What is #lim_(n -> oo) n ln((n + 1)/n)#?

1 Answer
Noah G
Dec 16, 2017

The limit equals #1#

Explanation:

Using logarithm laws:

#L = lim_(n -> oo) n(ln((n + 1)/n))#

#L = lim_(n->oo) n(ln(1 + 1/n))#

We can now do a little algebraic reworking:

#L = lim_(n-> oo) ln(1 + 1/n)/(1/n#

Now if you evaluate #n = oo# into both the numerator and the denominator, you will get #0/0#, so we can now apply l'Hospital's rule. By the chain rule, #d/(dn) ln(1 + 1/n) = -1/n^2 * 1/(1 + 1/n) = -1/(n^2 + n)#

#L = lim_(n-> oo) (-1/(n^2 + n))/(-1/n^2)#

#L = lim_(n->oo) n^2/(n^2 + n)#

#L = lim_(n-> oo) n^2/(n(n + 1))#

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

We now use partial fractions to decompose.

#n/(n + 1) = A/(n + 1) + B/1#

#n = A + B(n + 1)#

#n = A + Bn + B#

We can see instantly that #B = 1# and #A + B = 0#, therefore, #B = 1# and #A = -1#.

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

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

The first limit is clearly #0# and the second #1#.

#L = 0 + 1#

#L = 1#

If we check graphically, we can see this is evidently true.

enter image source here

Hopefully this helps!

Related questions
See all questions in Infinite Limits and Vertical Asymptotes
Impact of this question
2306 views around the world
You can reuse this answer
Creative Commons License