1) Calculer \(\lim\limits_{x \to + \infty} \frac {e^{7x-2}} {(e^{5x+3})^2} \)

2) Résoudre l'inéquation \(e^{2x^2+1} \times e^{x-5}>\frac {1} {e^{2x}} \)

1) Attention: forme indéterminée! Penser à simplifier l'expression \(\frac {e^{7x-2}} {(e^{5x+3})^2} \)

2) Ramener l'inéquation à une inéquation du type \(ax^2+bx+c>0\)

1) \(\lim\limits_{x \to + \infty} \frac {e^{7x-2}} {(e^{5x+3})^2} = \lim\limits_{x \to + \infty} \frac {e^{7x-2}} {e^{10x+6}} = \lim\limits_{x \to + \infty} {e^{7x-2-10x-6}}= \lim\limits_{x \to + \infty} {e^{-3x-8}} \)

\(\left. \begin{array}{rcr} \lim\limits_{x \to + \infty} {-3x-8} = -\infty \\ \lim\limits_{X \to - \infty} e^{X} = 0 \\ \end{array} \right\} \lim\limits_{x \to + \infty} e^{-3x-8} = 0 \)

\(\lim\limits_{x \to + \infty} \frac {e^{7x-2}} {(e^{5x+3})^2} = 0\)

2) \(e^{2x^2+1} \times e^{x-5}>\frac {1} {e^{2x}} \)

\(e^{2x^2+x-4} >e^{-2x} \)

\(\frac {e^{2x^2+x-4}} {e^{-2x}} >1\)

\(e^{2x^2+3x-4} >e^{0} \)

\(2x^2+3x-4>0\)

\(\Delta = 41\)

\(x_1= - \frac {3+\sqrt41} {4} \)  et   \(x_2= \frac {-3+\sqrt41} {4}\)

\(S= ]-\infty ; - \frac {3+\sqrt41} {4} [ \cup] \frac {-3+\sqrt41} {4};+\infty[\)

\(\frac{e^a}{e^b} = e^{a-b} \)

\((e^x)^n = e^{xn}\)

\(e^0=1\)

\(e^a\times e^b = e^{a+b} \)