I have this sequence of functions $$f_{n}(x)=\frac{\ln(nx)}{\sqrt{nx}}$$ with $x\in (0,\infty)$ and $n\geq 1$.
I want to check if it converges pointwise or uniformly.
I know I have to find a function for the first part, that it will be the limit of my sequence, and if it's not continuous this will show that it does converge pointwise but not uniformly. But I don't know how to find the limit of it.
Any help woudl be appreciated.
For $x>0$, by Hopital, $$\lim_{n\to\infty}\frac{\ln(nx)}{\sqrt{nx}}=\lim_{t\to\infty}\frac{\ln(t)}{\sqrt{t}}=\lim_{t\to\infty}\frac{1/t}{1/(2\sqrt{t})}=\lim_{t\to\infty}\frac{2}{\sqrt{t}}=0$$ which implies that the pointwise limit in $(0,+\infty)$ is $f=0$. Note that $f$ is continuous.
Now $$\sup_{x\in (0,+\infty)}|f_n(x)-f(x)|=\sup_{t\in (0,+\infty)}\left|\frac{\ln(t)}{\sqrt{t}}\right|\geq \lim_{t\to 0^+}\left|\frac{\ln(t)}{\sqrt{t}}\right|=+\infty$$ Which implies that $f_n$ does not converge uniformly in $(0,+\infty)$.
On the other hand $f_n$ converges uniformly in $[a,+\infty)$ with $a>0$. In fact, as $n$ goes to infinity, for $na>e^2$, $\frac{\ln(t)}{\sqrt{t}}$ is positive and decreasing (check the derivative) and $$\sup_{x\in [a,+\infty)}|f_n(x)-f(x)|=\sup_{t\in [na,+\infty)}\left|\frac{\ln(t)}{\sqrt{t}}\right|=\frac{\ln(na)}{\sqrt{na}} \to 0.$$