Difficult Sequence Convergence Show that $\lim\limits_{n \rightarrow \infty} n-n\sqrt{1-\frac{x}{n}}=\dfrac{x}{2}$. I tried different things but i don't get the point. Can someone help me?
 A: I love the method suggested by the hint, which leads to 
$$
\lim_{n\to\infty} \frac{x}{2+\frac{x}{n}}
$$
But another way is to expand the square root in the small quantity $\frac{x}{n}$, giving
$$
\lim_{n\to\infty}\left( n -(n-\frac{x}{2} + n\cdot O\left( (\frac{x}{n})^2\right) \right) = \frac{x}{2}
$$
A: It's called Generalized Binomial coefficient. From what you have, $(1-\frac{x}{n})^{\frac{1}{2}} = \sum_{j=0}^{\infty} \binom{\frac{1}{2}}{j}(-\frac{x}{n})^j$. This binomial coefficient looks daunting, but in fact you should approach it just like any other. For example, set it to $\alpha$ and start expanding $\binom{\alpha}{j} = \frac{\alpha!}{(\alpha-j)!j!}$ and plug in back $\frac{1}{2}$. You only need first three terms because the rest tend to $0$. Can you handle from here? 
A: $$\lim _{ n\rightarrow \infty  } n-n\sqrt { 1-\frac { x }{ n }  } =\lim _{ n\rightarrow \infty  } n-\sqrt { { n }^{ 2 }-nx } =\lim _{ n\rightarrow \infty  } \frac { n-\sqrt { { n }^{ 2 }-nx }  }{ n+\sqrt { { n }^{ 2 }-nx }  } \left( n+\sqrt { { n }^{ 2 }-nx }  \right) =\\ =\lim _{ n\rightarrow \infty  } \frac { { n }^{ 2 }-{ n }^{ 2 }+nx }{ n+\sqrt { { n }^{ 2 }-nx }  } =\lim _{ n\rightarrow \infty  } \frac { x }{ 1+\sqrt { 1-\frac { x }{ n }  }  } =\frac { x }{ 2 } \\ $$
A: In addition to other options provided in other answers, this can be done using L'Hopital's Rule.
Let $n=1/m$, and now you have
$$
\lim_{m\to0^+} \frac{1-\sqrt{1-xm}}m \overset{H}{=} \lim_{m\to0^+} \frac{x}{2\sqrt{1-xm}}=\frac{x}2
$$
