Ok. I know that the lim (n to inf) of the sequence [1+(1/n)]^n=e.
How would I find the lim (n to inf) of [1+(2/n)]^n using the above limit? I know the answer is e^2, but how would I show it?
Hi,
this one looks like a trick - but it works:
$\displaystyle \left(1+\frac{2}{n}\right)^n$ = $\displaystyle \left(1+\frac{1}{\frac{n}{2}}\right)^n$ =
$\displaystyle \left( \left(1+\frac{1}{\frac{n}{2}}\right)^\frac{n}{2}\r ight)^2$
Now substitute n/2 = y. Then you get:
$\displaystyle \left( \left(1+\frac{1}{y}\right)^y\right)^2$
$\displaystyle \lim_{y\rightarrow \infty}{\left(1+\frac{1}{y}\right)^y}=e$
Therefore $\displaystyle \left( \left(1+\frac{1}{\frac{n}{2}}\right)^\frac{n}{2}\r ight)^2=e^2$
EB
I am not going to write infinity because all limits of sequences are taken long infinite.
You have,
$\displaystyle \lim (3n)^{\frac{1}{2n}}=\lim 3^{\frac{1}{2n}}n^{\frac{1}{2n}}=\lim \sqrt{3}^{1/n}\cdot \sqrt{n}^{1/n}$
We divide the problem into two limit problems,
1)$\displaystyle \lim \sqrt{3}^{1/n}=\sqrt{3}^0=1$ (sequence composition rule).
2)$\displaystyle 1\leq \sqrt{n}^{1/n}\leq n^{1/n}$
Squeeze theorem, the limit,
$\displaystyle \lim n^{1/n}=1$ is well known.
Thus the answer is $\displaystyle 1\cdot 1 =1$