limiti fondamentali in R

Teoremi

$\underset{x\to 0}{\lim }\frac{sin(x)}{x}=1$

Dimostrazione
$\forall \alpha \in ]0,\frac{\pi }{2}[$,

$\frac{sin(\alpha )}{2}\le \frac{\alpha }{2}\le \frac{tan(\alpha )}{2}$

$1\le \frac{\alpha }{sin(\alpha )}\le \frac{1}{cos(\alpha )}$

$cos(\alpha )\le \frac{sin(\alpha )}{\alpha }\le 1$
$\forall \alpha \in ]0,\frac{\pi }{2}[$

$\underset{x\to 0}{\lim }cos(x)=1$

per il teorema dei due carabinieri e dato che la funzione è pari allora

$\underset{x\to 0}{\lim }\frac{sin(x)}{x}=1$

$\square$

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$\underset{x\to 0}{\lim }\frac{1-cos(x)}{x^2}=\frac{1}{2}$

$\underset{x\to 0}{\lim }\frac{1-cos(x)}{x^2}\cdot \frac{1+cos(x)}{1+cos(x)}=\underset{x\to 0}{\lim }\frac{1-co{s}^2(x)}{x^2(1+cos(x))}=\underset{x\to 0}{\lim }\frac{si{n}^2(x)}{x^2}\cdot \frac{1}{1+cos(x)}=1\cdot \frac{1}{1+1}=\frac{1}{2}$

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$\underset{x\to 0}{\lim }\frac{tg(x)}{x}=1$

$\underset{x\to 0}{\lim }\frac{sin(x)}{x}\cdot \frac{1}{cos(x)}=1\cdot \frac{1}{1}=1$

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$\underset{x\to 0}{\lim }\frac{tg(x)-sin(x)}{x^3}=\frac{1}{2}$

$\underset{x\to 0}{\lim }\frac{tg(x)-sin(x)}{x^3}=\underset{x\to 0}{\lim }\frac{\frac{sin(x)}{cos(x)}-sin(x)}{x^3}=\underset{x\to 0}{\lim }\frac{sin(x)\cdot \frac{1-cos(x)}{cos(x)}}{x^3}=\underset{x\to 0}{\lim }\frac{sin(x)}{x}\cdot \frac{1-cos(x)}{x^2}\cdot \frac{1}{cos(x)}=1\cdot \frac{1}{2}\cdot \frac{1}{1}=\frac{1}{2}$

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