a couple of fun limits...since the site is slow:)

[galactus]

Here are two fun 'challenge problems' if you wanna have a go. Chrisr?. BigGlenn?. Soroban?. or whomever. Right up your alley.

#1: Find the value of 'a' so that the following equation is true...WITHOUT L'HOPITAL.

\(\displaystyle \lim_{x\to \infty}\left(\frac{x+a}{x-a}\right)^{x}=e\)

#2: \(\displaystyle \lim_{n\to \infty}\frac{\sqrt[n]{(n+1)(n+2)\cdot \cdot \cdot (n+n)}}{n}\)

Once it is simplified, think about a Riemann sum.

 

[chrisr]

\(\displaystyle The\ value\ of\ 'a'\ wouldn't\ be\ 0.5\ for\ the\ first\ one\ by\ any\ chance?\)

 

[galactus]

Yep, it is Good job.

\(\displaystyle \left(\frac{x+a}{x-a}\right)^{x}=\left(1+\frac{2a}{x-a}\right)^{x}\)

 

[daon]

I don't have time for a proof, but if someone wants to verify, the second is asymptotic via

\(\displaystyle (n+1)(n+2)\dots (n+n) \approx \frac{1}{n^n}\cdot \left(\sum_{k=1}^n n+k \right)^n = \left(\frac{3n+1}{2}\right)^n\)

i.e. the original will have the same limit as

\(\displaystyle \frac{1}{n}\left(\frac{3n+1}{2}\right) \to \frac{3}{2}\)

 

[galactus]

#2: \(\displaystyle \lim_{n\to \infty}\frac{\sqrt[n]{(n+1)(n+2)\cdot \cdot \cdot (n+n)}}{n}\)

\(\displaystyle ln\left[\frac{\sqrt[n]{(n+1)(n+2)...(n+n)}}{n}\right]=\frac{1}{n}\left[ln(n+1)+ln(n+2)+....+ln(n+n)\right]-ln(n)\)

\(\displaystyle =\frac{1}{n}\left[ln(n(1+\frac{1}{n}))+ln(n(1+\frac{2}{n}))+....+ln(n(1+\frac{n}{n}))\right]-ln(n)\)

\(\displaystyle \frac{1}{n}\left[ln(1+\frac{1}{n})+ln(1+\frac{2}{n})+....+ln(1+\frac{n}{n})\right]+\frac{1}{n}\left[\underbrace{ln(n)+ln(n)+...+ln(n)}_{\text{n terms}}\right]-ln(n)\)

\(\displaystyle =\frac{1}{n}\left[ln(1+\frac{1}{n})+ln(1+\frac{2}{n})+....+ln(1+\frac{n}{n})\right]\)

If you notice, this is a Riemann sum:

\(\displaystyle \int_{0}^{1}ln(1+x)dx=\int_{1}^{2}ln(x)dx=2ln(2)-1\)

\(\displaystyle e^{2ln(2)-1}=\boxed{\frac{4}{e}}\)