Limiti neperiani

Ricevo da Santa la seguente domanda:
 
Gentile professore,
come si risolvono i seguenti limiti senza applicare De L’Hospital?
\[\underset{x\to +\infty }{\mathop{\lim }}\,{{\left( 1+\frac{x}{2{{x}^{2}}+1} \right)}^{x}}\quad \quad \underset{x\to +\infty }{\mathop{\lim }}\,{{\left( \frac{3x-1}{3x+2} \right)}^{\frac{x}{2}}}\quad .\]
Grazie.
 
Le rispondo così:
 
Cara Santa,
si tratta in entrambi i casi di forme indeterminate del tipo \({{1}^{\infty }}\), tipicamente riconducibili al limite notevole \[\underset{t\to \pm \infty }{\mathop{\lim }}\,{{\left( 1+\frac{1}{t} \right)}^{t}}=e\quad .\]
Infatti, poiché \[1+\frac{x}{2{{x}^{2}}+1}=1+\frac{1}{\left( 2{{x}^{2}}+1 \right)/x}\quad \quad \frac{3x-1}{3x+2}=1+\frac{1}{-\left( 3x+2 \right)/3}\] e \[\underset{x\to +\infty }{\mathop{\lim }}\,\left( \frac{2{{x}^{2}}+1}{x} \right)=+\infty \quad \quad \underset{x\to +\infty }{\mathop{\lim }}\,\left( -\frac{3x+2}{3} \right)=-\infty \] si ha:
\[\underset{x\to +\infty }{\mathop{\lim }}\,{{\left( 1+\frac{x}{2{{x}^{2}}+1} \right)}^{x}}=\underset{x\to +\infty }{\mathop{\lim }}\,{{\left( {{\left( 1+\frac{1}{\left( 2{{x}^{2}}+1 \right)/x} \right)}^{\frac{2{{x}^{2}}+1}{x}}} \right)}^{\frac{{{x}^{2}}}{2{{x}^{2}}+1}}}=\] \[={{\left( \underset{t\to +\infty }{\mathop{\lim }}\,{{\left( 1+\frac{1}{t} \right)}^{t}} \right)}^{\underset{x\to +\infty }{\mathop{\lim }}\,\frac{{{x}^{2}}}{2{{x}^{2}}+1}}}={{e}^{\frac{1}{2}}}=\sqrt{e}\]
\[\underset{x\to +\infty }{\mathop{\lim }}\,{{\left( \frac{3x-1}{3x+2} \right)}^{\frac{x}{2}}}=\underset{x\to +\infty }{\mathop{\lim }}\,{{\left( {{\left( 1+\frac{1}{-\left( 3x+2 \right)/3} \right)}^{-\frac{3x+2}{3}}} \right)}^{-\frac{3x}{2\left( 3x+2 \right)}}}=\] \[={{\left( \underset{t\to -\infty }{\mathop{\lim }}\,{{\left( 1+\frac{1}{t} \right)}^{t}} \right)}^{\underset{x\to +\infty }{\mathop{\lim }}\,\left( -\frac{3x}{6x+4} \right)}}={{e}^{-\frac{1}{2}}}=\frac{1}{\sqrt{e}}\quad .\]
Massimo Bergamini