\( \begin{align} \lim_{x \to 0} \left( \frac{\tan x}{x} \right) &= \lim_{x \to 0}\left(\frac{\sin x}{\cos x} \cdot \frac{1}{x}\right) =\\ \\ &= \lim_{x \to 0}\left(\frac{\sin x}{x} \cdot \frac{1}{\cos x}\right) =\\ \\ &= 1 \cdot 1=1\\ \end{align}\)

\( \displaystyle\lim_{x \to 0} \left( \frac{x}{\tan x} \right) = \lim_{x \to 0} \left( \frac{1}{\large\frac{\tan x}{x}} \right) = \frac{1}{1} = 1 \)

\( \begin{align} \lim_{x \to 0} \left( \frac{1-\cos x}{x} \right) &= \lim_{x \to 0}\frac{\left(1-\cos x \right) \cdot \left(1+\cos x\right)}{x \cdot \left(1+\cos x\right)} =\\ \\ &= \lim_{x \to 0}\frac{\left(1-\cos^2 x \right)}{x \cdot \left(1+\cos x\right)} = \lim_{x \to 0}\frac{\sin^2 x}{x \cdot \left(1+\cos x\right)} =\\ \\ &= \lim_{x \to 0}\left(\frac{\sin x}{x} \cdot \sin x \cdot \frac{1}{1+\cos x}\right) =\\ \\ &= 1 \cdot 0 \cdot \frac{1}{2} = 0\\ \end{align}\)

\( \begin{align} \lim_{x \to 0^\pm} \left( \frac{x}{1-\cos x} \right) &= \lim_{x \to 0^{\pm}} \left( \frac{1}{\large\frac{1-\cos x}{x}} \right) = \frac{1}{0^\pm} = \pm\infty\\ \end{align}\)

\( \begin{align} \lim_{x \to 0} \left( \frac{1-\cos x}{x^2} \right) &= \lim_{x \to 0}\frac{\left(1-\cos x \right) \cdot \left(1+\cos x\right)}{x^2 \cdot \left(1+\cos x \right)} =\\ \\ &= \lim_{x \to 0}\frac{\left(1-\cos^2 x \right)}{x^2 \cdot \left(1+\cos x\right)} =\\ \\ &= \lim_{x \to 0}\frac{\sin^2 x}{x^2 \cdot \left(1+\cos x\right)} =\\ \\ &= \lim_{x \to 0}\left(\frac{\sin x}{x} \cdot \frac{\sin x}{x} \cdot \frac{1}{1+\cos x}\right) =\\ \\ &= 1 \cdot 1 \cdot \frac{1}{2} = \frac{1}{2}\\ \end{align}\)

\( \begin{align} \lim_{x \to 0}\left(\frac{1-\cos x+\sin x}{1-\cos x-\sin x}\right) &= \lim_{x \to 0}\frac{\large\frac{1-\cos x}{x}+\frac{\sin x}{x}}{\large\frac{1-\cos x}{x}-\frac{\sin x}{x}} =\\ \\ &= \lim_{x \to 0}\frac{0+1}{0-1}=-1\\ \end{align} \)

\( \begin{align} \lim_{x \to 0}\left(\frac{1-\cos x+\sin x}{1-\cos x-\sin x}\right) &= \lim_{x \to 0}\left(\frac{1-\cos x+\sin x}{1-\cos x-\sin x} \cdot \frac{1+\cos x}{1+\cos x} \right) =\\ \\ &= \lim_{x \to 0}\left(\frac{\left(1-\cos x\right)\left(1+\cos x \right)+\sin x \left( 1+\cos x \right)}{\left(1-\cos x\right)\left(1+\cos x \right)-\sin x\left( 1+\cos x \right)} \right) =\\ \\ &= \lim_{x \to 0}\left(\frac{\left(1-\cos^2 x\right)+\sin x \left(1+\cos x \right)}{\left(1-\cos^2 x\right)-\sin x\left(1+\cos x \right)} \right) =\\ \\ &= \lim_{x \to 0}\left(\frac{\sin^2 x+\sin x \left(1+\cos x \right)}{\sin^2 x-\sin x\left(1+\cos x \right)} \right) =\\ \\ &= \lim_{x \to 0}\left(\frac{\cancel{\sin x} \cdot \left( \sin x + 1+ \cos x \right)}{\cancel{\sin x} \cdot \left( \sin x - \left(1+\cos x \right) \right)} \right) =\\ \\ &= \lim_{x \to 0}\left(\frac{0 + 1 + 1 }{0 - \left( 1 + 1 \right)} \right) = \frac{2}{-2} = -1\\ \end{align} \)

\( \begin{align} \lim_{x \to 0} \left( \frac{\tan x-\sin x}{2x^3} \right) &= \lim_{x \to 0} \left( \frac{\frac{\sin x}{\cos x}-\sin x}{2x^3} \right) =\\ \\ &= \lim_{x \to 0} \left( \frac{\frac{\sin x - \sin x \cdot \cos x}{\cos x}}{2x^3} \right) =\\ \\ &= \lim_{x \to 0} \left( \frac{\sin x \cdot \left( 1 - \cos x \right)}{2x^3 \cdot \cos x} \right) =\\ \\ &= \lim_{x \to 0}\left(\frac{1}{2 \cdot \cos x} \cdot \frac{\sin x}{x} \cdot \frac{1 - \cos x}{x^2}\right) =\\ \\ &= \frac{1}{2} \cdot 1 \cdot \frac{1}{2} = \frac{1}{4}\\ \end{align}\)

\( \begin{align} \lim_{x \to +\infty}\left( \frac{x}{x+1} \right)^x &= \lim_{x \to +\infty}\left[\left( \frac{x+1}{x} \right)^x\:\right]^{-1} =\\ \\ &= \left[\lim_{x \to +\infty}\left( 1 + \frac{1}{x} \right)^x\:\right]^{-1} = e^{-1} = \frac{1}{e}\\ \\ \end{align}\)

\( \begin{align} \lim_{x \to +\infty}\left[x\cdot\ln\left( 1 + \frac{1}{x} \right)\right] &= \lim_{x \to +\infty}\left[\ln\left( 1 + \frac{1}{x} \right)^x\:\right] =\\ \\ &= \ln \left[\lim_{x \to +\infty}\left( 1 + \frac{1}{x} \right)^x\:\right] = \ln e = 1\\ \\ \end{align}\)

\( \begin{align} \lim_{x \to +\infty}\left( \frac{x+1}{x-1} \right)^x &= \lim_{x \to +\infty}\left[\frac{\left( \frac{x+1}{x} \right)^x}{\left( \frac{x-1}{x} \right)^x}\right] =\\ \\ &= \frac{\displaystyle\lim_{x \to +\infty}\left( \frac{x+1}{x} \right)^x}{\displaystyle\lim_{x \to +\infty}\left( \frac{x-1}{x} \right)^x} = \frac{e}{e^{-1}} = e^2 \\ \\ \end{align}\)