# Use l'Hospital's Rule to find the exact value of the limit. limit goes to infinity. f(x)=(1+(5/x))^(x) This is `1^oo` so we take the logrithm.

`lim_(x->oo) f(x) = e^(lim_(x->oo) ln(f(x)))`

`lim_(x->oo) ln((1+5/x)^x) = lim_(x->oo) x*ln(1+5/x) = lim_(x->oo) (ln(1+5/x))/(1/x)`   Now this is 0/0 and I can use L'Hopital's rule.

`(d)/(dx) ln(1+5/x) = 1/(1+5/x) * (d)/(dx) (1+5/x) = 1/(1+5/x) * (-5/x^2) = -5/(x^2+x)`

`(d)/(dx)(1/x) = -1/x^2`

So our limit...

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This is `1^oo` so we take the logrithm.

`lim_(x->oo) f(x) = e^(lim_(x->oo) ln(f(x)))`

`lim_(x->oo) ln((1+5/x)^x) = lim_(x->oo) x*ln(1+5/x) = lim_(x->oo) (ln(1+5/x))/(1/x)`   Now this is 0/0 and I can use L'Hopital's rule.

`(d)/(dx) ln(1+5/x) = 1/(1+5/x) * (d)/(dx) (1+5/x) = 1/(1+5/x) * (-5/x^2) = -5/(x^2+x)`

`(d)/(dx)(1/x) = -1/x^2`

So our limit is now

`= lim_(x->oo) ((-5)/(x^2+x))/((-1)/x^2) = lim_(x->oo) (5x^2)/(x^2+x)`

And I can solve this by dividing top and bottom by x^2 to get

`= lim_(x->oo) (5/(1+1/x)) = 5`