题目

设曲线 $y=f(x)$ 与 $y=\sqrt{\dfrac{(1+x^2)\sqrt{x}}{e^{x-1}}} + \arctan\dfrac{x^2-1}{\sqrt{1+x^2}}$ 在点 $(1,\sqrt{2})$ 处相切

，则 $\lim\limits_{x\to1}(f(x)+1-\sqrt{2})^{\frac{1}{\ln x}} = $ ______

解答

两曲线在一点处 相切，说明在该点处：

  1. 坐标相同
  2. 切线斜率相同

利用这两点建立方程即可

坐标相同：
$$
f(1) = \sqrt{2}
$$

切线斜率相同：令 $y = y_1 + y_2$

$$
\begin{aligned}
\ln y_{1}(x) &= \dfrac{1}{2} [\ln(1+x^2) + \dfrac{1}{2}\ln x - x + 1] \\
\dfrac{y_{1}’}{y_1} &= \dfrac{1}{2} [\dfrac{2x}{1+x^2} + \dfrac{1}{2x} - 1] \\
\dfrac{y_{1}’(1)}{y_1(1)} &= \dfrac{1}{4} \\
y_{1}’(1) &= \dfrac{\sqrt{2}}{4} \\
\end{aligned}
$$

$$
y_{2}’(1) = \lim_{x\to1}\dfrac{\arctan\dfrac{x^2-1}{\sqrt{1+x^2}}}{x - 1} =
\lim_{x\to1}\dfrac{(x-1)(x+1)}{(x - 1) \sqrt{1+x^2}} = \sqrt{2}
$$

故：$f’(1) = \dfrac{\sqrt{2}}{4} + \sqrt{2} = \dfrac{5}{4}\sqrt{2}$

$$
A = \lim_{x\to1}\dfrac{\ln (f(x) + 1 - \sqrt{2})}{\ln x} =
\lim_{x\to1}\dfrac{f(x)-\sqrt{2}}{x-1} =
\lim_{x\to1}f’(x) = \dfrac{5}{4}\sqrt{2}
$$

故：$\lim\limits_{x\to1}(f(x)+1-\sqrt{2})^{\frac{1}{\ln x}} = e^{\frac{5\sqrt{2}}{4}}$