Prove that every sequence $a_n$, $n \in \mathbb{N}, a_n\neq 0$, that converges to 0 (The german word for this kind of sequence is "Nullfolge", I don't know how to translate this) satisfies the following:
$\lim_{n \to \infty} \frac{\sqrt{1+a_n}-1}{a_n}=\frac{1}{2}$
If the sequence is "positive", $a_n$ seems to be greater than$ \sqrt{1+a_n}-1$ because $\sqrt{1+a_n}$ is bigger than 1, but smaller than $1+a_n$, and subtracting 1 doesn't really change something, but I don't see if this could help me ???
Thanks for your help
Since $f(x)=\frac{1}{\sqrt{1+x}+1}$ is a continuous function in $x=0$, we obtain: $$\lim_{n \to \infty} \frac{\sqrt{1+a_n}-1}{a_n}=\lim_{n \to \infty} \frac{(\sqrt{1+a_n}-1)(\sqrt{1+a_n}+1)}{a_n(\sqrt{1+a_n}+1)}=$$ $$=\lim_{n \to \infty} \frac{a_n}{a_n(\sqrt{1+a_n}+1)}=\lim_{n \to \infty} \frac{1}{\sqrt{1+a_n}+1}=\frac{1}{\sqrt{1+\lim\limits_{n\rightarrow\infty}a_n}+1}=\frac{1}{2}$$