Linear Approximation is not doing a good job

Question

I know that from L'Hospital that $$\lim_{x\to 1} \frac{x}{x-1}-\frac{1}{\ln x}=\frac{1}{2}$$ but if we try to approximate $\ln x$ near $x=1$ we get $\ln x \approx x-1$, so in that case the limit should be $$\lim_{x\to 1} \frac{x}{x-1}-\frac{1}{x-1}=1$$ What's wrong?

Answer 1

The idea that limits are / can be evaluated by approximation is a totally wrong/absurd/baseless idea. I see it being used in some questions here on limits and looks like a wrong trend is catching on.

In mathematics we can replace $A$ by $B$ only when $A = B$. So you can not replace $\log x$ by $x - 1$ unless $x = 1$ (but this is not allowed as $x \to 1$ implies $x \neq 1$). What you can do however is to use the identity $$\log x = (x - 1) + O((x - 1)^{2}) \tag{1}$$ and replace $\log x$ with $(x - 1) + O((x - 1)^{2})$ (since they are equal). This leads to $$\frac{1}{\log x} = \frac{1}{x - 1} + O(1)\tag{2}$$ (going from $(1)$ to $(2)$ requires a division and it needs some experience in dealing with series) and your expression simplifies to $$\frac{x}{x - 1} - \frac{1}{x - 1} - O(1)$$ and this does not help you because now you don't know how to deal with $\lim_{x \to 1}O(1)$.

Therefore we need some identity like $(1)$ but of a higher order (such identities of higher order exist for many well behaved functions thanks to Taylor's theorem). Thus we can use the following identity $$\log x = (x - 1) - \frac{(x - 1)^{2}}{2} + O((x - 1)^{3})\tag{3}$$ and proceed as in the answer by user stewbasic.

However I prefer to avoid any sort of division when applying identities like $(1)$ or $(3)$ and also use the small $o$ notation instead of big $O$ notation because the small $o$ notation is defined directly in terms of a limit whereas big $O$ notation is defined in terms of bounds and not limits. We can proceed as follows: \begin{align} L &= \lim_{x \to 1}\frac{x}{x - 1} - \frac{1}{\log x}\notag\\ &= \lim_{x \to 1}\frac{x\log x - x + 1}{(x - 1)\log x}\notag\\ &= \lim_{t \to 0}\frac{(1 + t)\log(1 + t) - t}{t\log(1 + t)}\text{ (putting }t = x - 1)\notag\\ &= \lim_{t \to 0}\frac{(1 + t)\log(1 + t) - t}{t^{2}}\cdot\frac{t}{\log(1 + t)}\notag\\ &= \lim_{t \to 0}\frac{(1 + t)\log(1 + t) - t}{t^{2}}\notag\\ &= \lim_{t \to 0}\frac{\log(1 + t) - t}{t^{2}} + \frac{\log(1 + t)}{t}\notag\\ &= 1 + \lim_{t \to 0}\frac{\log(1 + t) - t}{t^{2}}\notag\\ &= 1 + \lim_{t \to 0}\dfrac{\left(t - \dfrac{t^{2}}{2} + o(t^{2})\right) - t}{t^{2}}\notag\\ &= 1 - \frac{1}{2} = \frac{1}{2}\notag \end{align} Here we have used the standard limit $$\lim_{t \to 0}\frac{\log(1 + t)}{t} = 1$$ to get rid of the logarithm from denominator (and thereby avoid division of series expansions) and then used the identity $$\log (1 + t) = t - \frac{t^{2}}{2} + o(t^{2})\tag{4}$$ as $t \to 0$.

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Note: The equations $(1), (2), (3)$ and $(4)$ are not approximations but rather exact formulas and hence we use $=$ instead of $\approx$ in such equations. Do not ever use approximations for evaluation of limits, but rather use identities like $(1)-(4)$ on which such approximations are usually based.

Answer 2

You need higher order terms in your approximation. We have $$\begin{eqnarray*} \ln x&=&(x-1)-\frac12(x-1)^2+O((x-1)^3)\\ &=&(x-1)\left[1-\frac12(x-1)+O((x-1)^2)\right]. \end{eqnarray*}$$ Thus $$\begin{eqnarray*} \frac1{\ln x} &=&\frac1{(x-1)}\left[1+\frac12(x-1)+O((x-1)^2)\right]\\ &=&\frac1{(x-1)}+\frac12+O((x-1)). \end{eqnarray*}$$ Note that because of the division by $x-1$, the second order term became important. Now $$ \frac{x}{x-1}-\frac1{\ln x}=\frac12+O((x-1))\to\frac12 $$ as $x\to1$.

Answer 3

In the same spirit as previous answers, start defining $x=y+1$ which makes $$\frac{x}{x-1}-\frac{1}{\log (x)}=1+\frac{1}{y}-\frac{1}{\log (1+y)}$$ Now, from Taylor around $y=0$ $$\log(1+y)=y-\frac{y^2}{2}+\frac{y^3}{3}+O\left(y^4\right)$$ Now, long division to get $$\frac{1}{\log (1+y)}=\frac{1}{y}+\frac{1}{2}-\frac{y}{12}+O\left(y^2\right)$$ $$1+\frac{1}{y}-\frac{1}{\log (1+y)}=\frac{1}{2}+\frac{y}{12}+O\left(y^2\right)$$ which shows the limit and how it is approached.