Question #c86dc

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Answer 1 · 2017-01-06T23:29:24

#a=2#

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

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

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

The last equality follows from the continuity of the function #f(x)=e^x#

Focusing on the new limit:

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

Direct substitution leads to a #0/0# indeterminate form, and so we can apply L'Hopital's rule.

#=lim_(x->oo)(d/dxln(1+1/(ax)))/(d/dx1/x)#

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

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

#=1/a#

Substituting this back into our original limit, we have

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

#=e^(1/a)#

Setting this equal to #sqrt(e) = e^(1/2)#, we get

#e^(1/a) = e^(1/2)#

#=> ln(e^(1/a))=ln(e^(1/2))#

#=>1/a=1/2#

#:.a=2#

Answer 2 · 2017-01-06T23:36:54

#a=2#

An alternative answer, using algebra and the possible definition of #e# as #e=lim_(x->oo)(1+1/x)^x#:

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

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

Let #u = ax#, and note that #u->oo# as #x->oo# for any #a>0# (note that this assumes #a>0#).

#=lim_(u->oo)((1+1/u)^u)^(1/a)#

#=(lim_(u->oo)(1+1/u)^u)^(1/a)#

The above equality follows from the continuity of the function #f(x) = x^c, c in RR#

#=e^(1/a)#

Setting this result equal to #sqrt(e) = e^(1/2)#, we get

#e^(1/a) = e^(1/2)#

#:. a = 2#