Then: lny = (1/x) [ln (ax-1)-ln (a-1)-lnx]

lim[x→+∞] lny

= lim[x→+∞][ln(a^x- 1)-ln(a- 1)-lnx]/x

L'Hospital's rule

= lim[x→+∞][a^xlna/(a^x- 1)- 1/x]

= lim[x→+∞][(a^x- 1+ 1)lna/(a^x- 1)- 1/x]

= lim[x→+∞][lna+lna/(a^x- 1)- 1/x]

When a> is at 1, the limit of the above formula is lna.

When 0

Therefore, when a> is at 1, the limit value is e (lna) = a.

When 0

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