上册 1.2 函数极限 第11题

\section*{解题过程：} （1） $\displaystyle \lim _{x \rightarrow 0^{-}} \frac{\int_{0}^{x^{2}}(\sin t)^{\frac{3}{2}} \mathrm{~d} t}{\int_{0}^{x} t(t-\sin t) \mathrm{d} t}=\lim _{x \rightarrow 0^{-}} \frac{2 x \cdot \sin ^{\frac{3}{2}} x^{2}}{x(x-\sin x)}=\lim _{x \rightarrow 0^{-}} \frac{2 x^{3}}{x-\left(x-\frac{x^{3}}{6}+o\left(x^{5}\right)\right)}=12$ ． （2） $\displaystyle \lim _{x \rightarrow 0^{+}} \frac{\int_{0}^{\sin x} \sqrt{\tan t} \mathrm{~d} t}{\int_{0}^{\tan x} \sqrt{\sin t} \mathrm{~d} t}=\lim _{x \rightarrow 0^{+}} \frac{\sqrt{\tan (\sin x)} \cos x}{(\tan x)^{\prime} \sqrt{\sin (\tan x)}}=\lim _{x \rightarrow 0^{+}} \cos ^{3} x \frac{\sqrt{\sin x}}{\sqrt{\tan x}}=\lim _{x \rightarrow 0^{+}} \cos ^{3} x \sqrt{\cos x}=1$ ． （3） $\displaystyle \lim _{x \rightarrow \infty} \frac{\left(\int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d} t\right)^{2}}{\int_{0}^{x} \mathrm{e}^{2 t^{2}} \mathrm{~d} t}=\lim _{x \rightarrow \infty} \frac{2 \int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d} t \cdot \mathrm{e}^{x^{2}}}{\mathrm{e}^{2 x^{2}}}=\lim _{x \rightarrow \infty} \frac{2 \int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d} t}{\mathrm{e}^{x^{2}}}=\lim _{x \rightarrow \infty} \frac{2 \mathrm{e}^{x^{2}}}{2 x \mathrm{e}^{x^{2}}}=0$ ． （4） $\displaystyle \lim _{x \rightarrow+\infty} \frac{\int_{0}^{x} \mathrm{e}^{2 t^{2}} \mathrm{~d} t}{\int_{0}^{x} \sqrt{t} \mathrm{e}^{t^{2}} \mathrm{~d} t}=\lim _{x \rightarrow+\infty} \frac{\mathrm{e}^{2 x^{2}}}{\sqrt{x} \mathrm{e}^{x^{2}}}=\lim _{x \rightarrow+\infty} \frac{\mathrm{e}^{x^{2}}}{\sqrt{x}}=\infty$ ． （5） $\displaystyle \lim _{x \rightarrow \infty} \frac{x \int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d} t}{\int_{0}^{x} t \mathrm{e}^{t^{2}} \mathrm{~d} t}=\lim _{x \rightarrow \infty} \frac{\int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d} t+x \mathrm{e}^{x^{2}}}{x \mathrm{e}^{x^{2}}}=\lim _{x \rightarrow \infty} \frac{\int_{0}^{x} \mathrm{e}^{t^{2}} \mathrm{~d} t}{x \mathrm{e}^{x^{2}}}+1=\lim _{x \rightarrow \infty} \frac{\mathrm{e}^{x^{2}}}{\mathrm{e}^{x^{2}}+2 x \mathrm{e}^{x^{2}}}+1=\lim _{x \rightarrow \infty} \frac{1}{1+2 x}+1=1$ ． （6） $\displaystyle \lim _{x \rightarrow 0} \frac{x-\int_{0}^{x} \cos t^{2} \mathrm{~d} t}{\sin ^{5} x}=\lim _{x \rightarrow 0} \frac{x-\int_{0}^{x} \cos t^{2} \mathrm{~d} t}{x^{5}}=\lim _{x \rightarrow 0} \frac{1-\cos x^{2}}{5 x^{4}}=\lim _{x \rightarrow 0} \frac{\frac{1}{2} x^{4}}{5 x^{4}}=\frac{1}{10}$ ． （7）令 $u=\sqrt{x}$ ，则 $\displaystyle \lim _{x \rightarrow 0^{+}} \frac{\int_{0}^{\sqrt{x}}\left(1-\cos t^{2}\right) \mathrm{d} t}{\sqrt{x^{5}}}=\lim _{u \rightarrow 0^{+}} \frac{\int_{0}^{u}\left(1-\cos t^{2}\right) \mathrm{d} t}{u^{5}}=\lim _{u \rightarrow 0^{+}} \frac{1-\cos u^{2}}{5 u^{4}}=\lim _{u \rightarrow 0^{+}} \frac{\frac{u^{4}}{2}}{5 u^{4}}=\frac{1}{10}$ ． （8） $\displaystyle \lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2}} \cos t \mathrm{~d} t}{\ln \left(1+x^{2}\right)}=\lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2}} \cos t \mathrm{~d} t}{x^{2}}=\lim _{x \rightarrow 0} \frac{2 x \cos x^{2}}{2 x}=1$ ． （9） $\displaystyle \lim _{x \rightarrow 0} \frac{\int_{0}^{x} \sin t^{2} \mathrm{~d} t}{\ln \left(1+x^{3}\right)}=\lim _{x \rightarrow 0} \frac{\int_{0}^{x} \sin t^{2} \mathrm{~d} t}{x^{3}}=\lim _{x \rightarrow 0} \frac{\sin x^{2}}{3 x^{2}}=\frac{1}{3}$ ． （10） $\displaystyle \lim _{x \rightarrow 0} \frac{x^{2} \int_{0}^{x^{2}} t \mathrm{e}^{t^{2}} \sin t \mathrm{~d} t}{1-\cos x^{4}}=\lim _{x \rightarrow 0} \frac{x^{2} \int_{0}^{x^{2}} t \mathrm{e}^{t^{2}} \sin t \mathrm{~d} t}{\frac{1}{2} x^{8}}=2 \lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2}} t \mathrm{e}^{t^{2}} \sin t \mathrm{~d} t}{x^{6}}$ $$ =2 \lim _{x \rightarrow 0} \frac{2 x x^{2} \mathrm{e}^{x^{4}} \sin x^{2}}{6 x^{5}}=\frac{4}{6} \lim _{x \rightarrow 0} \frac{\mathrm{e}^{x^{4}} \sin x^{2}}{x^{2}}=\frac{2}{3} . $$ （11） $\displaystyle \lim _{x \rightarrow 0} \frac{\int_{0}^{x}(\arcsin t-t) \mathrm{d} t}{x\left(\mathrm{e}^{x}-1\right)^{3}}=\lim _{x \rightarrow 0} \frac{\int_{0}^{x}(\arcsin t-t) \mathrm{d} t}{x^{4}}=\lim _{x \rightarrow 0} \frac{\arcsin x-x}{4 x^{3}}=\frac{1}{4} \lim _{x \rightarrow 0} \frac{\frac{1}{\sqrt{1-x^{2}}}-1}{3 x^{2}}$ $$ =\frac{1}{12} \lim _{x \rightarrow 0} \frac{1}{\sqrt{1-x^{2}}} \lim _{x \rightarrow 0} \frac{1-\sqrt{1-x^{2}}}{x^{2}}=\frac{1}{24} . $$ （12） $\displaystyle \lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2}} \sin t^{2} \mathrm{~d} t}{x \int_{0}^{x} \ln \left(1+t^{4}\right) \mathrm{d} t}=\lim _{x \rightarrow 0} \frac{2 x \sin x^{4}}{\int_{0}^{x} \ln \left(1+t^{4}\right) \mathrm{d} t+x \ln \left(1+x^{4}\right)}$ $$ =\frac{2}{\lim _{x \rightarrow 0} \frac{\int_{0}^{x} \ln \left(1+t^{4}\right) \mathrm{d} t}{x^{5}}+\lim _{x \rightarrow 0} \frac{x^{5}}{x^{5}}}=\frac{2}{\lim _{x \rightarrow 0} \frac{\ln \left(1+x^{4}\right)}{5 x^{4}}+1}=\frac{5}{3} . $$ （13） $\displaystyle \lim _{x \rightarrow 0} \frac{\int_{0}^{\sin ^{2} x} \ln (1+t) \mathrm{d} t}{\sqrt{1+x^{4}}-1}=\lim _{x \rightarrow 0} \frac{\int_{0}^{\sin ^{2} x} \ln (1+t) \mathrm{d} t}{x^{4}}\left(\sqrt{1+x^{4}}+1\right)$ $$ =2 \lim _{x \rightarrow 0} \frac{2 \sin x \cos x \ln \left(1+\sin ^{2} x\right)}{4 x^{3}}=\lim _{x \rightarrow 0} \frac{x \sin ^{2} x}{x^{3}}=1 . $$