Respuesta :

LammettHash
[tex]\displaystyle\lim_{t\to\infty}\dfrac{\sqrt t+t^2}{4t-t^2}=\lim_{t\to\infty}\dfrac{\dfrac1{t^{3/2}}+1}{\dfrac4t-1}=\lim_{t\to\infty}\dfrac1{-1}=-1[/tex]

Or, if you actually did mean to find the limit as [tex]x\to\infty[/tex], then the result would simply be the limand [tex]\dfrac{\sqrt t+t^2}{4t-t^2}[/tex], since this expression doesn't depend on [tex]x[/tex] (as far as can be seen).

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