solve limit with exponential function using maclaurin series Its a bit long so I am sorry,
I tried to solve the following limit:
$$\lim_{x \rightarrow 1} \left(\frac{x^{x+1}(\ln x+1)-x}{1-x} + x^{1/(x-1)} \right)$$
1.
$$\left[t=1-x ; x=1-t\right]$$
2.
$$\lim_{t \rightarrow 0}(\frac{(1+(-t))^{2-t}(\ln(1-t)+1)-1+t}{t}+(1+(-t))^{\frac{1}{-t}})$$
3.
$$
 = \lim_{t \rightarrow 0}\frac{(1+(-t))^{2-t}(\ln(1-t)+1)-1+t}{t}+\lim_{t \rightarrow 0}((1+(-t))^{\frac{1}{-t}})$$
4.
$$
= \lim_{t \rightarrow 0}\frac{(1+(-t))^{2-t}(\ln(1-t)+1)-1+t}{t}+e$$
5. Using maclaurin series for ln:
$$ =\lim_{t \rightarrow 0}\frac{(1+(-t))^{2-t}(-t+O(t)+1)-1+t}{t}+e$$
6.
$$ =\lim_{t \rightarrow 0}\frac{1(-t+O(t)+1)-1+t}{t}+e$$
7.
$$ =e$$
At this point I was pretty happy! But to my deepest disappointment I realized that I made a mistake and the real answer is $e-2$.
I had went over this many times and didn't find any arithmetical mistake so I guess the problem is with my logic(which is disturbing). 
I'd like to get some advice what is wrong with this path (I know that I can solve this via l'hopital's rule but I want to practice using maclaurin) 
And another question can some one explain to me how to use the O(x) sign I mean I know that this is a reminder but what are the ground rules? when I can ignore her and what can I do if I have two maclaurin series- can I add the reminders together or subtract them?  
 A: For the last, $O(\cdot)$ question: you want to read up on Landau and big-Oh notations.

Your mistake starts at 5, but it's so far a misunderstanding of $O(\cdot)$ vs. $o(\cdot)$. Note that $\ln(1+u) = u+o(u)$ (little $o(\cdot)$, not "big" $O(\cdot)$: they have different meanings!).
Now, your actual mistake is at 6, when you replace $(1-t)^{2-t}$ by $1$. You cannot do that: the low-order terms of this quantity will matter, as we will see below. More generally, when confronted to something like that, I encourage you to rewrite it in the exponential form
$$
(1-t)^{2-t} = e^{(2-t)\ln(1-t)}
$$
and then apply the Taylor expansions of $\ln(1+u)$ and $e^u$ around $0$, in that order. (Doing so, and we will do so below, you would get $(1-t)^{2-t} = 1-2t + o(t)$, and the second term will be important: this basically changes a term $(1-t+o(t))$ to $(1-2t)(1-t+o(t)))= 1-3t+o(t).$
Let us focus on the first term, call it $A$, since the limit $e$ for the second was correct.
$$\begin{align}
A &= \frac{(1+(-t))^{2-t}(\ln(1-t)+1)-1+t}{t} \\
&= \frac{(1-t)^{2-t}(1-t+o(t))-1+t}{t}\\
&= \frac{e^{(2-t)\ln(1-t)}(1-t+o(t))-1+t}{t}
\\
&= \frac{e^{(2-t)(-t+o(t))}(1-t+o(t))-1+t}{t} \tag{Expansion for $\ln(1-u)$}\\
&= \frac{e^{-2t+o(t)}(1-t+o(t))-1+t}{t} \\
&= \frac{e^{-2t+o(t)}(1-t+o(t))-1+t}{t} \tag{as $-2t+t^2+o(t)=-2t+o(t)$}
\\
&= \frac{(1-2t+o(t))(1-t+o(t))-1+t}{t} \tag{Expansion for $e^u$}\\
&= \frac{1-3t+o(t)-1+t}{t} \tag{as $1-3t+2t^2+o(t)=1-3t+o(t)$}\\
&= \frac{-2t+o(t)}{t}\\
&= -2+o(1) \xrightarrow[t\to0]{} -2
\end{align}$$
giving the result you want.
