# Evaluating Limit Without L'Hopital

Can anybody solve this limit without using L'Hopital? $$\lim_{x \to 1} \left(\frac{x}{x-1}-\frac{1}{\log(x)} \right)$$ 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!

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

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

• – lab bhattacharjee Mar 28 '17 at 1:40
• @labbhattacharjee Interesting to see a host of approaches. – Mark Viola Mar 28 '17 at 2:16
• @JackD'Aurizio Thank you sir! Much appreciated. – Mark Viola Mar 28 '17 at 2:21

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 $1$.

• 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. – blackhole1511 Mar 27 '17 at 21:40
