Confusion on where my error is in my solution of $\lim_{x\to0^{+}}\frac{x}{\ln(x)}$ without the use of l'hospitals

Question

I would like to ask something with respect to the limit $$\lim_{x\to0}\frac{x}{\ln x}$$ so if I do it with a calculator I get $L=0$ but my question is why I can't do this $$\lim_{x\to0}\frac{x}{\ln x}=\left(\frac{\ln x }{x}\right)^{-1} = \frac{\ln((x-1)+1)}{x-1}\frac{(x-1)}{x}\to1\times1=1$$ by using the notable limit $\lim_{x\to0}\frac{ln(x+1)}{x}=1$

where does this logic fall down

Edit: after reading the comments i realized that i was messing everything up, and confusing myself with the limt with basic problems. Thank you

Answer 1

It really looks like your argument is trying to say: $$\frac{\log x}{x}=\frac{\log(1+(x-1))}{x-1}\cdot \frac{x-1}{x}\tag 1$$ and the two terms go to $1$ as $x\to 0.$

But the two terms don’t go to $1.$ In the first term, it is true that $$\lim_{h\to0}\frac{\log(1+h)}{h}=1.\tag2$$ But if $h=x-1,$ then as $x\to 0,$ $h\to-1.$ So $(2)$ does not apply.

Similarly, $\frac{x-1}x$ does not converge to $1.$

You might be misapplying L’Hopital. L’Hopital requires $\frac{f(x)}{g(x)}$ to converge to an “indeterminate form,” here requiring either both $f(x),g(x)\to 0$ or $f(x),g(x)\to\pm\infty.$

But $x-1\to-1$ as $x\to 0,$ so neither term qualifies for L’Hopital.

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A quick argument that the limit of $\frac{\log(x)}x$ can’t be $1$ (or positive) is that for $0<x<1,$ $\log(x)<0,$ so $\frac{\log x}x<0,$ and the limit, if it exists, can’t be positive.

Answer 2

Well, you can't apply l'Hôpital to this limit. The numerator has limit $0$ and the denominator has limit $-\infty$, hence you have $$ \lim_{x\to0}\frac{1}{\ln x}=0 $$ and your limit is $$ \lim_{x\to0}\frac{x}{\ln x}=\lim_{x\to0}x\frac{1}{\ln x}=0\cdot0=0 $$

Let's see where you went wrong. Transforming $$ \dfrac{\ln x}{x}=\frac{\ln((x-1)+1}{x-1}\frac{x-1}{x} $$ does not really simplify things, because for the first fraction you get $$ \lim_{y\to-1}\frac{\ln(1+y)}{y} $$ which is definitely not $1$, but $\infty$ (it's a limit from the right, of course). Similarly, $$ \lim_{x\to0^+}\frac{x-1}{x}=-\infty $$ so you can conclude that $$ \lim_{x\to0}\frac{\ln x}{x}=-\infty $$