3. 设当 $x\rightarrow 0$ 时, $\sec x-\cos x$ 与 $\sqrt{a+x^n}-\sqrt{a}$ 是等价无穷小, 求常数 $a$ 与 $n$.
6. 利用等价无穷小代换法求下列极限:
(4) $\lim\limits_{x\rightarrow 0}\dfrac{x\ln(1-2x)}{1-\sec x}$
(5) $\lim\limits_{x\rightarrow 0}\dfrac{\sin x-\tan x}{x\ln(1+x^2)}$
(7) $\lim\limits_{x\rightarrow\infty}x\ln\frac{x+1}{x}$
(8) $\lim\limits_{x\rightarrow 0}\dfrac{\sqrt{1+x^2}-1}{e^{x^2}-1}$
1
3.
解:
(1) 若 $a=0$, 则由条件 $\lim\limits_{x\rightarrow 0}\dfrac{\sec x-\cos x}{\sqrt{x^n}}=1$. 于是
\[
\begin{split}
1&=\lim_{x\rightarrow 0}\frac{\frac{1}{\cos x}-\cos x}{x^{n/2}}\\
&=\lim_{x\rightarrow 0}\frac{1-\cos^2 x}{x^{n/2}}\\
&=\lim_{x\rightarrow 0}\frac{\sin^2 x}{x^{n/2}}\\
&=\lim_{x\rightarrow 0}\frac{x^2}{x^{n/2}}\\
\end{split}
\]
这推出 $n=4$.
(2) 若 $a > 0$. 则
\[
\begin{split}
1&=\lim_{x\rightarrow 0}\frac{\frac{1}{\cos x}-\cos x}{\sqrt{a}\cdot(\sqrt{1+\frac{x^n}{a}}-1)}\\
&=\frac{1}{\sqrt{a}}\cdot\lim_{x\rightarrow 0}\frac{1-\cos^2 x}{\frac{1}{2}\cdot\frac{x^n}{a}}\\
&=\frac{2a}{\sqrt{a}}\cdot\lim_{x\rightarrow 0}\frac{\sin^2 x}{x^n}\\
&=2\sqrt{a}\cdot\lim_{x\rightarrow 0}\frac{x^2}{x^n}\\
\end{split}
\]
于是推出
\[
\begin{cases}
2\sqrt{a}=1\\
n=2
\end{cases}
\]
即 $a=\frac{1}{4}$, $n=2$.
2
6. 利用等价无穷小代换法求下列极限:
(4) $\lim\limits_{x\rightarrow 0}\dfrac{x\ln(1-2x)}{1-\sec x}$
解:
\[
\begin{split}
\text{原式}&=\lim_{x\rightarrow 0}\frac{x\cdot(-2x)}{1-\frac{1}{\cos x}}\\
&=\lim_{x\rightarrow 0}\frac{-2x^2\cdot\cos x}{\cos x-1}\\
&=\lim_{x\rightarrow 0}\frac{2x^2}{1-\cos x}\\
&=\lim_{x\rightarrow 0}\frac{2x^2}{2\sin^2\frac{x}{2}}\\
&=\lim_{x\rightarrow 0}\frac{2x^2}{2\cdot(\frac{x}{2})^2}\\
&=\lim_{x\rightarrow 0}\frac{x^2}{\frac{x^2}{4}}\\
&=4
\end{split}
\]
当然, 如果熟悉 $1-\cos x\ \sim\ \frac{1}{2}x^2$ (当 $x\rightarrow 0$ 时), 可以直接代换, 计算会更简洁.
(5) $\lim\limits_{x\rightarrow 0}\dfrac{\sin x-\tan x}{x\ln(1+x^2)}$
解:
\[
\begin{split}
\text{原式}&=\lim_{x\rightarrow 0}\frac{\sin x\cdot(1-\frac{1}{\cos x})}{x\cdot x^2}\\
&=\lim_{x\rightarrow 0}\frac{\frac{\cos x-1}{\cos x}}{x^2}\\
&=\lim_{x\rightarrow 0}\frac{-(1-\cos x)}{x^2}\\
&=\lim_{x\rightarrow 0}\frac{-2\sin^2\frac{x}{2}}{x^2}\\
&=\lim_{x\rightarrow 0}\frac{-2\cdot(\frac{x}{2})^2}{x^2}\\
&=-\frac{1}{2}
\end{split}
\]
(7) $\lim\limits_{x\rightarrow\infty}x\ln\frac{x+1}{x}$
解:
\[
\text{原式}=\lim_{x\rightarrow\infty}\frac{\ln(1+\frac{1}{x})}{\frac{1}{x}}=1
\]
或
\[
\text{原式}=\lim_{x\rightarrow\infty}\ln\bigl((1+\frac{1}{x})^x\bigr)=\ln\lim_{x\rightarrow\infty}(1+\frac{1}{x})^x=\ln e=1
\]
(8)
\[
\lim_{x\rightarrow 0}\frac{\sqrt{1+x^2}-1}{e^{x^2}-1}=\lim_{x\rightarrow 0}\frac{\frac{1}{2}x^2}{x^2}=\frac{1}{2}.
\]