How do you determine the limit of #[(n+5)/(n-1)]^n# as n approaches #oo#?

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Answer 1 · 2017-01-25T01:58:31

#lim_(n->oo) ((n+5)/(n-1))^n = e^6#

Write the expression as:

#[(n+5)/(n-1)]^n = [(n-1+6)/(n-1)]^n = (1+ 6/(n-1))^n#

Now take the logarithm:

#ln ((n+5)/(n-1))^n = n ln (1+ 6/(n-1))#

and divide and multiply by #6/(n-1)#:

#ln ((n+5)/(n-1))^n = 6(n/(n-1))( (ln (1+ 6/(n-1)))/(6/(n-1)))#

Now consider the succession: #b_n = 6/(n-1)#. Clearly:

#lim_(n->oo) b_n = 0#

#lim_(n->oo) (ln (1+ 6/(n-1)))/(6/(n-1)) = lim_(x->0) ln(1+x)/x = 1#

As:

#lim_(n->oo) n/(n-1) = 1#

as well, we have then:

#lim_(n->oo) ln ((n+5)/(n-1))^n = 6#

Now, using the continuity of the exponential function:

#lim_(n->oo) ((n+5)/(n-1))^n = lim_(n->oo) e^(ln ((n+5)/(n-1))^n) = e^(lim_(n->oo) ln ((n+5)/(n-1))^n) = e^6#