How do you find the limit of #x^(7/x)# as x approaches infinity?

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Answer 1 · 2016-03-23T00:03:53

#lim_(x->oo)x^(7/x)=1#

First, we will use the following:

#e^ln(x) = x#
Because #e^x# is continuous on #(-oo,oo)#, we have #lim_(x->oo)e^f(x) = e^(lim_(x->oo)f(x))#

With these:

#lim_(x->oo)x^(7/x) = lim_(x->oo)e^(ln(x^(7/x)))#

#=lim_(x->oo)e^(7/xln(x))#

#=e^(lim_(x->oo)7/xln(x))#

Next, we will use L'Hopital's rule:

#lim_(x->oo)7/xln(x) = lim_(x->oo)(7ln(x))/x#

#=lim_(x->oo)(d/dx7ln(x))/(d/dxx)#

#=lim_(x->oo)(7/x)/1#

#=lim_(x->oo)7/x#

#=0#

Putting these together, we get our final result:

#lim_(x->oo)x^(7/x) = e^(lim_(x->oo)7/xln(x)) = e^0 = 1#