Calculate $\lim_{n\to\infty} (n - \sqrt {{n^2} - n} )$ Calculate limit: 
$$\lim_{n\to\infty} (n - \sqrt {{n^2} - n})$$
My try:
$$\lim_{n\to\infty} (n - \sqrt {{n^2} - n} ) = \lim_{n\to\infty} \left(n - \sqrt {{n^2}(1 - \frac{1}{n}} )\right) = \lim_{n\to\infty} \left(n - n\sqrt {(1 - \frac{1}{n}})\right)$$
$$\sqrt {(1 - \frac{1}{n}} ) \to 1$$
$$\lim_{n\to\infty} \left(n - n\sqrt {(1 - \frac{1}{n}})\right) = \lim_{n\to\infty} (n - n) = 0$$
Ok, it's not correct. In articles i found method:
$$\lim_{n\to\infty} \frac{{(n - \sqrt {{n^2} - n} )(n + \sqrt {{n^2} - n} )}}{{n + \sqrt {{n^2} - n} }} = \lim_{n\to\infty} \frac{n}{{n + \sqrt {{n^2} - n} }} = \lim_{n\to\infty} \frac{1}{{1 + \sqrt {1 - \frac{1}{n}}}} = \frac{1}{2}$$
Why the second method is valid, but the first one not? Whats difference?

Given $\displaystyle \lim_{n\rightarrow \infty}\left[n-\sqrt{n^2-n}\right] = \lim_{n\rightarrow \infty}\left[n-n\cdot \left(1-\frac{1}{n}\right)^{\frac{1}{2}}\right]$
Now Using $\displaystyle (1+x)^{n} = 1+nx+\frac{n(n-1)}{2}x^2+.....$
So we get $\displaystyle \lim_{n\rightarrow \infty}\left[n-n\left(1-\frac{n}{2n}+\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2n^2}+.......\infty\right)\right] = \frac{1}{2}$
