极限例题(二)

第一题

$$求\lim_{x \to \infty} \frac{x^{7}(1-2x)^{8}}{(3x+3)^{15}} \\ 思路: 分子分母同时除以x的最高次幂x^{15} \\ \frac{x^{7}(1-2x)^{8}}{(3x+3)^{15}} \Rightarrow \frac{x^{7}[x(\frac{1}{x}-x)]^{8}} {[x(3+\frac{3}{x})]^{15}} \\ \Rightarrow \frac{x^{7} \cdot x^{8}(\frac{1}{x}-x)^{8}} {x^{15}(3+\frac{3}{x})^{15}} \Rightarrow \frac{x^{15} (\frac{1}{x}-x)^{8}} {x^{15}(3+\frac{3}{x})^{15}} \\ \\ 约去x^{15}: \frac{(\frac{1}{x}-2)^{8}} {(3+\frac{3}{x})^{15}} \\ \because \frac{1}{x}(x\to \infty)=\frac{3}{x}(x\to \infty)=0 \\ \therefore \lim_{x\to \infty} \frac{(0-2)^{8}}{(3+0)^{15}}=\frac{2^{8}}{3^{15}}$$

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第二题

$$求 \lim_{x\to +\infty} x(\sqrt[]{x^{2}+1} - x) \\ \sqrt[]{x^{2}+1} - x \Rightarrow \frac{(\sqrt[]{x^{2}+1}-x) (\sqrt[]{x^{2}+1}+x)}{(\sqrt[]{x^{2}+1}+x)} \\ 其中分子(\sqrt[]{x^{2}+1}-x) (\sqrt[]{x^{2}+1}+x)可化为: \\ \Rightarrow (\sqrt[]{x+1})^{2}+x\sqrt[]{x^{2}+1}-x\sqrt[]{x^{2}+1}-x^{2} \\ \frac{x^{2}+1-x^{2}}{\sqrt[]{x^{2}+1}+x} \Rightarrow \frac{x}{x\sqrt[]{1+\frac{1}{x^{2}}} + 1} \Rightarrow \frac{1}{\sqrt[]{1+\frac{1}{x^{2}}} + 1} \\ \because \frac{1}{x^{2}}\to 0(x\to +\infty) \\ \therefore \lim_{x\to +\infty} x(\sqrt[]{x^{2}+1} - x) \Rightarrow \lim_{x\to +\infty} \frac{1}{\sqrt[]{1+0} +1} = \frac{1}{2}$$

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第三题

$$求 \lim_{x\to \infty}\frac{\sqrt[]{4x^{2}+x-1}}{\sqrt[]{x^{2} +sinx}} \\ 提取根号内的x(即\sqrt[]{x^{2}}) : \frac{x\sqrt[]{4+\frac{1}{x} -\frac{1}{x^{2}} }+1+x} {x\sqrt[]{1+\frac{sinx}{x^2} } } \\ 再提取外层x: \frac{x(\sqrt[]{4+\frac{1}{x}-\frac{1}{x^{2}} }+\frac{1}{x} +1)} {x\sqrt[]{1+\frac{sinx}{x^{2}} } } \\ \frac{\sqrt[]{4+\frac{1}{x}-\frac{1}{x^{2}} }+\frac{1}{x} +1} {\sqrt[]{1+\frac{sinx}{x^{2}} } } \\ \\ \because \frac{1}{x} \to 0(x\to \infty), \quad \frac{1}{x^{2}} \to 0(x\to \infty) \\ sinx \to 0(x\to \infty), \quad x^{2} \to 0(x\to \infty) \\ \\ \therefore \Rightarrow \frac{\sqrt[]{4+0-0} +0+1}{\sqrt[]{1+0} } =3 \\ \\ \therefore lim_{x\to \infty}\frac{\sqrt[]{4x^{2}+x-1}}{\sqrt[]{x^{2} +sinx}} =3$$

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第四题

$$求\lim_{n\to \infty } (\frac{1}{n^{2}+1} + \frac{1}{n^{2}+2}+...+\frac{n}{n^{2}+n}) \\ 思路:本题考查迫敛准则 \\ \because \frac{1}{n^{2}+n} \ll \frac{1}{n^{2}+k} \ll \frac{1}{n^{2}+1},\quad k = 1,2,...,n \\ \\ \therefore \frac{1+2+...+n}{n^{2}+n} \ll \frac{1}{n^{2}+1} + \frac{2}{n^{2}+2}+...+\frac{n}{n^{2}+n} \ll \frac{1+2+...+n}{n^{2}+1} \\ \\ \lim_{n\to \infty}\frac{1+2+...+n}{n^{2}+n}= \lim_{n\to \infty}\frac{\frac{1}{2}n(n+1)} {n^{2}+n} = \frac{1}{2} \\ \\ \lim_{n\to \infty}\frac{1+2+...+n}{n^{2}+1}= \lim_{n\to \infty}\frac{\frac{1}{2}n(n+1)} {n^{2}+1} \\ \Rightarrow \frac{1}{2} \cdot (\lim_{x\to \infty} \frac{n}{n^{2}+1} + \lim_{x\to \infty} \frac{1}{n^{2}+1} ) \\ =\frac{1}{2}(1+0)= \frac{1}{2} \\ \\ 根据迫敛准则得出:\\ \lim_{n\to \infty } (\frac{1}{n^{2}+1} + \frac{1}{n^{2}+2}+...+\frac{n}{n^{2}+n}) =\frac{1}{2}$$

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第五题

$$求\lim_{x \to 0} \frac{tan2x}{x}$$

$$分子分母均乘以2:\quad \lim_{x \to 0} \frac{2tan2x}{2x}$$

$$\because tanx=\frac{sinx}{cosx}$$

$$\therefore \lim_{x \to 0}\frac{2tan2x}{2x}= \lim_{x \to 0}\frac{2sin2x}{2x} \cdot \lim_{x \to 0} \frac{1}{cos2x} \Rightarrow 2\lim_{x \to 0}\frac{sin2x}{2x} \cdot \lim_{x \to 0} \frac{1}{cos2x}$$

$$\because x \to 0 ==2x\to 0, 且cosx(x\to 0)=1$$

$$\therefore \lim_{x \to 0} \frac{1}{cos2x} =1$$

$$\because \lim_{x \to 0} \frac{sinx}{x} =1 (第一个重要极限)$$

$$\therefore \lim_{x \to 0} \frac{sin2x}{2x} =1$$

$$= 2\cdot 1\cdot 1=2$$

$$\lim_{x \to 0} \frac{tan2x}{x}=2$$