Calculate this limit without l'hospital's rule

Ln(x) / Ln((x^2) - 1) = Ln(x) / [Ln(x+1) + Ln (x-1)] = 1/ [(Ln(x+1)/Ln(x)) + (Ln(x-1)/Ln(x))]

We first prove that lim x -> +inf (Ln(x+1)/Ln(x)) = 1.

To prove this we use the sandwich theorem. Observe that for large x, Ln(x) < Ln(x+1) < Ln (e. x) = Ln(x) + 1.

So, 1 < Ln(x+1)/Ln(x) < 1 + 1/Ln(x).
Taking limits as x -> +inf, we get 1 <= lim x -> +inf (Ln(x+1)/Ln(x)) <= 1,
i.e., lim x -> +inf (Ln(x+1)/Ln(x)) = 1.

A similar argument shows that lim x -> +inf (Ln(x-1)/Ln(x)) = 1

Consequently,

lim x -> +inf Ln(x) / Ln((x^2) - 1)
= lim x -> +inf 1/ [(Ln(x+1)/Ln(x)) + (Ln(x-1)/Ln(x))]
= 1/(1+1)
= 1/2

let f(x) = Ln(x) / Ln((x² -1)

and lt x-> ∞: Ln(x ± c) = L ->∞, c is a constant

Ln(x² -1) = Ln(x+1)+Ln(x-1)

and lt x-> ∞: f(x) -> L/(L + L) = 1/2

Without lhopital's rule, you cannot compute it.

inf/inf is inappropriate form. So,you should use the rule

I agree with some parts