Evaluate the limit $\lim_{n\to\infty} n(\sqrt{n}-\sqrt{n+1})$

Question

Evaluate the limit $\lim_{n\to\infty} n(\sqrt{n}-\sqrt{n+1})$

I am trying to evaluate this limit. First I multiplied by the conjugate to obtain: $a_n=\dfrac{-n}{\sqrt{n}+\sqrt{n+1}}$

I was able to show the limit by taking $f(x)=a_x$ and then applying l'Hôpital's rule. I was primarily wondering if there is a certain trick that can be used to find the limit without resorting to this.

\begin{align} \lim_{n\to\infty} n(\sqrt{n}-\sqrt{n+1})=-\infty \end{align}

Thanks.

Answer 1

Or we can look at $n(\sqrt{n+1}-\sqrt{n})=\dfrac{n}{\sqrt{n+1}+\sqrt{n}}\geq\dfrac{n}{2\sqrt{n+1}}=\dfrac{\sqrt{n}}{2\sqrt{1+\dfrac{1}{n}}}\rightarrow\infty$ as $n\rightarrow\infty$, so the limit is $-\infty$.

Answer 2

Start from

$${-n\over \sqrt{n}+\sqrt{n+1}}={-\sqrt{n}\over 1+\sqrt{1+{1\over n}}}={N\over D}$$

We have $D\to 1$ and $N\to -\infty$