Limits do exist

The limit is: $\lim_{x\to\infty} \left[x \ln \left(x \sin^2 \left(\frac{1}{\sqrt{x}} \right)\right)\right]$

$= \lim_{x\to\infty}x\left(x \sin^2 \left(\frac{1}{\sqrt{x}} \right)-1\right)$

$(\tiny{\color{#3D99F6}{\text{since } \displaystyle\lim_{n\to \infty} \dfrac{\ln \left(f(n)-1+1\right)}{f(n)-1}\times(f(n)-1)=f(n)-1 ~~\because \displaystyle\lim_{n\to \infty} f(n)=1}})$ $=\lim_{x\to\infty} \dfrac x 2\left(x \left\{1-\cos \left(\frac{2}{\sqrt{x}} \right)\right\}-2\right)$ Now using Maclaurin series for $\cos$ :

$\left(\small{\cos\left(\frac{2}{\sqrt x}\right)=1-\frac{2}{x}+\frac{2}{3x^2}-\cdots}\right)$

We get the value of limit as $\dfrac{\color{#D61F06}{-}1}{3}\approx \color{#D61F06}{-}0.33$ .

Its really a nice problem loved working on it. initially the limit was seemingly non existent to me .

Well to begin with i love limits approaching zero hence it was quite natural for me to put x = 1/x^2

Doing so the limit becomes

2(lnsinx-lnx)/x^2

Or 2(lnsinx/x)/x^2

Its a zero/zero form using L Hospital's Rule and writing sinx terms as x coz x is approaching zero

the limit becomes

xcosx-sinx/x^3 now use Maclaurin's series of sinx and cosx to get the limit as -1/3

Or for shortening you could have also used maclaurin's series directly of lnsinx/x