How to find the limit of this function We have the function $$\dfrac{\sqrt{n^4 + 100}}{4n}$$
I think the best method is by dividing by $n$, but I have no idea what that yields, mainly because of the square root.
 A: Dividing by $n$ is a great idea: Note that $$\sqrt{n^4 + 100}\cdot\frac 1n = \sqrt{(n^4 + 100)\cdot\frac 1{n^2}} = \sqrt{n^2 + \frac {100}{n^2}}$$
So your limit is, after dividing numerator and denominator by $n$:
$$\lim_{n\to \infty} \frac{\sqrt{n^2 + \frac {100}{n^2}}}{4} \to \infty$$
A: I assume you mean the limit as $n\to \infty$.  In that case, the limit is easily seen to be infinity as follows:
The numerator scales as $\sim n^2$ for large $n$, but the denominator scales as $n$ only.
The ratio thus scales as $\sim n$.
So the limit as $n\to \infty$ will be $\infty$.
A: $\text{Note that $\sqrt{n^4+100} > \sqrt{n^4}$. Hence, we have}$
$$\dfrac{\sqrt{n^4+100}}{4n} > \dfrac{\sqrt{n^4}}{4n} = \dfrac{n^2}{4n} = \dfrac{n}4$$
$\text{I trust you can finish it from here.}$
A: You have that 
$$\frac{\sqrt{n^4 + 100}}{4n} = \frac{n^2\sqrt{1 + \frac{100}{n^4}}}{4n}=\frac{n\sqrt{1 + \frac{100}{n^4}}}{4}$$
and $\frac{\sqrt{1 + \frac{100}{n^4}}}{4}\xrightarrow[n\to\infty]{}\frac{1}{4}$.
A: Perhaps another method I overlooked, stupidly enough:
$$ \dfrac{\sqrt{n^4 + 100}}{4n} = \dfrac{n^4 + 100}{4n^5 + 400n} = \dfrac{ n^4 + 100}{4n \cdot n^4 + 400n} = \dfrac{101}{404n} = \dfrac{1}{4n}$$
I'm not sure it is correct, I played around a bit with the algebra. 
