Limit questions

1. Oct 24, 2005

Nerpilis

find the $$\lim_{n\rightarrow\infty}(\frac{n+1}{n})^{n+1} =$$
so far what i have is
$$\lim_{n\rightarrow\infty}(\frac{n+1}{n})^{n+1} = \lim_{n\rightarrow\infty}(\frac{n}{n} + \frac{1}{n})^{n+1} = \lim_{n\rightarrow\infty}(1 + \frac{1}{n})^{n+1}\\$$

I know this has got to go to $$e$$ or something very close, my question lies in if I should be concerned with the +1 in the n+1 in the exponet.
There is another one which I'm totaly stumped on for a starting point, my gut tells me the limit is 1 or it may not even exist altogether.
$$\lim_{n\rightarrow\infty}(\frac{n+3}{n+1})^{n+4} =$$
but then again it could be +$$\infty$$ if you realize that in the () the numerator is larger than the denominator such that as the exponet is applied it goes to infinity instead od zero.

Last edited: Oct 24, 2005
2. Oct 24, 2005

Galileo

You said you already have a suspicion that the limit should have to do with e. So I assume you know the limit definition of e which has a very similar form?

The second has a very similar form. Can you rewrite it so the form resembles a limit you already know? (Like that of exp(x) or something).

3. Oct 24, 2005

HallsofIvy

Did you notice that
$$\left(1+\frac{1}{n}\right)^{n+1}$$
is the same as
$$\left(1+\frac{1}{n}\right)^n \left(1+ \frac{1}{n}\right)$$
?

4. Oct 24, 2005

Nerpilis

no I did not realize that $$\left(1+\frac{1}{n}\right)^n \left(1+ \frac{1}{n}\right)$$ I guess that would make the limit equal to e^2?

as far as the second problem I don't know what exp(x) is but i recognized that I can get things to look more like e

$$\lim_{n\rightarrow\infty}(\frac{n+3}{n+1})^{n+4} =$$

$$\lim_{n\rightarrow\infty}((\frac{n + 3}{n + 1})(\frac{1/n}{1/n}))^{n+4} =$$
$$\lim_{n\rightarrow\infty}\frac{(1 + (3/n))^{n+4}}{(1 + (1/n))^{n+4}} =$$

from here the denominator looks similar to e and the numerator almost looks like e, however I’m a little hesitant to call the limit e/(e^4) = 1/(e^3) therefore I applied bernouli’s inequality to the numerator :

$$\geq \lim_{n\rightarrow\infty}\frac{1 + (n + 4)(3/n)}{e^{4}} \geq \frac{\lim_{n\rightarrow\infty}(1 + 3 + (12/n))}{e^{4}} \geq \frac{4}{e^{4}}$$

Is my thinking correct in operating this way?

Last edited: Oct 24, 2005
5. Oct 25, 2005

Galileo

No. Why don't you just write it out and use some limit rules. This looks like the limit of a product of 2 factors, both with limits you know.

Don't just guess what you think the limit should be. What makes you think that:
$$\lim_{n\to \infty} \left(1+\frac{1}{n}\right)^{n+4}=e^4$$
? You can solve this limit in the same way as the previous one by considering it a product of 2 limits.

For a limit like: $\lim_{n\to \infty} (1+x/n)^n$ which looks alot like the one for e, what method would be useful? (Hint: substitution)

Another way is to write:
$$\frac{n+3}{n+1}=\frac{n+1+2}{n+1}=1+\frac{2}{n+1}$$

Last edited: Oct 25, 2005
6. Oct 26, 2005

Nerpilis

thank you , i have found my errors
$$\lim_{n\rightarrow\infty}(\frac{n+1}{n})^{n+1} = \lim_{n\rightarrow\infty}(\frac{n}{n} + \frac{1}{n})^{n+1} = \lim_{n\rightarrow\infty}(1 + \frac{1}{n})^{n+1}\\$$
$$\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^n \left(1+ \frac{1}{n}\right) = e \times \lim_{n\rightarrow\infty}\left(1+ \frac{1}{n}\right) = e$$
for the other one:
$$\lim_{n\rightarrow\infty}\left(\frac{n+3}{n+1}\right)^{n+4}= \lim_{n\to \infty} \left(1+\frac{2}{n+1}\right)^{n+4} = \lim_{n\rightarrow\infty}\left(1 + \frac{2}{n+1}\right)^{n+1}\left(1 + \frac{2}{n+1}\right)^{3} = \lim_{n\rightarrow\infty}\left(1 + \frac{2}{n+1}\right)^{n+1} \times 1$$

now i have found that this actually goes to $$e^{2}$$ but i did get some help but i don't know how to prove it other than writing out the sequence values for e and then for this sequence $$\lim_{n\rightarrow\infty}\left(1 + \frac{2}{n+1}\right)^{n+1}$$
and noticing that e is a subsequence of the above

Last edited: Oct 26, 2005
7. Oct 26, 2005

Galileo

You can make a substitution: m=n/x. That makes:

$$\lim_{n\to \infty} \left(1+\frac{x}{n}\right)^n$$
look like:
$$\lim_{m\to \infty}\left(1+\frac{1}{m}\right)^{xm}$$
Or m-> -infinity if x is negative.