If the value of lim n- infinity (2017^n+ 2018^n)^1/n is L, then the value of L/1009 is - Maths -

Question

If the value of lim n->infinity (2017^n+ 2018^n)^1/n is L, then
the value of L/1009 is

Lovina Kansal · Accepted Answer

Dear student

limn→∞2017n+2018n1n=limn→∞e1xln2017x+2018xIf limu→bf(u)=L and limx→ag(x)=b and f(x) is continuous at x=bThen limx→af(g(x))=LSo, g(x)=1xln2017x+2018x,f(u)=euSo, limx→∞1xln2017x+2018xlimx→∞ln2018x20172018x+1xApply L'Hopital's rule , we getlimx→∞12018x20172018x+12018xln201820172018x+1+20172018xln201720182018x=limx→∞20172018xln20172018+ln201820172018x+120172018x+1=ln2018So, limu→ln(2018)eu=2018So,L=2018and L1009=20181009=2

If the value of lim n->infinity (2017^n+ 2018^n)^1/n is L, then the value of L/1009 is.