Spiegare gentilmente i ragionamenti, i passaggi e argomentare.
$ f(x) = \sqrt{x^2+1} $
$ f'(x) = \frac{x}{\sqrt{x^2+1}} $
$ f$"$(x) = \frac{1}{\sqrt{(x^2+1)^3}} $
a.
i) $ \displaystyle\lim_{x \to +\infty} f(x) = \displaystyle\lim_{x \to +\infty}\frac{1}{\sqrt{1+\frac{1}{x^2}}} = 1$
ii) $ \displaystyle\lim_{x \to -\infty} f(x) = \displaystyle\lim_{x \to -\infty}\frac{1}{-\sqrt{1+\frac{1}{x^2}}} = -1$
b.
$ f$"$(x) = \frac{f(x)-xf'(x)}{x^2+1} $
$ \frac{1}{\sqrt{(x^2+1)^3}} = \frac{f(x)-xf'(x)}{x^2+1} $
$ \frac{1}{(x^2+1)\sqrt{(x^2+1)}} = \frac{f(x)-xf'(x)}{x^2+1} $
$ \frac{1}{\sqrt{(x^2+1)}} = f(x)-xf'(x) $
$ \frac{1}{\sqrt{(x^2+1)} }= \sqrt{x^2+1} - \frac{x^2}{\sqrt{x^2+1}} $
$ \frac{1}{\sqrt{(x^2+1)} }= \frac{x^2+1 - x^2}{\sqrt{x^2+1} }$
$ \frac{1}{\sqrt{(x^2+1)} }= \frac{1}{\sqrt{x^2+1} }$
O.K.