Salut,
[tex]\lim\limits_{x\to-\infty}x^3\left(e^{\frac{1}{x}}-e^{\frac{1}{x+1}}\right)=\lim\limits_{x\to-\infty}x^3\cdot e^{\frac{1}{x+1}}\left(\dfrac{e^{\frac{1}{x}}}{e^{\frac{1}{x+1}}}-1\right)=\\\\\\=\lim\limits_{x\to-\infty}x^3\cdot e^{\frac{1}{x+1}}\left(e^{\frac{1}{x}-\frac{1}{x+1}}-1\right)=\lim\limits_{x\to-\infty}x^3\cdot e^{\frac{1}{x+1}}\left(e^{\frac{1}{x^2+x}}-1\right)=\\\\\\=\lim\limits_{x\to-\infty}x^3\cdot e^{\frac{1}{x+1}}\cdot\dfrac{e^{\frac{1}{x^2+x}}-1}{\frac{1}{x^2+x}}\cdot\frac{1}{x^2+x}=\lim\limits_{x\to-\infty} e^{\frac{1}{x+1}}\cdot\dfrac{e^{\frac{1}{x^2+x}}-1}{\frac{1}{x^2+x}}\cdot\frac{x^3}{x^2+x}=\\\\\\=1\cdot 1\cdot (-\infty)=-\infty.[/tex]
Am folosit limita (eᵇ -- 1) / b tinde la 1, dacă b(x) tinde la 0.
Green eyes.
