Determine o lim(x → ∞) (raiz(x^2 + 1) - raiz(x))

Passo 1

Aplicamos a regra de L'Hôpital, multiplicando inicialmente pelo termo conjugado de sqrt(x^2 + 1) - sqrt(x) que e sqrt(x^2 + 1) + sqrt(x) . Assim:

underset(x -> oo)(l i m) sqrt(x^2 + 1) - sqrt(x) = underset(x -> oo)(l i m) sqrt(x^2 + 1) - sqrt(x) * (sqrt(x^2 + 1) + sqrt(x))/(sqrt(x^2 + 1) + sqrt(x)) = underset(x -> oo)(l i m) (x^2 - x + 1)/(sqrt(x^2 + 1) + sqrt(x)) =

underset(x -> oo)(l i m) (x^2 - x + 1)/(sqrt(x^2 + 1) + sqrt(x)) = underset(x -> oo)(l i m) ((d)/(d x) (x^2 - x + 1))/((d)/(d x) (sqrt(x^2 + 1)) + (d)/(d x) (sqrt(x))) = underset(x -> oo)(l i m) (2 x - 1)/((1)/(2) (2 x)/((sqrt(x^2 + x))) - (1)/(2) (1)/((sqrt(x)))) = underset(x -> oo)(l i m) (2 x - 1)/((1)/(2) (2 x)/(x (sqrt(1 + 1 // x))) - (1)/(2) (1)/((sqrt(x))))

underset(x -> oo)(l i m) (2 x - 1)/((1)/(2) (2 x)/(x (sqrt(1 + 1 // x))) - (1)/(2) (1)/((sqrt(x)))) = underset(x -> oo)(l i m) (2 x - 1)/((1)/(2) (2)/((1)) - (1)/(2) (1)/((sqrt(x)))) = underset(x -> oo)(l i m) (2 x - 1)/(2 - 0) = oo

Resposta

underset(x -> oo)(l i m) sqrt(x^2 + 1) - sqrt(x) = oo

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