\[\begin{align}
& \underset{x\to \infty }{\mathop{\lim }}\,{{e}^{x}}-x-1=\underset{x\to \infty }{\mathop{\lim }}\,{{e}^{x}}\left( 1-\frac{x}{{{e}^{x}}}-\frac{1}{{{e}^{x}}} \right) \\
& \underset{x\to \infty }{\mathop{\lim }}\,\frac{x}{{{e}^{x}}}\text{(suntem in cazul de nedeterminare }\frac{0}{0}\,\text{si aplicam {l}'Hopital)} \\
& \text{avem }\underset{x\to \infty }{\mathop{\lim }}\,\frac{x}{{{e}^{x}}}=\frac{{{x}'}}{{{\left( {{e}^{x}} \right)}^{\prime }}}=\frac{1}{{{e}^{x}}}=0,\text{ atunci }\underset{x\to \infty }{\mathop{\lim }}\,{{e}^{x}}\left( 1-\frac{x}{{{e}^{x}}}-\frac{1}{{{e}^{x}}} \right)={{e}^{x}}=\infty \\
& \, \\
& \text{Cat face }\underset{x\to -\infty }{\mathop{\lim }}\,{{e}^{x}}-x-1\,\text{?} \\
& \text{hmmm}...\text{ sa vedem:} \\
& \text{la cat tinde }{{e}^{x}}\text{?}\,\text{dar }x\,\text{?} \\
\end{align}\]