题目查询

设函数 $f(x)$ 在区间 $[0,+\infty)$ 上可导, 则

$\lim_{x \to 0^+} \frac{x^x - 1}{\ln x \cdot \ln(1 - x)} =$

$\lim\limits_{x \to 0}{\left(\frac{1+e^{x}}{2}\right)}^{\cot x}=\underline{\qquad}$.

$\lim\limits_{x \to 0}\left(\dfrac{1 + e^x}{2}\right)^{\cot x} = \underline{\qquad}$

$\lim\limits_{x \to \infty} x^2 \left(2 - x \sin \dfrac{1}{x} - \cos \dfrac{1}{x}\right) = \underline{\qquad}$

$x \to 0$时，$\int_{0}^{x} \dfrac{(1+t^2)\sin t^2}{1+\cos^2 t} dt$与$x^k$是同阶无穷小，则$k=$ $\underline{\qquad}$

设$\lim\limits_{x\to 0}\left(\dfrac{1}{x}-\dfrac{\ln(1+x)}{x\sin x}\right) = \underline{\qquad}$

$\lim\limits_{x\to0} \dfrac{1}{x}\left(\dfrac{\sqrt{1+x^2}}{\sin x} - \dfrac{1}{\tan x}\right) = \underline{\qquad}$

设 $\int_{1}^{+\infty} \frac{\ln(x+1)}{x^2} dx = \underline{\qquad}.$

求极限$\lim_{x\to 0}\left(\frac{1+\int_{0}^{x}e^{t^{2}}\,dt}{e^{x}-1}-\frac{1}{\sin x}\right)$.