Prove by definition $\lim_{n\to\infty} \sqrt{\frac{n}{n+1}}=1$ 
Prove by definition $\lim_{n\to\infty} \sqrt{\frac{n}{n+1}}=1$.

I got to $\left|\sqrt{\frac{n}{n+1}}-1\right|< ε$, but I don't know how to proceed.
 A: $$\begin{array}{rcl}
\left| \sqrt{\dfrac n {n+1}} - 1 \right|
&=& \left| \dfrac {\sqrt n - \sqrt{n+1}} {\sqrt{n+1}} \right| \\
&=& \left| \dfrac 1 {\sqrt{n+1} (\sqrt n + \sqrt {n+1})} \right| \\
&\le& \left| \dfrac 1 {\sqrt{n} (\sqrt n + \sqrt {n})} \right| \\
&=& \left| \dfrac 1 {2n} \right| \\
\end{array}$$
A: Let me explain Kenny Lau's answer a bit. The first step, he does
$$\left|\sqrt{\frac{n}{n+1}}-1\right|=\left|\frac{\sqrt{n}}{\sqrt{n+1}}-1\right|=\left|\frac{\sqrt{n}}{\sqrt{n+1}}-\frac{\sqrt{n+1}}{\sqrt{n+1}}\right|=\left|\frac{\sqrt{n}-\sqrt{n+1}}{\sqrt{n+1}}\right|$$
Now he proceeds to multiply both sides of the fraction with $\sqrt{n}+\sqrt{n+1}$; note that $(a-b)(a+b)=a^2-b^2$ so that the numerator becomes
$$(\sqrt{n}-\sqrt{n+1})(\sqrt{n}+\sqrt{n+1})=\sqrt{n}^2-\sqrt{n+1}^2=n-(n+1)=-1$$
so that we get
$$\left|\frac{\sqrt{n}-\sqrt{n+1}}{\sqrt{n+1}}\right|=\left|\frac{-1}{{\sqrt{n+1}(\sqrt{n}+\sqrt{n+1})}}\right|$$
He discards the $-$-sign because we're taking the absolute value anyways. Now he makes the point that $\sqrt{n}<\sqrt{n+1}$ so that
$$\sqrt{n+1}(\sqrt{n}+\sqrt{n+1})>\sqrt{n}(\sqrt{n}+\sqrt{n})=2n$$
hence,
$$\frac{1}{\sqrt{n+1}(\sqrt{n}+\sqrt{n+1})}<\frac{1}{\sqrt{n}(\sqrt{n}+\sqrt{n})}=\frac{1}{2n}$$
so in the end
$$\left|\sqrt{\frac{n}{n+1}}-1\right|<\left|\frac{1}{2n}\right|$$
the point he makes with this, is that you can always pick $n$ big enough so that it gets smaller than a certain $\epsilon$, because you can just make $\frac1{2n}$ smaller and then
$$\left|\sqrt{\frac{n}{n+1}}-1\right|$$
must be smaller too.
