Ensino Superior ⇒ Limite Exponencial Fundamental Tópico resolvido

[aprendiz123]

[tex3]\lim_{x \to 0} \frac{1}{x} \ln\sqrt{\frac{1+x}{1-x}}[/tex3]

Resposta:
1

[Karl Weierstrass]

[tex3]\lim_{x \to 0}\, \frac{1}{x} \,\cdot\,\ell n\,\sqrt{\frac{1\,+\,x}{1\,-\,x}}[/tex3]

[tex3]\hspace{70pt}\lim_{x \to 0}\, \frac{1}{x} \,\cdot\, \ell n\,\sqrt{\frac{1\,+\,x}{1\,-\,x}}\,=\,\lim_{x \to 0}\, \frac{1}{x}\,\cdot\, \, \ell n\,\left(\frac{1\,+\,x}{1\,-\,x}\right)^{\frac{1}{2}}[/tex3]

[tex3]\hspace{163pt}=\,\lim_{x \to 0}\, \ell n\,\left(\frac{1\,+\,x}{1\,-\,x}\right)^{\frac{1}{2x}}[/tex3]

[tex3]\hspace{163pt}=\,\ell n\,\lim_{x \to 0}\, \left(\frac{1\,+\,x}{1\,-\,x}\right)^{\frac{1}{2x}}[/tex3]

[tex3]\hspace{163pt}=\,\ell n\,\left[\lim_{x \to 0}\, \left(\frac{1\,+\,x}{1\,-\,x}\right)^{\frac{1}{x}}\right]^{\frac{1}{2}}[/tex3]

[tex3]\hspace{163pt}=\,\ell n\,\left[\lim_{x \to 0}\, \frac{(1\,+\,x)^{\frac{1}{x}}}{(1\,-\,x)^{\frac{1}{x}}}\right]^{\frac{1}{2}}[/tex3]

[tex3]\hspace{163pt}=\,\ell n\, \left\{\frac{\lim_{x \to 0}\,(1\,+\,x)^{\frac{1}{x}}}{\left[\lim_{x \to 0}\,(1\,+\,(-x))^{\frac{1}{(-x)}}\right]^{-1}}\right\}^{\frac{1}{2}}[/tex3]

[tex3]\hspace{163pt}=\,\ell n\, \left(\frac{e}{e^{-1}}\right)^{\frac{1}{2}}[/tex3]

[tex3]\hspace{163pt}=\,\ell n\, e[/tex3]

[tex3]\hspace{163pt}=\,1.[/tex3]


[tex3](*)\,\,\lim_{y \to 0}\,(1\,+\,y)^{\frac{1}{y}}\,=\,e[/tex3]

Demonstração:

[tex3]\hspace{70pt}y\,=\,\frac{1}{x} \,\Longrightarrow\,x\,=\,\frac{1}{y}[/tex3]

[tex3]\hspace{70pt}y\,\to\,0\,\Longrightarrow\,x\,\to\,\infty[/tex3]

[tex3]\hspace{70pt}\lim_{y \to 0}\,(1\,+\,y)^{\frac{1}{y}}\,=\,\lim_{x \to \infty}\,\left(1\,+\,\frac{1}{x}\right)^x\,=\,e.[/tex3]





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[aprendiz123]

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