Limit as n approaches infinity involving roots $$\lim_{n\to\infty}\frac{n}{1+2\sqrt{n}}$$
Given my understanding of how to solve these problems, I need to take the highest power of $n$ in the denominator and then divide both the numerator and denominator by that power. But the highest power in the denominator is lower than the one in the numerator, so I will still be left with $n^{1/2}$ in the numerator, which gives $\frac{\infty}{2}=\infty$. Is that the correct answer or am I missing something? I would have thought that since the power of the denominator is smaller than the numerator that you could factor it by long division, but I'm not sure I know how to do that.
 A: It's correct, but I'd avoid writing things like $\dfrac{\infty}{2}=\infty$ for anything other than intuition.
You could use long division if you really wanted to:
$$\dfrac{n}{1+2\sqrt{n}} = \dfrac{1}{2}\sqrt{n} + \dfrac{1}{4} - \frac{1}{4} \cdot \dfrac{1}{1+2\sqrt{n}}$$
This clearly gives you the same answer.
A: $\forall$  $M >0$ exist $n_0$  such that $\forall n >n_0$ $|\frac{n}{1+2\sqrt n}|>|\frac{ n}{4\sqrt n}|>|\sqrt n|>\sqrt {n_0}>\sqrt{M}$ so$ |\frac{n}{1+2\sqrt n}|$ will go to infinite for n enough large
A: Yes, it is correct. Another way to demonstrate it is to multiply the numerator and denominator by the conjugate:
$$\frac{n}{1+\sqrt{n}}=\frac{n}{1+\sqrt{n}}\frac{1-\sqrt{n}}{1-\sqrt{n}}$$
Doing the multiplication, the ratio becomes
$$\frac{n-n\sqrt{n}}{1-n}=\frac{1-\sqrt{n}}{-1+\frac{1}{n}}$$
The denominator converges to -1, while the numerator diverges to $\infty$.
A: Informally, as $n$ gets really large, the $1$ in $1+2\sqrt n$ becomes negligible. So, as $n \rightarrow \infty$
$$\frac{n}{1+2\sqrt n} 
   \approx \frac{n}{2\sqrt n} 
    = \frac 12 \sqrt n  
    \rightarrow \infty$$
Formally, the other answers confirm this.
