What is the limit of #x(sqrt(x^(2)+1)) # as x approaches infinity?

2 Answers
Jun 10, 2018

The limit #x -> oo, f(x) -> oo#

Explanation:

As #x# approaches infinity the +1 will lose all significance so we can just consider:

#x(sqrt(x^2))= x(x)= x^2#

we know #x -> oo, x^2 -> oo# so we can conclude:

#f(x) = x(sqrt(x^(2)+1))#

The limit #x -> oo, f(x) -> oo#

Jun 10, 2018

#lim_(x to oo) x(sqrt(underbrace(x^(2))_("humungous")+underbrace(1)\_("just 1")))#

#= lim_(x to oo) x* underbrace(sqrt(x^2))_("a positive number") #

#= lim_(x to oo) x^2 = oo#

Explanation: