What is #lim_(x->oo) (x^3/(x-1)-x^2)# ?

2 Answers
Eddie
Mar 12, 2017

See below

Explanation:

#lim_(x to oo) x^3 / (x-1) - x^2 #

#= lim_(x to oo) x^2 ( x / (x-1) - 1 )#

#= lim_(x to oo) x^2 ( 1 / (1-1/x) - 1 )#

#= lim_(x to oo) x^2 ( (1-1/x)^(-1) - 1 )#

Using Binomial Series

#= lim_(x to oo) x^2 ( (1+1/x + mathcal O (1/x)^2 ) - 1 )#

#= lim_(x to oo) x^2 ( (1/x + mathcal O (1/x)^2 ) )#

#= lim_(x to oo) x + mathcal O (1/x)^0 = oo#

George C.
Mar 12, 2017

#lim_(x->oo) (x^3/(x-1)-x^2) = oo#

Explanation:

When assessing:

#lim_(x->oo) (x^3/(x-1)-x^2)#

you cannot simply substitute #x=oo# since this results in the indeterminate form:

#oo - oo#

Instead, we can simplify the expression first, like this:

#lim_(x->oo) (x^3/(x-1)-x^2) = lim_(x->oo) ((x^3-x^2+x^2-x+x-1+1)/(x-1)-x^2)#

#color(white)(lim_(x->oo) (x^3/(x-1)-x^2)) = lim_(x->oo) (((x^2+x+1)(x-1)+1)/(x-1)-x^2)#

#color(white)(lim_(x->oo) (x^3/(x-1)-x^2)) = lim_(x->oo) ((color(red)(cancel(color(black)(x^2)))+x+1)+1/(x-1)-color(red)(cancel(color(black)(x^2))))#

#color(white)(lim_(x->oo) (x^3/(x-1)-x^2)) = lim_(x->oo) (x+1+1/(x-1))#

#color(white)(lim_(x->oo) (x^3/(x-1)-x^2)) >= lim_(x->oo) x = oo#

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