da *pizzaf40 » 28/09/2007, 01:40
Per \( \displaystyle {x}_{{0}}\cdot{y}_{{0}}={1} \) risulta:
\( \displaystyle {y}=\frac{{1}}{{x}} \)
quindi
\( \displaystyle {f{{\left({x},{y}\right)}}}=\frac{{1}}{{\sqrt{{x}}-\frac{{1}}{\sqrt{{y}}}}}={f{{\left({x}\right)}}}=\frac{{1}}{{\sqrt{{x}}-\frac{{1}}{\sqrt{{\frac{{1}}{{x}}}}}}}=\frac{{1}}{{\sqrt{{x}}-\sqrt{{x}}}}= \)infinito
Per \( \displaystyle {y}={0} \) devi distinguere i due casi:
1) \( \displaystyle {y}\to{{0}}^{+} \)
\( \displaystyle {f{{\left({x},{{0}}^{+}\right)}}}=\frac{{1}}{{\sqrt{{x}}-\frac{{1}}{{{{0}}^{+}}}}}={{0}}^{{-}} \)
2) \( \displaystyle {y}\to{{0}}^{{-}} \)
\( \displaystyle {f{{\left({x},{{0}}^{{-}}\right)}}}=\frac{{1}}{{\sqrt{{x}}-\frac{{1}}{{{{0}}^{{-}}}}}}={{0}}^{+} \)