Derivata, regula lui l'Hospital

victoria94 Mar Ian 03, 2012 4:33 pm

calculati dupa regula lui l'Hospital:
lim(x → +∞) (ln(1+e^2x)/ln(1+e^3x))

**Admin** Mar Ian 03, 2012 7:46 pm

victoria94 a scris:calculati dupa regula lui l'Hospital:lim(x → +∞) (ln(1+e^2x)/ln(1+e^3x))

$\begin{array}{l} \mathop {{\rm{lim}}}\limits_{x \to \infty } \frac{{ln\left( {1 + {e^{2x}}} \right)}}{{ln\left( {1 + {e^{3x}}} \right)}} = \mathop {{\rm{lim}}}\limits_{x \to \infty } \frac{{\frac{d}{{dx}}ln\left( {1 + {e^{2x}}} \right)}}{{\frac{d}{{dx}}ln\left( {1 + {e^{3x}}} \right)}} = ... = \mathop {{\rm{lim}}}\limits_{x \to \infty } \frac{{2\left( {{e^{3x}} + 1} \right)}}{{3\left( {{e^{2x}} + 1} \right){e^x}}} = \frac{2}{3}\mathop {{\rm{lim}}}\limits_{x \to \infty } \frac{{{e^{3x}} + 1}}{{{e^{3x}} + {e^x}}}\\ = \frac{2}{3}\left( {\mathop {{\rm{lim}}}\limits_{x \to \infty } \frac{{{e^{3x}}}}{{{e^{3x}} + {e^x}}} + \underbrace {\mathop {{\rm{lim}}}\limits_{x \to \infty } \frac{1}{{{e^{3x}} + {e^x}}}}_{ = 0}} \right) \end{array}$

$\begin{array}{l} \mathop {{\rm{lim}}}\limits_{x \to \infty } \frac{{{e^{3x}}}}{{{e^{3x}} + {e^x}}} = \mathop {{\rm{lim}}}\limits_{x \to \infty } \frac{{{e^{3x}}}}{{\left( {{e^{2x}} + 1} \right){e^x}}} = \mathop {{\rm{lim}}}\limits_{x \to \infty } \frac{{{e^{2x}}}}{{{e^{2x}} + 1}} = \mathop {{\rm{lim}}}\limits_{x \to \infty } \frac{{\frac{d}{{dx}}{e^{2x}}}}{{\frac{d}{{dx}}\left( {{e^{2x}} + 1} \right)}}\\ = \mathop {{\rm{lim}}}\limits_{x \to \infty } \left[ {\frac{1}{{\left( {{e^{3x}} + {e^x}} \right)}} + 1} \right] = 1\\ \mathop {{\rm{lim}}}\limits_{x \to \infty } \frac{{ln\left( {1 + {e^{2x}}} \right)}}{{ln\left( {1 + {e^{3x}}} \right)}} = \frac{2}{3} \end{array}$

Ps: semnul $\frac{d}{{dx}}$ înseamnă derivata după x.

victoria94 Mier Ian 04, 2012 12:28 am

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