洛必达法则

已知定义在\((0,+\infty)\)上的函数\(f(x)=\ln(x+1),g(x)=\sqrt{x}\)

(1)证明:\(f(x)<g(x)\)

(2)设\(\varphi(x)=\left[\dfrac{4}{g^2(x)}+t\right]f(x)\)在\((0,+\infty)\)上存在极值点，求\(t\)取值范围

解

(1)令\(\sqrt{x}=t,t>0\)，则\(f(x)-g(x)\xlongequal[]{\sqrt{x}=t}\ln(1+t^2)-t\)

记\(h(t)=\ln(1+t^2)-t,h^{\prime}(t)=\dfrac{2t}{1+t^2}-1=\dfrac{t^2-2t+1}{1+t^2}=\dfrac{(t-1)^2}{1+t^2}>0\)

则\(h(t)\)单调递增，从而\(h(t)>h(0)=0\)

得证

(2)\(\varphi(x)=\left(\dfrac{4}{x}+t\right)\ln(x+1),\varphi^{\prime}(x)=-\dfrac{4\ln(x+1)}{x^2}+\left(\dfrac{4}{x}+t\right)\dfrac{1}{x+1}\)

使得\(\varphi(x)\)有极值点，则要有\(\varphi(x)\)有零点

进而\(-\dfrac{4\ln(x+1)}{x^2}+\left(\dfrac{4}{x}+t\right)\dfrac{1}{x+1}=0\)

整理有\(t=\dfrac{4(x+1)\ln(x+1)-4x}{x^2}\)

记\(\gamma(x)=\dfrac{4(x+1)\ln(x+1)-4x}{x^2}\)

\(\gamma^{\prime}(x)=\dfrac{-4x\ln(x+1)-8\ln(x+1)+8x}{x^3}\)

记$ \beta(x)=-4x\ln(x+1)-8\ln(x+1)+8x,\beta^{\prime}(x)=-4\ln(x+1)-\dfrac{4x}{x+1}-\dfrac{8}{x+1}+8$

不难得到\(\beta^{\prime}(x)\)单调递增

从而\(\beta^{\prime}(x)<\beta(0)=0\)

从而\(\beta(x)\)单调递减

则\(\beta(x)<\beta(0)=0\)

则\(\gamma(x)\)单调递减

\(\lim\limits_{x\to 0}\gamma(x)=\lim\limits_{x\to 0}\dfrac{4\ln(x+1)}{2x}=\lim\limits_{x\to 0}\dfrac{\dfrac{4}{x+1}}{2}=2\)

从而\(\gamma(x)<2\)，而不难得到\(\gamma(x)>0\)

则\(t\in(0,2)\)