Meaning of the $\{1 + o(1)\}$

Question

Being a software developer, I have the basic understanding of big-O and small-o notation. But currently I've faced set of mathematical problems, where they operate with asymptotics on much more advanced level.

Getting to the point: What intuition should I bear behind the following notation? $$\{1 + o(1)\}$$ Just $o(1)$ is pretty straight for me - it's such function $f(n)$ , so $\lim_{n->\infty}{f(n)} = 0$. But this $1+$ messes everything.
I've worked through the couple of sections in asymptotics book, and found a simple problem which contains my dire enemy:
Check if following is true: $(1+o(1)) \cdot ch(x) - (1+o(1)) \cdot sh(x) = (1+o(1)) \cdot e^{-x}$
They recommend to substitute right part with $o(e^{x})$ and using that prove expression false.
But I'm not getting right answer as well as I'm not getting why insert this $(1 +o(1))$ structures in equation, since $o(1)$ members asymptotically dominated.
My solution
$$\lim_{n->\infty} {\frac {(1+o(1)) \cdot {\frac {e^x + e^{-x}} {2}} - (1+o(1)) \cdot {\frac {e^x - e^{-x}} {2}}} {e^x}} = 0$$
$$\lim_{n->\infty} {\frac {(1+o(1))e^{-x}} {e^x}} = 0$$
$$\lim_{n->\infty} {(1+o(1))e^{-2x}} = 0$$
$$\lim_{n->\infty} {e^{-2x}} = 0$$
So equation turned out to be true, while answers section suggests otherwise.
Honestly, I even don't see the difference between $(1+o(1)) \cdot e^{-x}$ and $e^{-x}$.
I would appreciate explanations and corrections.
Getting better in this field became essential for me.
Thanks!

I suppose that the second $\cosh$ is $\sinh$ – Claude Leibovici Oct 18 '14 at 14:48 — Claude Leibovici, Oct 18 '14 at 14:48
@ClaudeLeibovici Yep, thank you. I mistyped. Corrected. – Dmitri K Oct 18 '14 at 14:56 — Dmitri K, Oct 18 '14 at 14:56
This happens to every one ! Cheers :-) – Claude Leibovici Oct 18 '14 at 14:57 — Claude Leibovici, Oct 18 '14 at 14:57

Did · Accepted Answer · 2014-10-18T15:16:31.303

The notation $o(1)$ refers to any unspecified term going to $0$ hence you are asked to show that every $$(1+a(x))\cosh x-(1+b(x))\sinh(x)),$$ such that $a(x)\to0$ and $b(x)\to0$ can be written as $$(1+c(x))\mathrm e^{-x},$$ for some $c(x)\to0$. Since $\cosh x-\sinh x=\mathrm e^{-x}$, this is asking whether $$a(x)\cosh x-b(x)\sinh(x)=c(x)\mathrm e^{-x},$$ for some $c(x)\to0$, that is, whether $$\frac{a(x)\cosh x-b(x)\sinh(x)}{\mathrm e^{-x}}\to0.$$ But counterexamples to that are legion...

...Say, $a(x)=1/\cosh(x)$, $b(x)=0$, then the ratio is $\mathrm e^x$, which does not go to $0$.

Meaning of the $\{1 + o(1)\}$

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