[tex]f(x)=\frac{2}{\sqrt{x}}-\frac{1}{x}[/tex]
a)
Iti atasez tabelul de derivate si integrale
[tex]f'(x)=(2x^{\frac{-1}{2}} )'+\frac{1}{x^2} =-x^{\frac{-3}{2}}+ \frac{1}{x^2}=\frac{1}{x^2}-\frac{1}{x\sqrt{x} } =\frac{1-\sqrt{x} }{x^2}[/tex]
b)
Calculam limita spre +∞
[tex]\lim_{x\to +\infty} \frac{2}{\sqrt{x} } -\frac{1}{x} =\frac{2}{\infty} -\frac{1}{\infty}=0-0=0[/tex]
Dreapta de ecuatie y=0 este asimptota orizontala spre +∞
c)
[tex]\\\\ \lim_{x\to1} \frac{\frac{1-\sqrt{x} }{x^2} }{x-1} =\frac{0}{0}\\\\ \lim_{x\to1} \frac{\frac{1-\sqrt{x} }{x^2} }{x-1} = \lim_{x\to1}\frac{1-\sqrt{x} }{x^2(x-1)} = \lim_{x\to1}\frac{1-\sqrt{x} }{x^2(\sqrt{x} -1)(\sqrt{x} +1)} =\lim_{x\to1}\frac{-1}{x^2(\sqrt{x} +1)}=\frac{-1}{1(1+1)}=-\frac{1}{2}[/tex]
Un alt exercitiu cu ecuatia asimptotei gasesti aici: https://brainly.ro/tema/1030418
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