Evaluating Limit Without L'Hopital Can anybody solve this limit without using L'Hopital?
\begin{equation}
\lim_{x \to 1} \left(\frac{x}{x-1}-\frac{1}{\log(x)} \right)
\end{equation}
It is indeed easy to solve with L'Hopital's rule but I needed a solution with only algebraic manipulation or notable cases and I cannot find it. Thanks!
 A: 
Herein, we will invoke the limit definition of the exponential function 
$$e^x=\lim_{n\to \infty}\left(1+\frac xn\right)^n \tag 1$$

Note that making the substitution $x=e^{-y}$, we have
$$\begin{align}
\lim_{x\to 1}\left(\frac{x}{x-1}-\frac{1}{\log(x)}\right)&=\lim_{y\to 0}\left(\frac{1}{y}-\frac{1}{e^y-1}\right)\\\\
&=\lim_{y\to 0}\left(\frac{e^y-1-y}{y^2\frac{e^y-1}{y}}\right)
\end{align}$$
Recall the $\lim_{y\to 0}\frac{e^y-1}{y}=1$.  Then, the problem boils down to evaluating the limit
$$\lim_{y\to 0}\left(\frac{e^y-1-y}{y^2}\right)$$

From the Binomial Theorem, we have
$$\left(1+\frac {y}{n}\right)^n-1-y=\frac{n-1}{2n}y^2+\sum_{k=3}^n\binom{n}{k}\frac{y^k}{n^k}$$
Therefore, we have
$$\frac{\left(1+\frac {y}{n}\right)^n-1-y}{y^2}=\frac{n-1}{2n}+\sum_{k=3}^n\binom{n}{k}\frac{y^{k-2}}{n^k} \tag 2$$
For $|y|<1$, we can use the following estimates for the series on the right-hand side of $(2)$: 
$$\begin{align}
\left|\sum_{k=3}^n\binom{n}{k}\frac{y^{k-2}}{n^k} \right|&\le \sum_{k=3}^\infty\frac{|y|^{k-2}}{k!}\\\\
&\le \sum_{k=3}^\infty \left(|y|\right)^{k-2}\\\\
&=\frac{|y|}{1-|y|}\tag 3
\end{align}$$
Using $(1)$ and $(3)$, taking the limit as $n\to \infty$ of both sides of $(2)$ reveals
$$\lim_{n\to \infty}\frac{\left(1+\frac {y}{n}\right)^n-1-y}{y^2}=\frac{e^{y}-1-y}{y^2}=\frac12+O(|y|)\tag 4$$
whence taking the limit as $y\to 0$ of $(4)$ yields
$$\lim_{y\to 0}\frac{e^y-1-y}{y}=\frac12$$
Finally, we have

$$\lim_{x\to 1}\left(\frac{x}{x-1}-\frac{1}{\log(x)}\right)=\frac12$$

A: By substituting $x=e^t$ we are left with
$$ \lim_{t\to 0}\left(1+\frac{1}{e^t-1}-\frac{1}{t}\right)=\lim_{t\to 0}\frac{1}{t}\left(\frac{t}{e^t-1}-1+t\right)=B_1+1=\color{red}{\frac{1}{2}}$$
due to the generating function of Bernoulli numbers. As an alternative,
$$ \frac{t}{e^{t}-1} = \frac{t}{2}\coth\left(\frac{t}{2}\right)-\frac{t}{2} = 1+g(t^2)-\frac{t}{2} $$
with $g(z)$ being an analytic function in a neighbourhood of the origin, with $g(0)=0$, leads to the same result without directly involving Bernoulli numbers: $\frac{t}{2}\coth\left(\frac{t}{2}\right)$ is an even function whose limit as $t\to 0$ equals $1$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\lim_{x \to 1}\bracks{{x \over x - 1} - {1 \over \log\pars{x}}} & =
\lim_{x \to 0}\bracks{1 + {1 \over x} - {1 \over \log\pars{1 + x}}} =
\lim_{x \to 0}\bracks{%
1 + {1 \over x} - {1 \over x - x^{2}/2 + \,\mrm{O}\pars{x^{3}}}}
\\[5mm] & =
\lim_{x \to 0}\bracks{%
1 + {1 \over x} - {1 \over x}\,{1 \over 1 - x/2 + \,\mrm{O}\pars{x^{2}}}}
\\[5mm] & =
\lim_{x \to 0}\braces{%
1 + {1 \over x} - {1 \over x}\,\bracks{1 + {x \over 2} + \,\mrm{O}\pars{x^{2}}}} =\
\bbox[10px,#ffe,border:1px dotted navy]{\ds{1 \over 2}}
\end{align}
