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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?

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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.

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  • $\begingroup$ ok that's awesome but I cant couple of things : what happend to t^2 and o(t) in e^(-2t+t^2+o(t))? can I create a Taylor series around 0 when i have o(t) in the power? $\endgroup$ – misha312 Aug 1 '17 at 19:03
  • $\begingroup$ o(t) means "things negligible compared to t." Since $t^2$ is negligible compared to $t$ when $t\to 0$, it is "swallowed into" the "$o(t)$. $\endgroup$ – Clement C. Aug 1 '17 at 19:06
  • $\begingroup$ As for the second question -- yes, why not? Remember the definition of $o(t)$ (can you write it to see what it means?) @misha312 $\endgroup$ – Clement C. Aug 1 '17 at 19:08
  • $\begingroup$ well to be honest I don't quite know the definition... but I think now I finally starting to understand we can "remove" all the expressions that are going faster to zero than the element we reached (for example if we reached x^3 we can ignore all power greater than 3) $\endgroup$ – misha312 Aug 1 '17 at 19:18
  • $\begingroup$ @misha312 yes, that's the idea. As for the definition: o(t) means "something (a function) that you can write g(t)*t, where g(t) goes to 0 when $t\to0$." $\endgroup$ – Clement C. Aug 1 '17 at 19:20

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