洛必达
洛必达法则练习
1.求\(\lim_{x\rightarrow +\infty} \frac{e^{x}}{x^{100}}\)；\(\lim_{n\rightarrow +\infty} \frac{e^{n}}{n^{100}}\)
\[ \lim_{x\rightarrow +\infty} \frac{e^{x}}{x^{100}}\\=\lim_{x\rightarrow +\infty} \frac{(e^{x})^{(100)}}{100!}=+\infty. \]
由归结原则，数列极限也为0.
2.求\(\lim_{x\rightarrow +\infty} \frac{\ln^{20}(x)}{x}\);\(\lim_{n\rightarrow +\infty} \frac{\ln^{20}(n)}{n}\)
\[ \lim_{x\rightarrow +\infty} \frac{\ln^{20}(x)}{x}\\=\lim_{x\rightarrow +\infty} \frac{(\ln^{20}(x))^{'}}{1}=\lim_{x\rightarrow +\infty}(20/x)\ln^{19}(x) \]
重复求导易得极限为0.
3.求\(\lim_{x\rightarrow 0} \frac{1+(1+3x)^{a}-(1+2x)^{a}-(1+x)^{a}}{x^{2}}\)
\[ \lim_{x\rightarrow 0} \frac{1+(1+3x)^{a}-(1+2x)^{a}-(1+x)^{a}}{x^{2}}\\=\lim_{x\rightarrow 0} \frac{3a(1+3x)^{a-1}-2a(1+2x)^{a-1}-a(1+x)^{a-1}}{2x}\\=\lim_{x\rightarrow 0} \frac{9a(a-1)(1+3x)^{a-2}-4a(a-1)(1+2x)^{a-2}-a(a-1)(1+x)^{a-2}}{2}\\=(9a^{2}-9a-4a^{2}+4a-a^{2}+a)/2=2a^{2}-2a. \]
4.若 \(\lim_{x\rightarrow 0^{+}} \frac{\ln x-\ln(\sin x)}{ax^{2}}=1\)，求a(a不为0)
\[ \lim_{x\rightarrow 0^{+}} \frac{\ln{x}-\ln(\sin x)}{ax^{2}}\\=\lim_{x\rightarrow 0^{+}} \frac{\ln{(x/\sin x)}}{ax^{2}}\\=\lim_{x\rightarrow 0^{+}}\frac{(x-\sin x)/\sin x}{ax^{2}} \tag{1} \]
将\(\sin x\)展开到第二项即得
\[ {(x\rightarrow 0^{+})} \frac{x-\sin x}{x} \sim \frac{x^{2}}{6} \]
代入\((1)\)即得
\[ \lim_{x\rightarrow 0^{+}} \frac{\ln x-\ln(\sin x)}{ax^{2}}=\frac{1}{6a}=1 \Rightarrow a= \frac{1}{6}. \]
5.证明：\(\lim_{x\rightarrow 0^{+}} \frac{x^{x^{x}}}{x}=1\).
设\(k\)为正数，则
\[ \lim_{x\rightarrow 0^{+}} x\ln^{k}{x}=\lim_{x\rightarrow 0^{+}}\frac{\ln^{k}{x}}{1/x}=\lim_{x\rightarrow 0^{+}}\frac{k\ln^{k-1}{x}}{-1/x^{3}} \]
连续求导得
\[ lim_{x\rightarrow 0^{+}}\frac{k\ln^{k-1}{x}}{-1/x^{3}}=lim_{x\rightarrow 0^{+}}\frac{\prod_{i=1}^{[k]}(i+k-[k])\ln^{k-[k]}{x}}{-1/x^{1+2[k]}} \tag{1} \]
再对\((1)\)上下求一次导即得极限为0.
从而
\[ \lim_{x\rightarrow 0^{+}} \ln{x}(x^{x}-1)\\=\lim_{x\rightarrow 0^{+}} \ln x(e^{x\ln{x}}-1)\\=\lim_{x\rightarrow 0^{+}} \ln x(x\ln x) =0. \]
进而有
\[ \lim_{x\rightarrow 0^{+}} \frac{x^{x^{x}}}{x}=\lim_{x\rightarrow0^{+}}e^{\ln x(x^{x}-1)}=1.~~~\square \]