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

Question

Emily Ammerman · Accepted Answer

#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#

so that :

#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#

Charles Anctil · Answer

undefined To determine the limit of [(n+5)/(n-1)]^n as n approaches infinity, we can use the concept of limits. By simplifying the expression, we get [(1+5/n)/(1-1/n)]^n. As n approaches infinity, both 5/n and 1/n approach zero. Therefore, the expression simplifies to (1/1)^n, which is equal to 1. Hence, the limit of [(n+5)/(n-1)]^n as n approaches infinity is 1.