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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!

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    $\begingroup$ is $log(x)$ here natural logarith or base $10$? $\endgroup$ – haqnatural Mar 27 '17 at 20:35
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    $\begingroup$ It's natural logarithm $\endgroup$ – blackhole1511 Mar 27 '17 at 20:37
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    $\begingroup$ Is a series expansion permitted here? $\endgroup$ – Mark Viola Mar 27 '17 at 21:03
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    $\begingroup$ Define your "notable cases" . $\endgroup$ – DonAntonio Mar 27 '17 at 21:05
  • $\begingroup$ @DonAntonio , by notable cases I mean the following: $\lim_{x \to 0} \frac{e^x-1}{x}=1$ $\lim_{x \to 0} \frac{\log{x+1}}{x}=1$ $\lim_{x \to \infty} \frac{\log{x}}{x}=0$ $\lim_{x \to \infty} (1+\frac{1}{x})^x=e$ $\endgroup$ – blackhole1511 Mar 27 '17 at 21:10
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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

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

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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 $1$.

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  • $\begingroup$ That's an interesting solution, thanks! Yet, I was searching for something with simpler pre-calculus arguments... here, we must keep in mind a Taylor expansion and this was for pre-undergraduate students. $\endgroup$ – blackhole1511 Mar 27 '17 at 21:40
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

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