Question #2b749

2 Answers
Jan 7, 2018

#Lim_(xrarroo)ln(x^3+x+1)/ln(x+1)=3#

Explanation:

#Lim_(xrarroo)ln(x^3+x+1)/ln(x+1)=oo/oo#

Using L'Hopitals rule:

#Lim_(xrarroo)(1/(x^3+x+1)*(3x^2+1))/(1/(x+1))=Lim_(xrarroo)((3x^2+1)(x+1))/(x^3+x+1)#

#Lim_(xrarroo)(3x^3+3x^2+x+1)/(x^3+x+1)=Lim_(xrarroo)(x^3(3+3/x+x/x^2+1/x^3))/(x^3(1+1/x^2+1/x^3))#

#=(3+0+0+0)/(1+0+0)=3/1=3#

Jan 7, 2018

#lim_(x->oo) ln(x^3+x+1)/ln(x+1) = 3#

Explanation:

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#