Limit of Sequence: "2, $\frac{3}{2}, \frac{4}{3}, \frac{5}{4}$

[NoLimits]

Hello again,

I am having trouble with a particular limit problem and would appreciate any help/pointers you can offer. The question is asking for the nth term of the sequence $$2, \frac{3}{2}, \frac{4}{3}, \frac{5}{4}$$

.. and also asks for a limit of the sequence. My immediate guess was to apply l'hopital's rule, which would mean setting n to approach infinity and using something like this:

$$lim_n→∞ \frac{n+1}{n}$$

It seems to me like it could work, however I do not understand how an actual 'limit' value can be determined from a sequence of unknown and changing numbers ('n'). What I mean is, in order to make my limit work then the nth term would have to equal infinity, would it not?

** Edit **: According to an online limit solver the limit is 1, which I can see is possible if the n values are canceled out.

 

[LCKurtz]

Hello again,

I am having trouble with a particular limit problem and would appreciate any help/pointers you can offer. The question is asking for the nth term of the sequence $$2, \frac{3}{2}, \frac{4}{3}, \frac{5}{4}$$

.. and also asks for a limit of the sequence. My immediate guess was to apply l'hopital's rule, which would mean setting n to approach infinity and using something like this:

$$lim_n→∞ \frac{n+1}{n}$$

It seems to me like it could work, however I do not understand how an actual 'limit' value can be determined from a sequence of unknown and changing numbers ('n'). What I mean is, in order to make my limit work then the nth term would have to equal infinity, would it not?

** Edit **: According to an online limit solver the limit is 1, which I can see is possible if the n values are canceled out.

You don't "cancel out" the $n$'s. You write as$$
\frac{n+1} n = \frac n n + \frac 1 n = 1 + \frac 1 n$$and take the limit as $n\to\infty$ of that.