Why $\lim\limits_{n \to \infty} \frac{\frac1{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}} = \lim\limits_{n \to \infty}\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1}}$? I just need clarification about answer to this question:

$$\lim_{n \to \infty} \frac{\sum\limits_{i = 1}^{n} \frac{1}{\sqrt{i}}}{\sqrt{n}} = \lim_{n \to \infty} \frac{\sum\limits_{i = 1}^{n+1} \frac{1}{\sqrt{i}} - \sum\limits_{i = 1}^{n} \frac{1}{\sqrt{i}}}{\sqrt{n+1} - \sqrt{n}} = \lim_{n \to \infty} \frac{\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}} = \lim_{n \to \infty}\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1}} = 2$$
$$\lim_{n \to \infty} \left(\frac{1}{\sqrt{n}} \sum\limits_{i = 1}^{n} \frac{1}{\sqrt{i}}\right)^k = 2^k$$


In this case I don't understand why
$$\lim_{n \to \infty} \frac{\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}} = \lim_{n \to \infty}\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1}}$$
My reasoning. $$\frac{ \frac{1}{a} }{b - c} = \frac{1}{a(b-c)}$$Hence, we get:
$$\lim_{n \to \infty} \frac{\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}}=\lim_{n \to \infty} \frac{1}{n+1 - \sqrt{n}\sqrt{n+1}} = \lim_{n \to \infty} \frac{ \frac{1}{n} }{ \frac{n}{n} + \frac{1}{n} + \sqrt{\frac{n^2}{n^2}+\frac{n}{n^2} } } = \lim_{n \to \infty}\frac{0}{1+0+\sqrt{1+0}}=0$$
While $\lim_{n \to \infty}\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1}} = \lim_{n \to \infty}\frac{ \sqrt{\frac{n}{n} + \frac{1}{n}} + \sqrt{\frac{n}{n}} }{ \sqrt{\frac{n}{n} + \frac{1}{n} } } = \lim_{n \to \infty}\frac{\sqrt{1+0} + \sqrt{1} }{\sqrt{1+0}} = 2$
Any explanation, what's going on? Thank you

We have,
$$\lim_{n \to \infty} \frac{\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}} = \lim_{n \to \infty}\left(\frac{1}{\sqrt{n+1}}\cdot  \frac{1}{\sqrt{n+1}-\sqrt{n}}\cdot \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}}\right)$$
To rationalize the denominator $\sqrt {n+1}-\sqrt n$, we need to use the conjugate based on the formula: $a^2-b^2=(a+b)(a-b)$, which gives
$$\left(\sqrt{n+1} - \sqrt{n}\right)\cdot \left(\sqrt{n+1} + \sqrt{n}\right)=n+1-n=1.$$
Hence, we have
$$
\begin{align}\lim_{n \to \infty} \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1}}=\lim_{n \to \infty} \frac{\sqrt{n+1}}{\sqrt{n+1}}+\lim_{n \to \infty} \frac{\sqrt{n}}{\sqrt{n+1}}=1+1=2.\end{align}
$$
