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

Question

5973 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-10T21:36:43

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

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#

Answer 2 · 2018-06-10T21:38:19

#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#