Question #0f6bd

In this problem, we will make repeated use of the fact that due to the continuity of #e^x#, we have #lim_(x->oo)e^f(x) = e^(lim_(x->oo)f(x))#. We will also use the properties of logarithms that

• #e^(ln(x)) = x#
• #ln(a^x) = xln(a)#

along with L'Hopital's rule, the chain rule, and the derivatives #d/dxln(x) = 1/x# and #d/dxx^n = nx^(n-1)#

Proceeding...

#lim_(x->oo)e^((1+x^2)^(1/ln(x))) = e^[lim_(x->oo)(1+x^2)^(1/ln(x))]#

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

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

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

#lim_(x->oo)ln(1+x^2)/ln(x) = lim_(x->oo)(d/dxln(1+x^2))/(d/dxln(x))#

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

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

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

#=2#

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

#:.lim_(x->oo)e^((1+x^2)^(1/ln(x))) = e^(e^2)#

First making #y = x^2# we have

#(1+x^2)^(1/lnx)equiv(1+y)^(2/lny)#

calling now #u=lny#

#(1+y)^(2/lny)equiv(1+e^u)^(2/u) = (e^(-u)+1)e^2#

but #x->oo# implies in #y->oo# and
#y->oo# implies in #u->oo# so

#lim_(x->oo)(1+x^2)^(1/lnx)equivlim_(u->oo) (e^(-u)+1)e^2=e^2#

Finally

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