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.
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 $n$:

$$\lim_{n\to \infty} \frac{\sqrt{n^2 + \frac {100}{n^2}}}{4} \to \infty$$
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.