Evaluate:
#lim_(x->oo) ln(x^3+x+1)/ln(x+1)#
The limit is in the form #oo/oo# but we can use the properties of logarithms to simplify.
As the the function is defined for # x+1 > 0# we can divide and multiply the argument of the logarithm at the numerator by #(x+1)^3#:
#lim_(x->oo) ln(x^3+x+1)/ln(x+1) = lim_(x->oo) ln(((x^3+x+1)(x+1)^3)/(x+1)^3)/ln(x+1)#
Use now the properties:
#ln ( a xx b) = lna+lnb#
#ln a^n = n ln a#
to get:
#lim_(x->oo) ln(x^3+x+1)/ln(x+1) = lim_(x->oo) (ln((x^3+x+1)/(x+1)^3)+3ln(x+1))/ln(x+1)#
#lim_(x->oo) ln(x^3+x+1)/ln(x+1) = lim_(x->oo) (ln((x^3+x+1)/(x+1)^3))/ln(x+1) + 3#
Now:
#lim_(x->oo) (x^3+x+1)/(x+1)^3 = 1#
so:
#lim_(x->oo) ln((x^3+x+1)/(x+1)^3) = 0#
and then:
#lim_(x->oo) ln(x^3+x+1)/ln(x+1) = 3#