Show that (n+1)/(n^2+1) converges to 0

I have a question about finding the limit point and I have an answer to this question as well but I don't understand it, this is the answer they gave:

lim(n->infinity) (n+1)/(n^2+1) = lim (1/n+1/(n^2)) / (1+1/(n^2))= (0+0)/(1+0)=0 therefore (n+1)/(n^2+1) converges to 0.

I understand the part where lim (1/n+1/(n^2)) / (1+1/(n^2))= (0+0)/(1+0)=0

but i don't understand how to get lim (1/n+1/(n^2))/(1+1/(n^2)) from lim (n+1)/(n^2+1)

please explain what they did to lim (n+1)/(n^2+1) to get lim (1/n+1/(n^2))/(1+1/(n^2))

Thank you!! :)

On problems like this, we multiply by the reciprocal of the highest power in the denominator (n^2) for both the numerator and denominator, so we are multiplying by 1 and not changing the value of the limit.

Here, we are multiplying by (1/n^2)/(1/n^2) = 1. When we distribute in the numerator, we get n * 1/n^2 + 1 * 1/n^2 = 1/n +1/n^2, and when we distribute in the denominator, we get n^2 * 1/n^2 + 1 * 1/n^2 = 1 + 1/n^2.

Divide both the numerator and denominator across by n^2.