How to find limits for function $f_n = \sqrt{n}\left(\sqrt{x - \frac{1}{n}}- \sqrt{x}\right)$ if $Df \in (0,\infty)$
I think as $n \rightarrow \infty$ it would be $\infty$ and then no matter the $x$ the limit would be $\infty$. But in case when $x$ is also infinitely large, I can maybe apply L'Hospital rule? Then: $$\frac{\left(\sqrt{x - \frac{1}{n}}+\sqrt{x}\right)}{\frac{1}{{\sqrt{n}}}} = \left[ \frac{\infty}{\infty} \right] = \frac{\frac{1}{2}\left(x - \frac{1}{n}\right)^{-\frac{1}{2}}+\frac{1}{2}x^{-\frac{1}{2}}}{-\frac{1}{2}n^{-\frac{3}{2}}} = 0 \ \ [\text{as } n \rightarrow \infty]$$ Is this correct?
Ok, first I just take a derivative of $\frac{1}{n^{-\frac{1}{2}}}:$ $$\frac{1}{n^{-\frac{1}{2}}} = \frac{1}{2}\frac{n^{-\frac{3}{2}}}{n}=\frac{1}{2}n^{-\frac{5}{2}}$$
First of all, if $x >0$, then for large $n$, $\sqrt{x -\frac{1}{n}} \geqslant \sqrt{\frac{x}{2}} >0$. So your sequence will verify \begin{align} \sqrt{n}\left(\sqrt{x - \frac{1}{n}} + \sqrt{x}\right) \geqslant \sqrt{n}\left( \sqrt{\frac{x}{2}}+\sqrt{x}\right) \to_{n\to+\infty} +\infty \end{align}
But I think you miswrote the question and there is a "-" missing, and it would be \begin{align} \sqrt{n}\left(\sqrt{x - \frac{1}{n}} - \sqrt{x} \right) \end{align} maybe? In addition, I don't understand what means "$Df \in (0\infty)$": who is $f$, what is $D$ here, and is $x$ fixed or is it a variable?
Edit As you miswrote it, the thing is to look closer nd find for a quotient converging to a derivative: say $f$ is the square-root function. Then, for $n$ large enough so $x-\frac{1}{n} >0$ : \begin{align} \sqrt{n}\left(\sqrt{x -\frac{1}{n}} - \sqrt{x} \right) & = \sqrt{n}\left(f(x-\frac{1}{n}) - f(x) \right) \\ & = \dfrac{f(x-\frac{1}{n})-f(x)}{\frac{-1}{n}} \times \sqrt{n} \times (-\frac{1}{n}) \\ &= \dfrac{f(x-\frac{1}{n})-f(x)}{\frac{-1}{n}} \times \left(-\frac{1}{\sqrt{n}} \right) \end{align}
The first term in the product converges to $f'(x) = \dfrac{1}{2\sqrt{x}}$ while the second term converges to $0$. Then you can conclude