Approximation of the exponential of an operator to the second order The second-order approximation of the exponential function of a real variable $x$ is
$$e^{x}\approx 1+x+\frac{x^2}{2!}, \hspace{0.3cm}\text{for}\hspace{0.3cm}x^3\to 0$$
and we can write
$$e^{x}= 1+x+\frac{x^2}{2!}+\mathcal{O}(x^3), \hspace{0.3cm}\text{for}\hspace{0.3cm}x^3\to 0$$
The exponential of an operator $\hat{A}$ is defined as:
 $$e^{\lambda\hat{A}}=I+\lambda\hat{A}+\frac{(\lambda\hat{A})^2}{2!}++\frac{(\lambda\hat{A})^3}{3!}+\cdots$$
where $\lambda$ is some parameter (real or complex number). When we are allowed to write
$$e^{\lambda\hat{A}}\approx I+\lambda\hat{A}+\frac{(\lambda\hat{A})^2}{2!}$$
i.e. to approximate it to the second-order? We cannot just say that the operator $\hat{A}$ has to be small such that $\hat{A}^3\to 0$, because it doesn't make sense as with the real variable $x$. Is it enough to have $\lambda^3 \to 0$ ?
 A: The short answer is that you need to define a norm to measure the "size" of each operator.
In order to answer your question, we need to ask what it means to say that $e^{\hat{A} }  = I + \hat{A}  + \frac{ \hat{A}^2 }{2!} + \mathcal{O}(\hat{A}^3 )$.
This is where we need to define a norm on your space of operators. This is a function mapping from your space of operators to $[0,\infty)$. I'm going to denote the norm of an operator as $||\hat{A}||$. 
Once you've defined a norm, i.e. a way of measuring the "size" of an operator, then we can talk about things like $\mathcal{O}(\hat{A} )$.
So once you've defined (or been given) a norm $||\hat{A}||$, what you're looking for is basically the same as you have in the real variable $x$ as in your post, but you're sending the norm of the operator to zero, and saying that as this goes to zero, you want the norm of the higher order terms to go to zero faster than the lower order terms.
So the higher order terms are asymptotically negligible, and so for very small values of $||\hat{A}||$, we can ignore the higher order terms without it making too much of a difference. Formally, I'd write this as: $$||e^{\hat{A} }  - ( I + \hat{A}  + \frac{ \hat{A}^2 }{2!}) || = \mathcal{O}(||\hat{A}^3|| ),$$
which is just like how in the real variable case you can say $|e^x - (1+ x + x^2/2)| = O(x^3)$. Hopefully this helps you see the parallels between the real variable and the operator cases.
Also, it turns out that if you multiply by $\lambda$ and then send $\lambda$ to zero, this is going to be the same as sending the norm of the operator to zero. That is, sending $\lambda$ to zero sends $||\lambda \hat{A}||$ to zero. So you were on the right track with your suggestion of using a scalar multiplicative constant. :)
