$\lim_{x\rightarrow 0}(\ln x^{2})^{2x}$

Question

$\lim_{x\rightarrow 0}(\ln x^{2})^{2x}$

I felt the only approach to find the limits is taking log on both sides $y=\lim_{x\rightarrow 0}(\ln x^{2})^{2x}$

$\ln y=\lim_{x\rightarrow 0}2x \ln(\ln x^{2})$

Near $0$ , $\ln(\ln x^{2})$ will be $\ln(- \infty)$ . which is undefined.

So shall i conclude that the limit does not exist ?

Answer 1

As noticed in some comments, we have that

$$(\ln (x^{2}))^{2x}$$

is not a well defined expression as $x\to 0$ therefore in some context, as introductury calculus courses, we conclude suddenly that it is meaningless to consider the limit at that point.

Anyway, as observed from other users in the comments, when $x=\frac p q$ with $p,q \in \mathbb Z$ and $q$ odd we can algebrically evaluate the expression as $x\to 0$ and, within this restriction, the limit is $1$.

Indeed we have that

$$(\ln (x^{2}))^{2x}=\left|\ln (x^{2})\right|^{2x}=e^{2x \ln\left(\left|\ln (x^{2}\right)\right|)}$$

and by $x=\pm e^{-y}$ with $y \to \infty$

$$2x \ln\left(\left|\ln (x^{2}\right)\right|)=\pm \frac{2\ln (2y)}{e^y}\to 0$$

therefore

$$(\ln (x^{2}))^{2x}\to e^0=1$$

The problem with this more general approach is that when $2x=\frac p q$ with $p,q \in \mathbb Z$ both odds we have that

$$(\ln (x^{2}))^{2x}=-\left|\ln (x^{2})\right|^{2x}=-e^{2x \ln\left(\left|\ln (x^{2}\right)\right|)}$$

and the limit is $-1$ in this case.

Therefore to conclude we can say that the limit doesn't exists even for the rational restriction for $x$ in which the expression can be well defined.

Answer 2

Using $$\big[\log ^2(x)\big]^{2 x}=\exp\Big[2x \log\Big(\big[\log ^2(x)\big] \Big]$$

For small values of $x$

$$\big[\log ^2(x)\big]^{2 x}=1+2 x \log \left(\log ^2(x)\right)+2 x^2 \log ^2\left(\log ^2(x)\right)+\cdots$$

Trying for $x=\frac 1{e^4}$ the lhs is $256^{\frac{1}{e^4}}\approx1.10690$ while the truncated expansion gives $$1+\frac{2 \log ^2(16)}{e^8}+\frac{\log (256)}{e^4}\approx 1.10672$$

The limit is $1$.

Edit

If the problem is $\big[\log (x^2)\big]^{2 x}$, then, for small $x$ $$\big[\log (x^2)\big]^{2 x}=1+2 x \log ( \log (x^2))+2 x^2 \log ^2( \log (x^2))+\cdots$$ and the limit is $1$ again.