How do you evaluate the limit #(1+1/sqrtx)^x# as x approaches #oo#?

3 Answers
Sep 10, 2016

#(1+1/sqrt(x))^x -> oo# as #x->oo#

Explanation:

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

#=lim_(x->oo)e^(xln(1+1/sqrt(x)))#

#=e^(lim_(x->oo)xln(1+1/sqrt(x)))" (*)"#

(The previous step is true as #e^x# is a continuous function)

#lim_(x->oo)xln(1+1/sqrt(x)) = lim_(x->oo)ln(1+1/sqrt(x))/(1/x)#

#=lim_(x->oo)(d/dxln(1+1/sqrt(x)))/(d/dx1/x)# (by L'Hopital's rule)

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

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

#=oo#

Substituting this into #"(*)"#:

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

Thus #(1+1/sqrt(x))^x -> oo# as #x->oo#

Sep 10, 2016

#oo#

Explanation:

Another approach using Bernoulli's compounding formula ie

#\lim_{n to oo } (1+1/n)^{n} = e#

So here:
#lim_(x to oo) (1+1/sqrtx)^x#

#= lim_(x to oo) ( (1+1/sqrtx)^(sqrt x))^(sqrt x)#

with #u = sqrt x#

#= lim_(u to oo) ( (1+1/u)^u)^u#

#= lim_(u to oo) e^u = oo#

Sep 10, 2016

#oo#

Explanation:

Using the binomial expansion

#(1+u)^r = sum_(k=0)^oo ((r),(k))u^k#

making #y = sqrt(x)#

#z = (1+1/y)^(y^2) = 1+y^2(1/y)+(y^2(y^2-1))/(1 xx 2)(1/y)^2+cdots#

#z = 1 + y + f(y)#

#lim_(y->oo)z = oo# because #lim_(y->oo)f(y) ge 0#