Question #6c442

1 Answer
Feb 21, 2017

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

Explanation:

Write the function as:

#(1+1/x)^x = (e^ln(1+1/x))^x = e^(xln(1+1/x))#

We evaluate now:

#lim_(x->oo) xln(1+1/x)#

This is in the indeterminate form #0*oo#, but we can transform it to the form #0/0# and solve it using l'Hospital's rule:

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

#lim_(x->oo) xln(1+1/x) = lim_(x->oo) (d/dx ln(1+1/x))/(d/dx (1/x))#

#lim_(x->oo) xln(1+1/x) = lim_(x->oo) (1/(1+1/x)*(-1/x^2))/(-1/x^2)#

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

Now, as #e^x# is a continuous function we have:

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

So, in conclusion:

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