Ricevo da Antonio la seguente domanda:
Salve professore,
avrei bisogno del suo aiuto con questi limiti…
\[A)\quad\lim_{x \to +\infty }e^{-x^{2}-2x+\cos x}\arctan\left ( x^{3}-4x+\log x \right )\quad B)\quad\lim_{x \to 0 }\frac{\sin(x^{3})+2x^{2}}{\log(\sqrt{1+x^{3}})}\cdot (e^{x\cdot \cos x}-1)\]
\[C)\quad\underset{x\to 1}{\mathop{\lim }}\,\frac{\log x\cdot \log \left| x-1 \right|}{(x-1)}\cdot \sin \frac{1}{x}\quad D)\quad\lim_{x \to \frac{\pi }{2} }\tan x\cdot \left ( e^{x-\frac{\pi }{2}}-1 \right )\]
\[E)\quad\lim_{x \to+\infty }\frac{4x-3\cdot 2^{x}}{\left ( 5\cdot 2^{x-1}-3 \right )\left ( \sqrt{2x}-\sqrt{x-1} \right )}\quad F)\quad\lim_{x \to+\infty }\frac{e^{x}-3^{x}+x}{x^{2}-\log(x^{3})}+\sin x\]
\[G)\quad\lim_{x \to 0 }\frac{\log ( \cos x )}{| \sin^{3}x |}+\sin ( e^{\frac{1}{x}} )\quad H)\quad\lim_{x \to 0 }\frac{\log (2-e^{x^{3}})}{2x^{2}-\sin^{2}x}\cdot \arctan\left ( \sin\frac{1}{x} \right )\]
La ringrazio.
Gli rispondo così:
Caro Antonio,
cerco di riassumere i passaggi (supponendo che \(\log\) indichi il logaritmo naturale): nota che nei casi F), G) e H) si è fatto uso del teorema del confronto, e nel caso G) si sono calcolati separatamente i limiti destro e sinistro, che si sono rivelati coincidenti.
\[A)\quad \underset{x\to +\infty }{\mathop{\lim }}\,{{e}^{-{{x}^{2}}-2x+\cos x}}\arctan \left( {{x}^{3}}-4x+\log x \right)=\]\[=\underset{x\to +\infty }{\mathop{\lim }}\,{{e}^{-{{x}^{2}}\left( 1+2/x-\cos x/{{x}^{2}} \right)}}\cdot \underset{x\to +\infty }{\mathop{\lim }}\,\arctan \left( {{x}^{3}}\left( 1-4/{{x}^{2}}+\log x/{{x}^{3}} \right) \right)=\]\[={{e}^{-\infty }}\cdot \frac{\pi }{2}=0\quad .\]
\[B)\quad \underset{x\to 0}{\mathop{\lim }}\,\frac{\sin ({{x}^{3}})+2{{x}^{2}}}{\log (\sqrt{1+{{x}^{3}}})}\cdot ({{e}^{x\cdot \cos x}}-1)=\]\[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2\left( \sin ({{x}^{3}})/{{x}^{2}}+2 \right)}{\log (1+{{x}^{3}})/{{x}^{3}}}\cdot \underset{x\to 0}{\mathop{\lim }}\,\cos x\cdot \frac{{{e}^{x\cdot \cos x}}-1}{x\cos x}=\]\[=\frac{2\left( 0+2 \right)}{1}\cdot \left( 1\cdot 1 \right)=4\quad .\]
\[C)\quad \underset{x\to 1}{\mathop{\lim }}\,\frac{\log x\cdot \log \left| x-1 \right|}{(x-1)}\cdot \sin \frac{1}{x}\to \]\[\to t=x-1\to =\underset{t\to 0}{\mathop{\lim }}\,\frac{\log \left( 1+t \right)\cdot \log \left| t \right|}{t}\sin \frac{1}{1+t}=\]\[=1\cdot \left( -\infty \right)\cdot \sin \left( 1 \right)=-\infty \quad .\]\[D)\quad \underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,\tan x\cdot \left( {{e}^{x-\frac{\pi }{2}}}-1 \right)\to \]\[\to t=x-\frac{\pi }{2}\to =\underset{t\to 0}{\mathop{\lim }}\,\tan \left( t+\pi /2 \right)\cdot \left( {{e}^{t}}-1 \right)=\]\[=-\underset{t\to 0}{\mathop{\lim }}\,\frac{t\cos t}{\sin t}\cdot \underset{t\to 0}{\mathop{\lim }}\,\frac{\left( {{e}^{t}}-1 \right)}{t}=-1\cdot 1=-1\quad .\]
\[E)\quad \underset{x\to +\infty }{\mathop{\lim }}\,\frac{4x-3\cdot {{2}^{x}}}{\left( 5\cdot {{2}^{x-1}}-3 \right)\left( \sqrt{2x}-\sqrt{x-1} \right)}=\]\[=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{{{2}^{x}}\left( -3+4x/{{2}^{x}} \right)}{{{2}^{x}}\left( 5/2-3/{{2}^{x}} \right)}\cdot \underset{x\to +\infty }{\mathop{\lim }}\,\frac{1}{\sqrt{x}\left( \sqrt{2}-\sqrt{1-1/x} \right)}=\]\[=-\frac{6}{5}\cdot \frac{1}{+\infty \left( \sqrt{2}-1 \right)}=-\frac{6}{5}\cdot 0=0\quad .\]
\[F)\quad \underset{x\to +\infty }{\mathop{\lim }}\,\frac{{{e}^{x}}-{{3}^{x}}+x}{{{x}^{2}}-\log ({{x}^{3}})}+\sin x\to \]\[\to \frac{{{e}^{x}}-{{3}^{x}}+x}{{{x}^{2}}-\log ({{x}^{3}})}-1\le \frac{{{e}^{x}}-{{3}^{x}}+x}{{{x}^{2}}-\log ({{x}^{3}})}+\sin x\le \frac{{{e}^{x}}-{{3}^{x}}+x}{{{x}^{2}}-\log ({{x}^{3}})}+1\to \]\[\to g\left( x \right)=\frac{{{e}^{x}}-{{3}^{x}}+x}{{{x}^{2}}-\log ({{x}^{3}})}\to \underset{x\to +\infty }{\mathop{\lim }}\,g\left( x \right)=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{{{3}^{x}}\left( -1+{{(e/3)}^{x}}+x/{{3}^{x}} \right)}{{{x}^{2}}\left( 1-3\log x/{{x}^{2}} \right)}=\]\[=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{{{3}^{x}}}{{{x}^{2}}}\cdot \underset{x\to +\infty }{\mathop{\lim }}\,\frac{-1+{{(e/3)}^{x}}+x/{{3}^{x}}}{1-3\log x/{{x}^{2}}}=+\infty \cdot \left( -1 \right)=-\infty \to \]\[\to \underset{x\to +\infty }{\mathop{\lim }}\,g\left( x \right)-1=\underset{x\to +\infty }{\mathop{\lim }}\,g\left( x \right)+1=-\infty \to \underset{x\to +\infty }{\mathop{\lim }}\,g\left( x \right)+\sin x=-\infty \quad .\]
\[G)\quad \underset{x\to 0}{\mathop{\lim }}\,\frac{\log (\cos x)}{|{{\sin }^{3}}x|}+\sin ({{e}^{\frac{1}{x}}})\to \underset{x\to 0}{\mathop{\lim }}\,\left( g\left( x \right)+\sin ({{e}^{\frac{1}{x}}}) \right)\to \]\[\to \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,g\left( x \right)+\sin ({{e}^{\frac{1}{x}}})\to g\left( x \right)-1\le g\left( x \right)+\sin ({{e}^{\frac{1}{x}}})\le g\left( x \right)+1\to \]\[\to \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,g\left( x \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{\log (1+\cos x-1)}{\left( \cos x-1 \right)}\cdot \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{{{x}^{2}}\left( \cos x-1 \right)}{{{x}^{2}}{{\sin }^{2}}x}\cdot \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{1}{\sin x}=\]\[=1\cdot \left( -\frac{1}{2} \right)\cdot \frac{1}{{{0}^{+}}}=-\infty \to \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,g\left( x \right)-1=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,g\left( x \right)+1=-\infty \to \]\[\to \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{\log (\cos x)}{|{{\sin }^{3}}x|}+\sin ({{e}^{\frac{1}{x}}})=-\infty \quad ;\]\[\to \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,g\left( x \right)+\sin ({{e}^{\frac{1}{x}}})\to \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\frac{\log (1+\cos x-1)}{\left( \cos x-1 \right)}\cdot \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\frac{{{x}^{2}}\left( \cos x-1 \right)}{{{x}^{2}}{{\sin }^{2}}x}\cdot \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\frac{-1}{\sin x}+\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\sin ({{e}^{\frac{1}{x}}})=\]\[=1\cdot \left( -\frac{1}{2} \right)\cdot \frac{-1}{{{0}^{-}}}+\sin \left( {{e}^{-\infty }} \right)=-\infty +0=-\infty \to \]\[\to \underset{x\to 0}{\mathop{\lim }}\,\frac{\log (\cos x)}{|{{\sin }^{3}}x|}+\sin ({{e}^{\frac{1}{x}}})=-\infty \quad .\]
\[H)\quad \underset{x\to 0}{\mathop{\lim }}\,\frac{\log (2-{{e}^{{{x}^{3}}}})}{2{{x}^{2}}-{{\sin }^{2}}x}\cdot \arctan \left( \sin \frac{1}{x} \right)\to \]\[-\frac{\pi }{4}\cdot \frac{\log (2-{{e}^{{{x}^{3}}}})}{2{{x}^{2}}-{{\sin }^{2}}x}\le \frac{\log (2-{{e}^{{{x}^{3}}}})}{2{{x}^{2}}-{{\sin }^{2}}x}\cdot \arctan \left( \sin \frac{1}{x} \right)\le \frac{\pi }{4}\cdot \frac{\log (2-{{e}^{{{x}^{3}}}})}{2{{x}^{2}}-{{\sin }^{2}}x}\to \]\[\to \underset{x\to 0}{\mathop{\lim }}\,\frac{\log (2-{{e}^{{{x}^{3}}}})}{2{{x}^{2}}-{{\sin }^{2}}x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\log (1+1-{{e}^{{{x}^{3}}}})}{\left( 1-{{e}^{{{x}^{3}}}} \right)}\cdot \underset{x\to 0}{\mathop{\lim }}\,\frac{-x\left( {{e}^{{{x}^{3}}}}-1 \right)}{{{x}^{3}}\left( 2-{{\sin }^{2}}x/{{x}^{2}} \right)}=\]\[=1\cdot \frac{0\cdot 1}{\left( 2-1 \right)}=0\quad .\]
Massimo Bergamini
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Alcuni limiti
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