How to solve $\lim_{x\to1}\left[\frac{x}{x-1}-\frac{1}{\ln(x)}\right]$ without using L'Hospital's rule?

Question

$$\lim_{x\to1}\left[\frac{x}{x-1}-\frac{1}{\ln(x)}\right]$$

How can I solve this without using the L'Hopital's rule? Any tips or hints would be greatly appreciated. I tried using the substitution $x=e^y$, but to no avail.

Answer 1

One may recall that, as $u \to0$, by the Taylor expansion, $$ \frac1{1+u}=1-u+u^2+O(u^3)\tag1 $$ $$ \ln(1+u)=u-\frac12 u^2+O(u^3).\tag2 $$ Set $u:=x-1$ in your initial expression, we have $u \to 0$ as $x \to 1$, and we get $$ \frac{x}{x-1}-\frac{1}{\ln(x)}=\frac{1+u}u-\frac{1}{\ln(1+u)}=1+\frac1u-\frac{1}{\ln(1+u)}. \tag3 $$ Then, using $(2)$ and $(1)$,
$$ \begin{align} \frac{1}{\ln(1+u)}&=\frac{1}{u-\frac12 u^2+O(u^3)}\\ &=\frac1u\times\frac{1}{1-u/2+O(u^2)}\\ &=\frac1u\times\left( 1+\frac{u}2+O(u^2)\right)\\ &=\frac1u+\frac12+O(u) \tag4 \end{align} $$ From $(4)$, as $u \to0$, you deduce $$ \frac1u-\frac{1}{\ln(1+u)} \to \color{red}{-}\frac12 $$ inserting in $(3)$ gives

$$\lim_{x\to1}\left[\frac{x}{x-1}-\frac{1}{\ln(x)}\right]=\frac12.$$

Answer 2

Putting $x = 1 + h$ so that $h \to 0$ we see that the desired limit can be calculated as \begin{align} L &= \lim_{h \to 0}\frac{1 + h}{h} - \frac{1}{\log(1 + h)}\notag\\ &= \lim_{h \to 0}\frac{(1 + h)\log(1 + h) - h}{h\log(1 + h)}\notag\\ &= \lim_{h \to 0}\frac{(1 + h)\log(1 + h) - h}{h^{2}}\cdot\frac{h}{\log(1 + h)}\notag\\ &= \lim_{h \to 0}\frac{\log(1 + h) - h + h\log(1 + h)}{h^{2}}\notag\\ &= \lim_{h \to 0}\frac{\log(1 + h) - h}{h^{2}} + \frac{\log(1 + h)}{h}\notag\\ &= \lim_{h \to 0}\dfrac{h - \dfrac{h^{2}}{2} + o(h^{2}) - h}{h^{2}} + 1\notag\\ &= \lim_{h \to 0}-\frac{1}{2} + o(1) + 1\notag\\ &= \frac{1}{2}\notag \end{align}

It is possible to calculate the limit of $\dfrac{\log(1 + h) - h}{h^{2}}$ without using Taylor series expansions but it requires more work. For example we can start with the inequality$$1 - t < \frac{1}{1 + t} < 1 - t + t^{2}$$ for $ t > 0$ and integrate it in interval $[0, h]$ to get $$h - \frac{h^{2}}{2} < \log (1 + h) < h - \frac{h^{2}}{2} + \frac{h^{3}}{3}$$ and therefore $$-\frac{1}{2} < \frac{\log(1 + h) - h}{h^{2}} < -\frac{1}{2} + \frac{h}{3}$$ for $h > 0$. Letting $h \to 0^{+}$ and using Squeeze theorem we get the limit as $(-1/2)$. It is possible to do the similar thing for $h \to 0^{-}$.

Answer 3

Set $x=1+h$ and develop $\ln(1+h)$ at order $2$: \begin{align*} \frac{x}{x-1}-\frac{1}{\ln(x)}&=\frac{1+h}h-\frac 1{\ln(1+h)}=\frac1h\Biggl(1+h-\frac1{1-\cfrac h2+o(h)}\Biggr)\\ &=\frac1h\Bigl(1+h-\Bigl(1+\frac h2+o(h)\Bigr)\Bigr)=\frac12+o(1)\xrightarrow[h\to0]{}\frac12. \end{align*}

Answer 4

The limit follows by direct computation from the following basic inequalities for $\log$:

For $x\geq 1$:

$$\frac{2(x-1)}{x+1} \leq \log(x) \leq \frac{x^2-1}{2x}$$

For $0<x\leq 1$:

$$\frac{2(x-1)}{x+1} \geq \log(x) \geq \frac{x^2-1}{2x}$$

These inequalities (together with the weaker $1-\tfrac{1}{x}\leq \log(x) \leq x-1$) are very useful to have in your toolbox. All these inequalities are based on well chosen estimates for the integral in $$\log(x) = \int_1^x\frac{dt}{t}.$$