求极限：lim lim _(xarrow {0)^+}xln (sin (x)^2)

步骤 1：转换极限形式
将原极限转换为更易于处理的形式，即
$$\lim _{x\rightarrow {0}^{+}}x\ln (\sin {x}^{2})=\lim _{x\rightarrow {0}^{+}}\dfrac {\ln (\sin {x}^{2})}{\dfrac {1}{x}}$$
步骤 2：应用洛必达法则
由于分子和分母在$x\rightarrow 0^+$时都趋于0，可以应用洛必达法则，即
$$\lim _{x\rightarrow {0}^{+}}\dfrac {\ln (\sin {x}^{2})}{\dfrac {1}{x}}=\lim _{x\rightarrow {0}^{+}}\dfrac {f'(x)}{g'(x)}$$
其中$f(x)=\ln (\sin {x}^{2})$，$g(x)=\dfrac {1}{x}$，则
$$f'(x)=\dfrac {2x\cos {x}^{2}}{\sin {x}^{2}}$$
$$g'(x)=-\dfrac {1}{{x}^{2}}$$
步骤 3：再次应用洛必达法则
由于分子和分母在$x\rightarrow 0^+$时都趋于0，再次应用洛必达法则，即
$$\lim _{x\rightarrow {0}^{+}}\dfrac {f'(x)}{g'(x)}=\lim _{x\rightarrow {0}^{+}}\dfrac {f''(x)}{g''(x)}$$
其中$f'(x)=2x\cos {x}^{2}$，$g'(x)=-\dfrac {1}{{x}^{2}}$，则
$$f''(x)=2\cos {x}^{2}-4{x}^{2}\sin {x}^{2}$$
$$g''(x)=\dfrac {2}{{x}^{3}}$$
步骤 4：计算极限
将$f''(x)$和$g''(x)$代入极限中，得到
$$\lim _{x\rightarrow {0}^{+}}\dfrac {f''(x)}{g''(x)}=\lim _{x\rightarrow {0}^{+}}\dfrac {2\cos {x}^{2}-4{x}^{2}\sin {x}^{2}}{\dfrac {2}{{x}^{3}}}$$
$$=\lim _{x\rightarrow {0}^{+}}\dfrac {2{x}^{3}\cos {x}^{2}-4{x}^{5}\sin {x}^{2}}{2}$$
$$=\lim _{x\rightarrow {0}^{+}}({x}^{3}\cos {x}^{2}-2{x}^{5}\sin {x}^{2})$$
$$=0$$