Respuesta :

LammettHash
[tex]\displaystyle\lim_{x\to\infty}(5\e^{-x}+1)^2e^x=\lim_{x\to\infty}\frac{(5e^{-x}+1)^2e^{2x}}{e^x}=\lim_{x\to\infty}\frac{(5+e^x)^2}{e^x}[/tex]

Replacing [tex]y=e^x[/tex], the limit is equivalent to

[tex]\displaystyle\lim_{y\to\infty}\frac{(5+y)^2}y[/tex]

which clearly approaches [tex]\infty[/tex].

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