题目211
设
f(x)=1-\cos{x},求极限
\lim\limits_{x\to0}\dfrac{(1-\sqrt{\cos{x}})(1-\sqrt[3]{\cos{x}})(1-\sqrt[4]{\cos{x}})(1-\sqrt[5]{\cos{x}})}{f\\{f[f(x)]\\}}解答
当
x \to 0 时:
[
1-\sqrt{\cos x} \sim 1 - (1 - \frac{1}{2}x^2)^{\frac{1}{2}} \sim
-(-\frac{1}{4})x^2 = \frac{1}{4}x^2
]同理,由此推导可证得:
1 - \cos^\alpha x \sim \dfrac{1}{2} \alpha x^2[
\begin{aligned}
\text{原式} &=
\frac
{\dfrac{1}{4} \cdot \dfrac{1}{6} \cdot \dfrac{1}{8} \cdot \dfrac{1}{10} \cdot x^8}
{1 - \cos(f[f(x)])} \\\\
&=
\frac
{\dfrac{x^8}{1920}}
{\dfrac{1}{2}[1 - \cos f(x)]^2} \\\\
&=
\frac
{\dfrac{x^8}{1920}}
{\dfrac{1}{2}[\dfrac{1}{2}(1 - \cos x)^2]^2}
\\\\
&=
\frac
{\dfrac{x^8}{1920}}
{\dfrac{1}{128} \cdot x^8}
\\\\
&=
\frac{1}{15}
\\\\
\end{aligned}
]题目212
[
\text{求极限 } \lim_{x\to0}\frac{\tan(\sin x) - x}{\arctan x - \arcsin x}
]解答
该极限为
\dfrac{0}{0} 型极限,常用方法有:
- 洛必达法则
- 等价无穷小
- 泰勒展开
[
\tan x - x \sim \frac{1}{3}x^3
\quad\Rightarrow\quad
\tan(\sin x) - \sin x \sim \frac{1}{3}\sin^3x
]由以上 等价无穷小,我们可以考虑 加减交叉项 凑出需要的形式
[
\begin{aligned}
\text{原式}
&=
\lim_{x\to 0}
\frac{\tan(\sin x) - \sin x + \sin x - x}{-\dfrac{1}{3}x^3 - \dfrac{1}{6}x^3}
\\\\
&=
-2\lim_{x\to 0}
\frac{\tan(\sin x) - \sin x}{x^3}
-2\lim_{x\to 0}
\frac{\sin x - x}{x^3}
\\\\
&=
-2\lim_{x\to 0}
\frac{\dfrac{1}{3}\sin^3 x}{x^3}
-2\lim_{x\to 0}
\frac{-\dfrac{1}{6}x^3}{x^3}
\\\\
&=
-\frac{2}{3} + \frac{1}{3}
\\\\
&=
-\frac{1}{3}
\\\\
\end{aligned}
]题目213
[
\text{求极限 }
\lim_{x\to0}\frac{\sqrt[4]{1-\sqrt[3]{1-\sqrt{1-x}}} - 1}
{(1+x)^\frac{1}{\sqrt[3]{x^2}} - 1}
]解答
[
\sqrt[4]{1-\sqrt[3]{1-\sqrt{1-x}}} - 1
\sim
\sqrt[4]{1-\sqrt[3]{\frac{1}{2}x}} - 1
\sim
-\frac{1}{2^{\frac{7}{3}}}x^{\frac{1}{3}}
][
(1+x)^\frac{1}{\sqrt[3]{x^2}} - 1 \sim x^{\frac{1}{3}}
][
\text{原式} = \lim_{x\to0}\frac{-\dfrac{1}{2^{\frac{7}{3}}}x^{\frac{1}{3}}}{x^{\frac{1}{3}}} =
-\frac{1}{2^{\frac{7}{3}}}
]题目214
[
\lim_{x\to0}\frac{(3+\sin x^2)^x-3^{\sin x}}{x^3}
]解答
底数相同 的 幂指函数 相减,一般考虑 左提右式 或 拉格朗日中值定理(不推荐)
[
\begin{aligned}
\lim_{x\to0}\frac{(3+\sin x^2)^x-3^{\sin x}}{x^3}
&=
\lim_{x\to0}\frac{e^{x\ln(3+\sin x^2)} - e^{\sin x\ln 3}}{x^3}
\\\\
&=
\lim_{x\to0}\frac{e^{\sin x\ln 3}[x\ln(3+\sin x^2) - \sin x\ln 3]}{x^3}
\\\\
&=
\lim_{x\to0}\frac{x\ln(3+\sin x^2) - x\ln 3 - (\sin x\ln 3 - x\ln 3)}{x^3}
\\\\
&=
\lim_{x\to0}\frac{x\cdot [\ln(3+\sin x^2) - \ln 3]}{x^3} -
\ln 3\cdot \lim_{x\to0}\frac{\sin x - x}{x^3}
\\\\
&=
\lim_{x\to0}\frac{x\cdot \ln(1+\dfrac{\sin x^2}{3})}{x^3} + \frac{\ln 3}{6}
\\\\
&=
\lim_{x\to0}\frac{x \cdot \sin x^2}{3x^3} + \frac{\ln 3}{6}
\\\\
&=
\frac{1}{3} + \frac{\ln 3}{6}
\\\\
&=
\frac{2 + \ln 3}{6}
\\\\
\end{aligned}
]题目215
[
\text{求极限 }\lim_{x\to 1}\frac{x - x^x}{1 - x + \ln x}
]解答
先处理分子:
[
\lim_{x\to 1}\frac{x - x^x}{1 - x + \ln x} =
\lim_{x\to 1}x \cdot \frac{1 - x^{x - 1}}{1 - x + \ln x} =
-\lim_{x\to 1}\frac{e^{(x-1)\ln x} - 1}{1 - x + \ln x} =
-\lim_{x\to 1}\frac{(x-1)\ln x}{1 - x + \ln x}
]不妨换元,令
t = x - 1,则
x = 1 + t[
\text{原式}= -\lim_{t\to0}
\frac
{t\ln(1 + t)}
{ln(1 + t) - t} =
-\lim_{t\to0}
\frac
{t^2}
{\bigg(t - \dfrac{1}{2}t^2 + o(t^2)\bigg) - t} = 2
]题目216
[
\text{求极限 } \lim_{x\to0}\frac{(1+x)^{\frac{2}{x}} - e^2[1 - \ln(1+x)]}{x}
]解答
[
\begin{aligned}
\lim_{x\to0}\frac{(1+x)^{\frac{2}{x}} - e^2[1 - \ln(1+x)]}{x}
&=
\lim_{x\to0}\frac{e^{\frac{2\ln(1 + x)}{x}} - e^2[1 - \ln(1+x)]}{x}
\\\\
&=
e^2 \cdot \lim_{x\to0}\frac{e^{\frac{2\ln(1 + x) - 2x}{x}} - 1 + \ln(1+x)}{x}
\\\\
&=
e^2 \cdot \lim_{x\to0}\frac{e^{\frac{2\ln(1 + x) - 2x}{x}} - 1}{x} +
e^2 \cdot \lim_{x\to0}\frac{\ln(1+x)}{x}
\\\\
&=
e^2 \cdot \lim_{x\to0}\frac{2\ln(1 + x) - 2x}{x^2} +
e^2
\\\\
&=
-e^2 +
e^2
\\\\
&= 0
\end{aligned}
]题目217
[
\text{求极限 } \lim_{x\to0}\frac{(1+x)^{\frac{1}{x}} - (1 + 2x)^{\frac{1}{2x}}}{\sin x}
]解答
[
\begin{aligned}
\lim_{x\to0}\frac{(1+x)^{\frac{1}{x}} - (1 + 2x)^{\frac{1}{2x}}}{\sin x}
&=
\lim_{x\to0}\frac{e^{\frac{\ln(x + 1)}{x}} - e^{\frac{\ln(2x + 1)}{2x}}}{x}
\\\\
&=
\lim_{x\to0} e^{\frac{\ln(2x + 1)}{2x}} \cdot
\frac{e^{\frac{2\ln(x + 1) - \ln(2x + 1)}{2x}} - 1}{x}
\\\\
&=
e\cdot
\lim_{x\to0}
\frac{2\ln(x + 1) - \ln(2x + 1)}{2x^2}
\\\\
&=
e\cdot
\lim_{x\to0}
\frac{2x -x^2 - 2x + 2x^2 + o(x^2)}{2x^2}
\\\\
&=
\frac{e}{2}
\end{aligned}
]题目218
[
\text{求极限 }
\lim_{x\to0}\frac{1 + \dfrac{x^2}{2} - \sqrt{1+x^2}}{(\cos x - e ^{x^2})\sin x^2}
]解答
x \to 0 时,有如下推导:
[
\cos x = 1 - \frac{1}{2}x^2 + o(x^2), \quad
e^{x^2} = 1 + x^2 + o(x^2) \quad \Rightarrow \quad
\cos x - e^{x^2} \sim -\frac{3}{2}x^2
][
(1 + x^2)^{\frac{1}{2}} - 1 \sim \frac{1}{2}x^2 - \frac{1}{8}x^4
]利用上述推导解决问题:
[
\begin{aligned}
\lim_{x\to0}\frac{1 + \dfrac{x^2}{2} - \sqrt{1+x^2}}{(\cos x - e ^{x^2})\sin x^2}
&=
\lim_{x\to0}\frac{\dfrac{1}{8}x^4}{-\dfrac{3}{2}x^4}
&=
-\frac{1}{12}
\end{aligned}
]题目219
[
\text{求极限 }
\lim_{x\to0}\frac{x\sin x^2 - 2(1 - \cos x)\sin x}{x^5}
]解答
[
x\sin x^2 = x^3 - \dfrac{1}{6}x^6 + o(x^6) \quad
][
2(1 - \cos x) \sin x = (x^2 - \dfrac{1}{12}x^4 + o(x^4)) \cdot (x - \dfrac{1}{6}x^3 + o(x^3)) = x^3 - (\frac{1}{6} + \frac{1}{12}) x^5 + o(x^5)
][
\lim_{x\to0}\frac{x\sin x^2 - 2(1 - \cos x)\sin x}{x^5} =
\lim_{x\to0}\frac{x^3 - \dfrac{1}{6}x^6 - x^3 + \dfrac{1}{4}x^5 + o(x^5)}{x^5} = \frac{1}{4}
]题目220
设 \(f'(0) = 0\), \(f''(0)\)存在, 求极限 _{x0}
解答
“f-f” 型 同名函数 相减,考虑 拉格朗日中值定理
然后,通过题干给出的条件,建立等式
由于
f''(0) 存在,故
f'(x) 在点
x=0 处 连续 (可导
\Rightarrow 连续)
故
\lim\limits_{x\to 0}f'(x) = f'(0)先由
Lagrange 中值定理可得:
[
\begin{aligned}
\lim\limits_{x\to0}\dfrac{f(x) - f(\ln(1 + x))}{x^3}
&=
\lim\limits_{x\to0}\dfrac{f'(\xi)(x - \ln(1+x))}{x^3}
\end{aligned}
]其中
\ln(1 + x) < \xi < x,两侧同除 x 取极限 ,然后 夹逼,可得:
[
\lim\limits_{x\to 0}\dfrac{\ln(1 + x)}{x} < \lim\limits_{x\to 0} \dfrac{\xi}{x} < \lim\limits_{x\to 0}\dfrac{x}{x}
\quad\Rightarrow\quad
\lim\limits_{x\to 0} \dfrac{\xi}{x} = 1
\quad\Rightarrow\quad
\xi \sim x \quad(x\to 0)
]最后用 导数定义 收尾:
[
\begin{aligned}
\lim\limits_{x\to0}\dfrac{f'(\xi)(x - \ln(1+x))}{x^3}
&=
\dfrac{1}{2} \lim\limits_{x\to0} \dfrac{f'(\xi)}{x}
\\\\
&=
\dfrac{1}{2} \lim\limits_{x\to0} \dfrac{f'(\xi) - f'(0)}{\xi} \cdot \dfrac{\xi}{x}
\\\\
&=
\dfrac{1}{2} \lim\limits_{x\to0} \dfrac{f'(\xi) - f'(0)}{\xi - 0} \cdot 1
\\\\
&=
\dfrac{1}{2} f''(0)
\\\\
\end{aligned}
]题目221
[
\lim_{x\to0^+}\frac{\sqrt{a}\arctan\sqrt{\dfrac{x}{a}}-\sqrt{b}\arctan\sqrt{\dfrac{x}{b}}}{x\sqrt{x}}
]解答
[
\arctan x = x - \dfrac{1}{3}x^3 \quad \Rightarrow \quad
\arctan \sqrt{\dfrac{x}{a}} = \sqrt{\dfrac{x}{a}} - \dfrac{x^{\frac{3}{2}}}{3a^{\frac{3}{2}}} + o(x^{\frac{3}{2}})
]根据上述推导,可对等式中的分子进行如下变形:
[
\sqrt{a}\arctan\sqrt{\dfrac{x}{a}}-\sqrt{b}\arctan\sqrt{\dfrac{x}{b}} =
x^{\frac{1}{2}} - \dfrac{x^{\frac{3}{2}}}{3a} - x^{\frac{1}{2}} + \dfrac{x^{\frac{3}{2}}}{3b} + o(x^{\frac{3}{2}}) = \dfrac{a - b}{3ab} x^{\frac{3}{2}} + o(x^{\frac{3}{2}})
]刚好展开到分母对应的阶数,于是就做完了
[
\lim_{x\to0^+}\frac{\sqrt{a}\arctan\sqrt{\dfrac{x}{a}}-\sqrt{b}\arctan\sqrt{\dfrac{x}{b}}}{x\sqrt{x}} = \dfrac{a - b}{3ab}
]题目222
设函数
f(x) 连续, 且
f(0)\ne0, 求极限
\lim\limits_{x\to0}\dfrac{x\int^x_{0}f(x-t)dt}{\int_0^xtf(x-t)dt}解答
极限中有 变上限积分,考虑用 洛必达法则 求导消掉积分符号
然后分母的 被积函数 中含有
x,考虑对积分变量换元,从而分离出
x令
x - t = u,则
-dt = du,有
[
\int_0^xf(x-t)dt = \int_0^xf(u)du
][
\int_0^xtf(x-t)dt = \int_0^x(x - u)f(u)du = x\int_0^xf(u)du - \int_0^x uf(u)du
]拆分好后,按照先前给出的思路,洛必达 即可
[
\lim\limits_{x\to0}\dfrac{x\int_0^xf(u)du}{x\int_0^xf(u)du - \int_0^x uf(u)du} =
\lim\limits_{x\to0}\dfrac{\int_0^xf(u)du + xf(x)}{\int_0^xf(u)du}
]这里没说
f(x) 可导,再洛就寄了,考虑 积分中值定理 来去掉 积分符号
\(\exists \xi \in (0,x), s.t. xf(\xi) = \int_0^x f(u)du\),则原式 = {x0} = {x0}
又由于
f(x) 连续,且
f(0) \ne 0,故
\lim\limits_{\xi \to 0} f(\xi) = f(0) \ne 0于是,原式 = _{(, x) (0,0)} = = 2
题目223
设函数
f(x) 连续,且
\lim\limits_{x\to0}\dfrac{f(x)}{x} = 2, 求极限
\lim\limits_{x\to0}\dfrac{\displaystyle\int_0^xe^{xt}\arctan(x-t)^2dt}{\displaystyle\int_0^xtf(x-t)dt}解答
由函数
f(x) 连续 及
\lim\limits_{x\to0}\dfrac{f(x)}{x} = 2,易知:
\lim\limits_{x\to0}f(x)=f(0)=0再由 导数定义,可得:
f'(0) = \lim\limits_{x\to0}\dfrac{f(x) - f(0)}{x - 0} = 2变上限积分函数 求极限,考虑 洛必达法则 求导去积分符号
分子的 变上限积分函数 中,既有
xt 又有
x-t,换元法 不能同时消掉,故考虑 广义积分中值定理
由 广义积分中值定理
\displaystyle\int_a^b f(x)g(x)dx = f(\xi) \displaystyle\int_a^b g(x)dx,其中
g(x) 在
(a,b) 上不变号
易知在
x\to 0 时,
\arctan(x-t)^2 不变号,于是有:
\displaystyle\int_0^xe^{xt}\arctan(x-t)^2dt = e^{x\xi}\displaystyle\int_0^x\arctan(x-t)^2dt,其中
\xi\in(0,x)[
\begin{aligned}
\lim\limits_{x\to0}\frac{\displaystyle\int_0^xe^{xt}\arctan(x-t)^2dt}{\displaystyle\int_0^xtf(x-t)dt} &=
e^0 \cdot \lim\limits_{x\to0}\frac{\displaystyle\int_0^x\arctan(x-t)^2dt}{\displaystyle\int_0^xtf(x-t)dt} \\\\
&=
\lim\limits_{x\to0}\frac{\displaystyle\int_0^x\arctan u^2du}{x\displaystyle\int_0^xf(u)du - \displaystyle\int_0^x uf(u)du}
\\\\
&=
\lim\limits_{x\to0}\frac{\arctan x^2}{xf(x) + \displaystyle\int_0^xf(u)du - xf(x)}
\\\\
&=
\lim\limits_{x\to0}\frac{x^2}{\displaystyle\int_0^xf(u)du}
\\\\
&=
\lim\limits_{x\to0}\frac{2x}{f(x)} =
2 \cdot \lim\limits_{x\to0}\frac{1}{\dfrac{f(x) - f(0)}{x - 0}}
\\\\
&= \frac{2}{f'(0)} \\\\
&= 1
\end{aligned}
]