Value of the given limit. What is $$\lim _{ x\rightarrow 1 }{ \frac { x\log { \left( x \right) -x+1 }  }{ \left( x-1 \right) \log { \left( x \right)  }  }  } $$ 
 Note I have used $\lim _{ x\rightarrow 0 }{ \frac { \log { \left( 1+x \right)  }  }{ x }  } =1$ . So I wrote $\log(x)=\log(1+x-1)$ and hence I got $\frac{x^2-2x+1}{x-1}$ after cancelling I got the value as $0$. But the correct answer is $\frac{1}{2}$. Where's my mistake? Any hint. I don't want Lhospital rule or Taylor series if they aren't compulsory to work out the answer .
 A: Now with editing this gives a solution using only the limit 
$\lim\limits_{y\to 0}\frac{e^{y}-1}{y}=1$.
Start by setting $x=e^y$,  you get 
$$\lim\limits_{y\to 0}\frac{ye^y-e^y+1}{(e^y-1)y}$$ Multiply by $\lim\limits_{y\to 0}\frac{e^y-1}{y}=1$ to get
$$\lim\limits_{y\to 0}\frac{ye^y-e^y+1}{y^2}=
\lim\limits_{y\to 0}\frac{e^y-1}{y}+\frac{1+y-e^y}{y^2}=1-L$$
Where $$L=\lim\limits_{y\to 0}\frac{e^y-1-y}{y^2}$$
Now using l'Hopital twice (while we try to think of another way) we see $L=\frac{1}{2}$.
To avoid l'Hopital 
Write
$$\lim\limits_{y\to 0}\frac{e^y-1-y}{y^2}=L$$
Then 
$$\lim\limits_{y\to 0}\frac{e^{-y}-1+y}{y^2}=L$$
adding
$$\lim\limits_{y\to 0}\frac{e^{-y}+e^y-2}{y^2}=2L$$
Now on the other hand 
$$\lim\limits_{y\to 0}\frac{e^{y}-1}{y}=1$$
and 
$$\lim\limits_{y\to 0}\frac{e^{-y}-1}{y}=-1$$
and multiplying them together
$$\lim\limits_{y\to 0}\frac{2-e^{-y}-e^y}{y^2}=-1$$
It follows that $2L=1$.
A: HINT
Put $x-1=h $
As x goes to $1$, h goes to $0$. Then use expansion for $log(1+h)$ in numerator and result you stated in your attempt in denominator 
$lim_{h \to0} \frac{(1+h)log(1+h) - h}{hlog(1+h)}$
$lim_{h \to0} \frac{(1+h)log(1+h) - h}{h^2}$
$lim_{h \to0} \frac{(1+h)(h-\frac{h^2}{2} + ...) - h}{h^2}$
A: It makes sense substituting $x=t+1$, so the limit becomes
$$
\lim_{t\to0}\frac{(1+t)\log(1+t)-t}{t\log(1+t)}
$$
With a Taylor expansion it is quite easy:
$$
\log(1+t)=t-\frac{t^2}{2}+o(t^2)
$$
so
$$
(1+t)\log(1+t)-t=
(1+t)\left(t-\frac{t^2}{2}+o(t^2)\right)-t=
\frac{t^2}{2}+o(t^2)
$$
and the denominator can be written $t\log(1+t)=t(t+o(t))=t^2+o(t^2)$.
Therefore
$$
\lim_{t\to0}\frac{(1+t)\log(1+t)-t}{t\log(1+t)}
=
\lim_{t\to0}\frac{\dfrac{t^2}{2}+o(t^2)}{t^2+o(t^2)}=\frac{1}{2}
$$

You can also use l'Hôpital, but first rewrite the limit as
$$
\lim_{t\to0}\frac{(1+t)\log(1+t)-t}{t\log(1+t)}
=
\lim_{t\to0}\frac{(1+t)\log(1+t)-t}{t^2}\,\frac{t}{\log(1+t)}
$$
The limit of the first factor becomes
$$
\lim_{t\to0}\frac{(1+t)\log(1+t)-t}{t^2}\;\overset{\mathrm{(H)}}{=}\;
\lim_{t\to0}\frac{\log(1+t)}{2t}=\frac{1}{2}
$$
The second factor has limit $1$.
A: The problem is quite easy  using Taylor and L’Hôpital  but the O.P. wishes this way  forbidden so the question becomes, at least for me, something difficult.  Anyway, here a method of approximation to solve the question (especially when it is knowing that the limit is $\frac 12$):
Let $E$ be the given expression. We have
$$E=\frac{x}{x-1}-\frac{1}{\ln(x)}$$ Put $x=t+1$ so we have $$E=\frac{t+1}{t}-\frac{1}{\ln(t+1)}=1+f(t)$$ where $$f(t)=\frac{1}{t}-\frac{1}{\ln(1+t)}$$
The domain of $f$ is $(-1,0)\cup (0,\infty)$ however if the searched limit exists then prolonging by continuity we know the function $f$ can be considered as continuous on $(-1, \infty)$. The derivative is $$f’(t)=-\frac{1}{t^2}+\frac{1}{\ln^2(1+t)}\gt 0$$ so the function is deduced to be increasing on its domain. Now, calculating some values in the neighborhood of $t=0$
$$ f(0.01)\approx  -0.499170\\ f(0.001)\approx -0.499916\\f(0.0001)\approx -0.499991\\
f(-0.01)\approx  -0.500837\\f(-0.001)\approx  -0.500083\\f(-0.0001)\approx -0.500008$$
Knowing that the limit exists, that $f$ is continuous for $x\gt -1$ and increasing, we can conclude, that this limit indeed is equal to ${\frac {-1}{2}}$.
 Thus $$\lim_{t\to 0}E=1-\frac 12=\color{red}{\frac 12}$$
