx=1
通用公式: lim x → 0 s i n △ △ = 1 \lim_{x \to 0} \frac{sin△}{△} = 1 x→0lim△sin△=1
例1: lim x → 0 s i n 3 x x \lim_{x \to 0} \frac{sin3x}{x} x→0limxsin3x
解: lim x → 0 s i n 3 x x = lim x → 0 s i n 3 x 3 x ⋅ 3 = 1 ⋅ 3 = 3 \lim_{x \to 0} \frac{sin3x}{x} = \lim_{x \to 0} \frac{sin3x}{3x}·3 = 1·3 = 3 x→0limxsin3x=x→0lim3xsin3x⋅3=1⋅3=3
对于 0 0 型 \frac{0}{0} 型 00型后面可以使用洛必达法则求极限
2.2第二重要极限
lim x → ∞ ( 1 + 1 x ) x = e 或 lim x → 0 ( 1 + x ) 1 x = e \lim_{x \to ∞} (1+\frac{1}{x})^x=e 或 \lim_{x \to 0} (1+x)^\frac{1}{x}=e x→∞lim(1+x1)x=e或x→0lim(1+x)x1=e
通用公式: 1 x ⟹ lim △ → 0 ( 1 + △ ) 1 △ = e \frac{1}{x}\Longrightarrow \lim_{△ \to 0} (1+△)^\frac{1}{△}=e x1⟹△→0lim(1+△)△1=e
例1: lim x → 0 ( 1 − 2 x ) 1 x , 求其极限值? \lim_{x \to 0} (1-2x)^\frac{1}{x},求其极限值? x→0lim(1−2x)x1,求其极限值?,求其极限值?
解: 将 x = 0 代入:可判断为 1 ∞ 型 将x=0代入:可判断为1^∞型 将x=0代入:可判断为1∞型
所以要用第二极限求解:
lim x → 0 ( 1 − 2 x ) 1 x = lim x → 0 [ 1 + ( − 2 x ) ] 1 − 2 x ⋅ − 2 x ⋅ 1 x = lim x → 0 e − 2 x ⋅ 1 x = e − 2 \begin{align*} \lim_{x \to 0} (1-2x)^\frac{1}{x} & = \lim_{x \to 0} [1+(-2x)]^{\frac{1}{-2x}·-2x· \frac{1}{x}} \\ & = \lim_{x \to 0} e^{-2x· \frac{1}{x}} \\ & = e^{-2} \\ \end{align*} x→0lim(1−2x)x1=x→0lim[1+(−2x)]−2x1⋅−2x⋅x1=x→0lime−2x⋅x1=e−2