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.
