# How to show $\lim\limits_{k\to\infty}\frac{1}{k-2\sqrt{k}}=0$?

Can we say that $k$ grows faster than $\sqrt{k}$ when term is large? But what is the formal way write it ?

Note that, for all positive integer $k$, $$\frac{1}{k-2\sqrt{k}}=\frac{\frac{1}{k}}{1-2\frac{\sqrt{k}}{k}}=\frac{\frac{1}{k}}{1-2\frac{1}{\sqrt{k}}}$$
Thus, if you know that $$\lim_{k\to\infty}\frac{1}{k}=0\quad\text{and}\quad\lim_{k\to\infty}\frac{1}{\sqrt{k}}=0,$$ then you can conclude that $$\lim_{k\to\infty}\frac{1}{k-2\sqrt{k}}=\frac{0}{1-2\cdot 0}=0$$
You can factor a $\sqrt{k}$ out of the bottom to get: $\lim_{k \to \infty} \frac{1}{\sqrt{k}(\sqrt{k}-2)}$. Now it should be clear that the bottom goes to infinity.
We can restrict $k > 4$, then it suffices to show that $$\lim_{k \to \infty} (k - 2\sqrt{k}) = \infty$$ Using the fact that $2\sqrt{k} \in o(k)$, $$\lim_{k \to \infty} (k - 2\sqrt{k}) = \lim_{k \to \infty} k = \infty$$
• But how do you know $\lim_{k \to \infty} (k - 2\sqrt{k}) = \infty$? – CoolKid Mar 30 '16 at 0:09