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by FS Aid Quoted from 8 – Administrative Assistance, Email Request, 5-30 Administrative Requirements, Boarding Schools, 2-99, 2-166. Do you know where the girls are?
– Not.
Do you remember where they were yesterday?
Did you see them yesterday?
Have you ever seen them anywhere else?
– Why couldn’t you
by FS Aid Â· Quoted at 16 – Selection Interview, 5-6 – Have you ever been in another city in the last six months?
Have you ever traveled to another city in the last six months?
Have you ever traveled to another city since you’ve been here?
– No.

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.1.4 Title 15. Title IV, Part A, Sections 4007 to 4010, is effective as a statutory prohibition.Q:

Proving: $\lim_{n\to \infty} (1- \frac1n)^n=0$

I want to show that: $$\lim_{n\to \infty} (1- \frac1n)^n=0$$
I started to solve it this way:

I see the variable $x$ can be replaced by $\frac1n$ but when I try to:

I am stuck on what to do after $ln(1-x)$… I know that $ln$ is a logarithm, a function with many properties, but it is not a quotient, so:

What should I do next to reach the answer?
EDIT: I wanted to show that $\lim_{n\to \infty} (1- \frac1n)^n=0$

Hint
Note that
$$
\left( 1- \frac 1n \right)^n = \left( 1 – \frac 1n \right)^{\frac{n(n-1)}{2}}
$$
So
$$
\left( 1 – \frac 1n \right)^n = e^{ -1} \to 0
$$

First we know that: $\sum_{n=1}^\infty\frac1{n^k}=\frac1{k+1}\to0$ as $k\to\infty$, by comparison test. Now we know: $\frac{1}{n^k}=n^{\left(-k\right)}$ and $\left(1-n^{ -k}\right)^n=e^{n\log\left(1-n^{ -k}\right)}$. So we need to show that:
$\lim_{n\to\infty}e^{n\log\left(1-n^{ -k}\right)}\to0$. And: $\log\left(1-n^{ -k}\right)=n^{ -k}-\frac{n^{ -k}}{2}+…$ So:
c6a93da74d

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