## mathivh one year ago Limit (to infinity) question

1. mathivh

So when you have: $\lim_{x \rightarrow \infty}(\frac{ n^2 }{ n^2 +2n+1})$ you can divide this by $n^2$ so you would get $\lim_{x \rightarrow \infty}(\frac{ n^2/n^2 }{ n^2/n^2+ 2n/n^2 + 1/n^2 }) = (\frac{1}{1+0+0}) = 1$ But why can't you divide $\lim_{x \rightarrow \infty}(n+1)$ by n, so that you would get:$\lim_{x \rightarrow \infty}(n/n +1/n)= 1 + 0 = 1$

2. mathivh

Obviously I meant $\lim_{n \rightarrow \infty}$ instead of $\lim_{x \rightarrow \infty}$

3. Loser66

You CAN'T just divide by n. If you want to do that, you MUST multiply by n also. Like, $$a =a *\dfrac{n}{n}$$ to not change the original problem.

4. Loser66

In previous case, you divide by n BOTH numerator and denominator; it doesn't change the original problem, right? $$\dfrac{a}{b}=\dfrac{a/c}{b/c}$$

5. mathivh

Aah ok I understand now, pretty silly of me to have overseen that. Thanks for the help @Loser66 !

6. Loser66

np