Ricevo da Antonio la seguente domanda: Salve professore, avrei bisogno del suo aiuto con questi limiti... Si calcolino, se esistono, tramite l'utilizzo di limiti notevoli, i seguenti limiti: \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin {{x}^{3}}+2{{x}^{2}}}{\ln \left( 1+{{x}^{2}}\sin x \right)}\left( {{e}^{{{x}^{2}}}}-1 \right)\arctan \left( \sin \left( \frac{1}{x} \right) \right)\] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{1+{{x}^{2}}}-1}{\sin {{x}^{2}}+2x}\sin \left( \frac{1}{x} \right)\] \[\underset{x\to +\infty }{\mathop{\lim }}\,\ln \left( 1+\frac{1}{x} \right)\frac{x\cdot {{3}^{2x}}-\sin x+{{x}^{3}}}{{{9}^{x}}-\arctan x}\] \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{{{2}^{\sin \sqrt{\frac{x}{x+1}}}}-1}{1-\cos x-\sin \left( 2x \right)}\] \[\underset{x\to -\infty }{\mathop{\lim }}\,\left( \sqrt{{{x}^{2}}+x}+x \right)\frac{\ln \left( |x|+\sin x \right)}{x}\] \[\underset{x\to -\infty }{\mathop{\lim }}\,\left( \sqrt{{{x}^{2}}-x}+x \right)\frac{\ln \left( |x|+\cos x \right)}{{{x}^{2}}}\] La ringrazio. Gli rispondo così: Caro Antonio, tenendo presente che, qualora sia \(\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=0\), per il teorema del confronto si ha \(\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)\sin \left( 1/x \right)=0\), e anche, ad esempio, \(\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)\arctan \left( \sin \left( 1/x \right) \right)=0\), essendo \(-\pi /4\le \arctan \left( \sin \left( 1/x \right) \right)\le \pi /4\), e più in generale il fatto che una funzione limitata moltiplicata per un infinitesimo risulti in un infinitesimo, ecco come “svilupperei” ciascuno dei limiti che proponi… \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin {{x}^{3}}+2{{x}^{2}}}{\ln \left( 1+{{x}^{2}}\sin x \right)}\left( {{e}^{{{x}^{2}}}}-1 \right)\arctan \left( \sin \left( \frac{1}{x} \right) \right)=\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)\arctan \left( \sin \left( \frac{1}{x} \right) \right)=0\]in quanto \[\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to 0}{\mathop{\lim }}\,\left( x\cdot \frac{{{e}^{{{x}^{2}}}}-1}{{{x}^{2}}}\cdot \frac{{{x}^{2}}\sin x}{\ln \left( 1+{{x}^{2}}\sin x \right)}\cdot \frac{x}{\sin x}\cdot \left( \frac{\sin {{x}^{3}}}{{{x}^{3}}}+2 \right) \right)=\left( 0\cdot 1\cdot 1\cdot 1\cdot 3 \right)=0\quad .\] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{1+{{x}^{2}}}-1}{\sin {{x}^{2}}+2x}\sin \left( \frac{1}{x} \right)=\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)\sin \left( \frac{1}{x} \right)=0\]in quanto \[\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to 0}{\mathop{\lim }}\,\left( x\cdot \frac{\sqrt{1+{{x}^{2}}}-1}{{{x}^{2}}}\cdot \frac{1}{x\left( \sin {{x}^{2}}/{{x}^{2}} \right)+2} \right)=\left( 0\cdot \frac{1}{2}\cdot \frac{1}{0+2} \right)=0\quad .\] \[\underset{x\to +\infty }{\mathop{\lim }}\,\ln \left( 1+\frac{1}{x} \right)\frac{x\cdot {{3}^{2x}}-\sin x+{{x}^{3}}}{{{9}^{x}}-\arctan x}=\underset{x\to +\infty }{\mathop{\lim }}\,\ln \left( 1+\frac{1}{x} \right)x\frac{{{9}^{x}}\left( 1-\sin x/(x+{{9}^{x}})+{{x}^{3}}/{{9}^{x}} \right)}{{{9}^{x}}\left( 1-\arctan x/{{9}^{x}} \right)}=\] \[=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{\ln \left( 1+1/x \right)}{1/x}\underset{x\to +\infty }{\mathop{\lim }}\,\frac{\left( 1-\sin x/(x+{{9}^{x}})+{{x}^{3}}/{{9}^{x}} \right)}{\left( 1-\arctan x/{{9}^{x}} \right)}=1\cdot \frac{1-0+0}{1-0}=1\quad .\] \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{{{2}^{\sin \sqrt{\frac{x}{x+1}}}}-1}{1-\cos x-\sin \left( 2x \right)}=\] \[=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{{{2}^{\sin \sqrt{\frac{x}{x+1}}}}-1}{^{\sin \sqrt{\frac{x}{x+1}}}}\cdot \frac{\sin \sqrt{x/(x+1)}}{\sqrt{x/(x+1)}}\cdot \frac{1}{\left( \left( 1-\cos x \right)/\left( 2x \right)-\sin 2x/\left( 2x \right) \right)}\cdot \frac{1}{2\sqrt{x\left( x+1 \right)}}=\] \[=\ln 2\cdot 1\cdot \frac{1}{\left( 0-1 \right)}\cdot \left( +\infty \right)=-\infty \quad .\] \[\underset{x\to -\infty }{\mathop{\lim }}\,\left( \sqrt{{{x}^{2}}+x}+x \right)\frac{\ln \left( |x|+\sin x \right)}{x}=\underset{x\to -\infty }{\mathop{\lim }}\,\left( \frac{x}{\sqrt{{{x}^{2}}+x}-x} \right)\frac{\ln \left( |x|\left( 1+\sin x/|x| \right) \right)}{x}=\] \[=\underset{x\to -\infty }{\mathop{\lim }}\,\left( \frac{1}{\left( |x|/x \right)\sqrt{1+1/x}-1} \right)\cdot \left( \frac{\ln |x|}{x}+\frac{\ln \left( 1+\sin x/|x| \right)}{x} \right)=\frac{1}{\left( -1 \right)\cdot 1-1}\cdot \left( 0+\frac{0}{\infty } \right)=0\quad .\] \[\underset{x\to -\infty }{\mathop{\lim }}\,\left( \sqrt{{{x}^{2}}-x}+x \right)\frac{\ln \left( |x|+\cos x \right)}{{{x}^{2}}}=\underset{x\to -\infty }{\mathop{\lim }}\,\left( \frac{x}{\sqrt{{{x}^{2}}+x}-x} \right)\frac{\ln \left( |x|\left( 1+\cos x/|x| \right) \right)}{{{x}^{2}}}=\] \[=\underset{x\to -\infty }{\mathop{\lim }}\,\left( \frac{1}{\left( |x|/x \right)\sqrt{1+1/x}-1} \right)\cdot \left( \frac{\ln |x|}{{{x}^{2}}}+\frac{\ln \left( 1+\cos x/|x| \right)}{{{x}^{2}}} \right)=-\frac{1}{2}\cdot \left( 0+\frac{0}{\infty } \right)=0\quad .\] Massimo Bergamini