We are tempted to substitute and say that #7/x# goes to zero so it is #1# to the infinity that is #1#. But this is wrong because it is enough that the quantity in the parenthesis is slightly different from #1# and the big exponent does the rest.

So it is better to transform it in something that remove the #x# from the exponent. The best trick is to use the logarithm. So I study the limit of the logarithm and in the end I will do the exponential of the solution.

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

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

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

Now the limit is in the form of #0/0# and I can apply the rule of Hôpital evaluating the derivative:

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

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

#=lim_(x->oo) (7/((x-7)x))/(-x^-2)#

#=lim_(x->oo) -(7x^2)/((x-7)x)#

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

This form is still #0/0# so I re-apply Hôpital's rule

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

#=lim_(x->oo) -(d/dx7x)/(d/dx(x-7))#

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

This is the limit of the logarithm, so the final result is obtained doing the exponential of this:

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