Spot mistake in finding $\lim \limits_{x\to1}\left(\frac x {x-1} - \frac1 {\log(x)} \right)$

Question

This is the limit I'm trying to solve: $\lim \limits_{x\to1}\left(\frac x {x-1} - \frac1 {\log(x)} \right)$

I thought: let's define $x=k+1$, so that $k\to0$ as $x\to1$.

Then it becomes: $$\lim \limits_{k\to0}\left(\frac {k+1} {k} - \frac1 {\log(k+1)} \right)$$ and then, $$\lim \limits_{k\to0}\left(\frac {k+1} {k} - \frac1 {\frac {\log(k+1)\times k}k} \right)=\lim \limits_{k\to0}\left(\frac {k+1} {k} - \frac1 {k} \right).$$ Which results in $\frac k k$ , that should be 1, but wolfram says it's $\frac 1 2$...

Did I do something illegal?

Answer 1

Yes, the illegal part is this step: $$\lim_{k \to 0}\frac1{\frac{\log(k + 1)}k}\frac1k = \lim_{k \to 0}\frac1k$$

I see that you applied the known limit $$\lim_{t \to 0}\frac{\log(1 + t)}t = 1$$ but the fact is that $$\lim_{x \to \alpha}f(x)g(x) = \lim_{x \to \alpha}f(x)\times\lim_{x \to \alpha}g(x)$$ is only valid when both limit are finite. In your case you're left with $\lim\limits_{k \to 0}\frac1k$, which not only is not finite, but does not exist entirely.

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If you are looking for a way to evaluate your limit, I'd suggest MacLauring (that is, a Taylor expansion around $x =0$), which is the simplest and most elegant way. But since you said that you can't use Taylor yet, I fear your only possibility is going with L'Hospital.

Answer 2

Hint

When $k$ is small, you have by Taylor expansion $$\log(1+k)= k-\frac{k^2}{2}+O\left(k^3\right)$$ and long division gives $$\frac{1}{\log(1+k)}=\frac{1}{k}+\frac{1}{2}+O\left(k^1\right)$$

I am sure that you can take from here.

Answer 3

Combining the two fractions, you have $$\lim _{x\to 1} \frac {x\log (x)-x+1}{(x-1)\log (x)}$$

This goes to $\frac 0 0$, so you can use L'hospitals. Taking the derivative of the top and bottom, you get $$\lim _{x\to 1}\frac {\log (x)+1-1}{\log (x)+(x-1)x^{-1}}$$ Simplifying and multipling top and bottom by x, we get $$\lim _{x\to 1}\frac {x\log (x)}{x\log (x)+x-1}$$ Again we have $\frac 0 0$, so more L'hospitals and we get $$\lim _{x\to 1}\frac {\log (x)+1}{\log (x)+1+1}$$ Now we can evaluate, to $\frac 1 2$