How do you find the limit of #(1-[6/x])^x# as x approaches infinity using l'hospital's rule?

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Answer 1 · 2016-04-21T12:25:43

#lim_(x->oo)(1-6/x)^x= 1/e^6#

#lim_(x->oo)(1-6/x)^x = lim_(x->oo)e^ln((1-6/x)^x)#

#=lim_(x->oo)e^(xln(1-6/x))#

#=e^(lim_(x->oo)xln(1-6/x))" (*)"#

(Note that the above step holds as #e^x# is a continuous function)

#lim_(x->oo)xln(1-6/x) = lim_(x->0)ln(1-6x)/x#

As the above is a #0/0# indeterminate form , we can apply L'hopital's rule.

#lim_(x->0)ln(1-6x)/x = lim_(x->0)(d/dxln(1-6x))/(d/dxx)#

#=lim_(x->0)(-6/(1-6x))/1#

#=-6#

Substituting this back in for #lim_(x->oo)xln(1-6/x)# in #"(*)"# gives us

#lim_(x->oo)(1-6/x)^x = e^(-6) = 1/e^6#