1

I was trying to solve a question specifically

Find $lim_{n \rightarrow \infty}$ $\sqrt{n^2+n}-n$

After one manipulation that is $lim_{n \rightarrow \infty} \sqrt{n^2+n}+n =lim_{n \rightarrow \infty} \sqrt{n^2+n-n^2}$

Which then just becomes \infty

But by using $(a+b)(a-b) = a^2+b^2$ We get

$\sqrt{n^2+n}-n=\frac{n^2+n-n^2}{\sqrt{n^2+n}+n}=\frac{n}{\sqrt{n^2+n}+n}= \frac{1}{\sqrt{1+\frac{1}{n}}+1}$

Now taking that limit makes it $\frac{1}{2}$ this was in [this answer on MSE][1]

My question is why does the limit change and how to know if the result is the desired one/correct one because it does not come naturally to me to use the difference of squares identity in this case, is there any rigorous theory on such cases or specifically limits

Edit 1: this problem was in Walter Rudin, Principles of Mathematical Analysis if that matters Regards, Aditya [1]: https://math.stackexchange.com/a/783576/838062

Sam Hoffman
  • 610
  • 9
    Where does the first manipulation come from? Note that $\sqrt{a} - b \ne \sqrt{a-b^2}$ in general. – dxiv Jan 31 '22 at 03:44
  • Ohh well it was a really silly mistake then but in general do two different manipulations which do not change the function change the limit(to a finite point or $\infty$) and is there a rigorous theory for limits? – Sam Hoffman Jan 31 '22 at 03:50
  • 6
    If the limit exists then it is unique, pretty much by definition. If you get different limits using different methods, then either you made a mistake, or the limit does not exist. – dxiv Jan 31 '22 at 03:52
  • Ok got it thanks! – Sam Hoffman Jan 31 '22 at 04:05

0 Answers0