Question

what is the lim of (8n)/((3n^2)+5)?

Answer 1

$\lim \limits_{n \to \infty} ~\dfrac{8n}{3n^2 +5} \\\\\\=8\lim \limits_{n \to \infty} ~ \dfrac{\tfrac n{n^2}}{\tfrac{3n^2}{n^2} + \tfrac 5{n^2}}\\\\\\=8 \lim \limits_{n \to \infty} ~\dfrac{ \tfrac 1n}{ 3 + \tfrac 5{n^2}}}\\\\\\=8 \cdot \dfrac{ 0}{3 +0} = 8 \cdot 0 = 0$

Answer 2

We want to evaluate the limit of $\frac{8n}{3n^{2}+5}$ as n tends toward infinity (+∞)

The limit of this expression is the same as the limit of $\frac{8n}{3n^{2}}$

But this last expression can be simplified, we can remove one n up and down which gives us : $\frac{8}{3n}$

We know the limit of this one, it is simply Zero 0, by definition of the class lesson.

So to conclude the limit of $\frac{8n}{3n^{2}+5}$ as n tends toward infinity is 0.

Good Luck