0

I'm trying to express the following function in terms of a limit L plus either big O and little o. In other words,

$$ f(h) = \frac{1+h-e^h}{h^2} $$

should be expressed in the form $f(h) = L+O(h^n) = L+o(h^m)$ as $h \to 0$.

I have evaluated the limit using L'Hopital's rule, and obtain:

$$ \lim_{h \to 0} \frac{1+h-e^h}{h^2} = -\frac{1}{2} $$

However, this does not have any remaining big O or little o term. Another option I explored is to Taylor expand $f(h)$ around $h=0$, but then $O(h^n)$ will be arbitrarily chosen based on where I choose to truncate.

I read an answer to a similar problem here, but it's not clear how the answer was reached, and doesn't say anything about a little o term.

Thanks in advance!

hopper19
  • 121
  • https://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lh%c3%b4pital-rule-or-series-expansion – lab bhattacharjee Oct 10 '19 at 09:56
  • The asymptotics depend on the slowest decaying term, not the fastest. For example $\epsilon^4 + \epsilon^5 + \epsilon^6 = O(\epsilon^4)$. – Michael Oct 10 '19 at 09:59
  • Thanks for the replies guys. I've checked out the link @labbhattacharjee sent, and it was useful for expressing it in terms of big O. I'm still unsure how to deal with the little o part of the problem. Edit: I reworded the problem to reflect that both expressions for big O and little o are required. – hopper19 Oct 10 '19 at 11:42

0 Answers0