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

Question

Luke Alvarez · Accepted Answer

#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#

Samantha Adkison · Answer

undefined To find the limit of (1-[6/x])^x as x approaches infinity using L'Hospital's rule, we can rewrite the expression as e^(x * ln(1-[6/x])). Taking the natural logarithm of (1-[6/x]) and applying L'Hospital's rule, we get the limit of ln(1-[6/x]) as x approaches infinity. Differentiating the numerator and denominator, we have (-6/x^2) / (-6/x^2). Simplifying this expression, we get 1. Therefore, the limit of ln(1-[6/x]) as x approaches infinity is 1. Finally, taking the exponential of both sides, we find that the limit of (1-[6/x])^x as x approaches infinity is e^1, which is equal to e.