Limits of functions

We list some useful limits .

For $ x\in (-\frac{\pi}{2},0)\cup(0,\frac{\pi}{2}) $ we have

$$|\sin x| <|x|<| \tan x|=\frac{|\sin x|}{\cos x}$$
$$\frac{\cos x}{|\sin x|}<\frac{1}{| x|}<\frac{1}{|\sin x|}$$
$$\cos x<|\frac{\sin x}{ x}|<1$$
$$\cos x<\frac{\sin x}{ x}<1$$
$$-1<-\frac{\sin x}{x}<-\cos x$$
$$0<1-\frac{\sin x}{x}<1-\cos x$$
$$|\frac{\sin x}{x}-1|<1-\cos x$$

as

$$\lim_{x\to 0}1-\cos x=0$$
$$ \lim_{x\to 0}\frac{\sin x}{x}=1\:;\lim_{x\to 0}\frac{\sin x-x}{x^3}=-\frac{1}{6}$$
$$ \lim_{x\to 0}\frac{\cos x-1}{x^2}=-\frac{1}{2}$$
$$\lim_{x\to 0}\frac{e^x-1}{x}=1\\:\lim_{x\to 0}\frac{e^x-1-x}{x^2}=\frac{1}{2}$$
$$\lim_{x\to 0}\frac{\ln (1+x)}{x}=1\:;\lim_{x\to 1}\frac{\ln x}{x-1}=1$$
$$\lim_{x\to 0}\frac{(1+x)^a-1}{x}=a$$
$$ a>0\:\:\lim_{x\to 0}\frac{a^x-1}{x}=\ln a$$
$$ \mathop{\lim}\limits_{x \to \infty}(a^\frac{1}{x}-1)x =\lim_{y\to 0}\frac{a^y-1}{y}=\lim_{y\to 0}\frac{e^{\ln a^y}-1}{y}=\lim_{y\to 0}\frac{e^{y\ln a}-1}{y}=\ln a$$
$$\lim_{x \to 0}(1+x)^{\frac{1}{x}}=e$$
$$\lim_{x \to 0}(1-x)^{\frac{1}{x}}=\frac{1}{e}$$
$$\lim_{x \to 0}(1+\sin x)^{\frac{1}{x}}=e$$
$$\lim_{x \to 0}(\cos x)^{\frac{1}{x^2}}=\lim_{x \to 0}(1-(1-\cos x))^{\frac{1}{1-\cos x}\frac{1-\cos x}{x^2}}=e^{-\frac{1}{2}}$$
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