极限例题(二)
第一题
\[
求\lim_{x \to \infty} \frac{x^{7}(1-2x)^{8}}{(3x+3)^{15}}
\\
思路: 分子分母同时除以x的最高次幂x^{15}
\\
\frac{x^{7}(1-2x)^{8}}{(3x+3)^{15}}
\Rightarrow
\frac{x^{7}[x(\frac{1}{x}-x)]^{8}} {[x(3+\frac{3}{x})]^{15}}
\\
\Rightarrow
\frac{x^{7} \cdot x^{8}(\frac{1}{x}-x)^{8}}
{x^{15}(3+\frac{3}{x})^{15}}
\Rightarrow
\frac{x^{15} (\frac{1}{x}-x)^{8}} {x^{15}(3+\frac{3}{x})^{15}}
\\
\\
约去x^{15}: \frac{(\frac{1}{x}-2)^{8}}
{(3+\frac{3}{x})^{15}}
\\
\because \frac{1}{x}(x\to \infty)=\frac{3}{x}(x\to \infty)=0
\\
\therefore \lim_{x\to \infty}
\frac{(0-2)^{8}}{(3+0)^{15}}=\frac{2^{8}}{3^{15}}
\]
第二题
\[求 \lim_{x\to +\infty} x(\sqrt[]{x^{2}+1} - x)
\\
\sqrt[]{x^{2}+1} - x
\Rightarrow
\frac{(\sqrt[]{x^{2}+1}-x) (\sqrt[]{x^{2}+1}+x)}{(\sqrt[]{x^{2}+1}+x)}
\\
其中分子(\sqrt[]{x^{2}+1}-x) (\sqrt[]{x^{2}+1}+x)可化为:
\\
\Rightarrow
(\sqrt[]{x+1})^{2}+x\sqrt[]{x^{2}+1}-x\sqrt[]{x^{2}+1}-x^{2}
\\
\frac{x^{2}+1-x^{2}}{\sqrt[]{x^{2}+1}+x}
\Rightarrow
\frac{x}{x\sqrt[]{1+\frac{1}{x^{2}}} + 1}
\Rightarrow
\frac{1}{\sqrt[]{1+\frac{1}{x^{2}}} + 1}
\\
\because \frac{1}{x^{2}}\to 0(x\to +\infty)
\\
\therefore \lim_{x\to +\infty} x(\sqrt[]{x^{2}+1} - x)
\Rightarrow
\lim_{x\to +\infty} \frac{1}{\sqrt[]{1+0} +1} = \frac{1}{2}
\]
第三题
\[求 \lim_{x\to \infty}\frac{\sqrt[]{4x^{2}+x-1}}{\sqrt[]{x^{2} +sinx}}
\\
提取根号内的x(即\sqrt[]{x^{2}}) :
\frac{x\sqrt[]{4+\frac{1}{x} -\frac{1}{x^{2}} }+1+x}
{x\sqrt[]{1+\frac{sinx}{x^2} } }
\\
再提取外层x:
\frac{x(\sqrt[]{4+\frac{1}{x}-\frac{1}{x^{2}} }+\frac{1}{x} +1)}
{x\sqrt[]{1+\frac{sinx}{x^{2}} } }
\\
\frac{\sqrt[]{4+\frac{1}{x}-\frac{1}{x^{2}} }+\frac{1}{x} +1}
{\sqrt[]{1+\frac{sinx}{x^{2}} } }
\\
\\
\because \frac{1}{x} \to 0(x\to \infty),
\quad
\frac{1}{x^{2}} \to 0(x\to \infty)
\\
sinx \to 0(x\to \infty),
\quad
x^{2} \to 0(x\to \infty)
\\
\\
\therefore \Rightarrow
\frac{\sqrt[]{4+0-0} +0+1}{\sqrt[]{1+0} }
=3
\\
\\
\therefore
lim_{x\to \infty}\frac{\sqrt[]{4x^{2}+x-1}}{\sqrt[]{x^{2} +sinx}}
=3
\]
第四题
\[
求\lim_{n\to \infty }
(\frac{1}{n^{2}+1} + \frac{1}{n^{2}+2}+...+\frac{n}{n^{2}+n})
\\
思路:本题考查迫敛准则
\\
\because \frac{1}{n^{2}+n} \ll
\frac{1}{n^{2}+k} \ll
\frac{1}{n^{2}+1},\quad
k = 1,2,...,n
\\
\\
\therefore
\frac{1+2+...+n}{n^{2}+n} \ll
\frac{1}{n^{2}+1} + \frac{2}{n^{2}+2}+...+\frac{n}{n^{2}+n} \ll
\frac{1+2+...+n}{n^{2}+1}
\\
\\
\lim_{n\to \infty}\frac{1+2+...+n}{n^{2}+n}=
\lim_{n\to \infty}\frac{\frac{1}{2}n(n+1)}
{n^{2}+n} = \frac{1}{2}
\\
\\
\lim_{n\to \infty}\frac{1+2+...+n}{n^{2}+1}=
\lim_{n\to \infty}\frac{\frac{1}{2}n(n+1)}
{n^{2}+1}
\\
\Rightarrow
\frac{1}{2} \cdot
(\lim_{x\to \infty} \frac{n}{n^{2}+1} +
\lim_{x\to \infty} \frac{1}{n^{2}+1}
)
\\
=\frac{1}{2}(1+0)= \frac{1}{2}
\\
\\
根据迫敛准则得出:\\
\lim_{n\to \infty }
(\frac{1}{n^{2}+1} + \frac{1}{n^{2}+2}+...+\frac{n}{n^{2}+n})
=\frac{1}{2}
\]
第五题
\[求\lim_{x \to 0} \frac{tan2x}{x}
\]
\[分子分母均乘以2:\quad \lim_{x \to 0} \frac{2tan2x}{2x}
\]
\[\because tanx=\frac{sinx}{cosx}
\]
\[ \therefore \lim_{x \to 0}\frac{2tan2x}{2x}=
\lim_{x \to 0}\frac{2sin2x}{2x}
\cdot
\lim_{x \to 0} \frac{1}{cos2x}
\Rightarrow
2\lim_{x \to 0}\frac{sin2x}{2x}
\cdot
\lim_{x \to 0} \frac{1}{cos2x} \]
\[\because x \to 0 ==2x\to 0, 且cosx(x\to 0)=1
\]
\[\therefore \lim_{x \to 0} \frac{1}{cos2x} =1
\]
\[\because \lim_{x \to 0} \frac{sinx}{x} =1 (第一个重要极限)
\]
\[\therefore \lim_{x \to 0} \frac{sin2x}{2x} =1
\]
\[= 2\cdot 1\cdot 1=2
\]
\[\lim_{x \to 0} \frac{tan2x}{x}=2
\]
