What is the limit of #(x-ln(x^2+1))# as x approaches #oo#?

Question

11744 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-09T11:22:47

#oo#

#lim_(x to oo) x-ln(x^2+1)#

#lim_(x to oo) ln (e^x)-ln(x^2+1)#

#lim_(x to oo) ln ((e^x)/(x^2+1))#

the natural log function is continuous at this limit so...

#= ln( lim_(x to oo) (e^x)/(x^2+1))#

we can already that the exponential term will grow more quickly but we can also apply L'Hopital as it is #oo/oo# indeterminate

#= ln( lim_(x to oo) (e^x)/(2x))#

#= ln( lim_(x to oo) (e^x)/(2)) = ln (oo) = oo#