How do you evaluate #[ ( 1 + ( r/x ) ) ^ x ]# as x approaches infinity?

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Answer 1 · 2016-12-26T14:36:38

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

Write #f(x)# as:

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

Now we have:

#lim_(x->oo) ln(1+r/x)/(r/x)= lim_(y->0) ln(1+y)/y#

To evaluate this last limit we note that:

#lim_(y->0) ln(1+y)/y=lim_(y->0) (ln(1+y)-ln1)/(y-0)#

is by definition the value for #y=0# of the derivative of #ln(1+y)#, so that:

#lim_(y->0) ln(1+y)/y = [1/(1+y]]_(y=0) = 1#

And thus:

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