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!

  • 2
    $\begingroup$ is $log(x)$ here natural logarith or base $10$? $\endgroup$ – haqnatural Mar 27 '17 at 20:35
  • 1
    $\begingroup$ It's natural logarithm $\endgroup$ – blackhole1511 Mar 27 '17 at 20:37
  • 1
    $\begingroup$ Is a series expansion permitted here? $\endgroup$ – Mark Viola Mar 27 '17 at 21:03
  • 1
    $\begingroup$ 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

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

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


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 $1$.

  • $\begingroup$ 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

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.