Does $f_n :$ $ ]0;1[ \to \Bbb R, f_n = \frac{nx}{nx+1}$ converge uniformly?

Question

Let $f_n :$ $ ]0;1[ \to \Bbb R, f_n = \frac{nx}{nx+1}$ be a sequence of functions.

Is the convergence pointwise or uniform?

It is easy to show, for $x$ fixed, that the numeric series defined by $f_n(x)$ converges to $1$. So $f_n$ converges pointwise to $f = 1$.

Now my questions is, is the convergence uniform?

I've tried to compute $$\lim \limits_{n \to \infty} \left\lVert f_n - f \right\rVert_\infty = \lim \limits_{n \to \infty} \sup_{x \in ]0, 1[} \left| \frac{1}{nx+1} \right|$$ but I don't know how to do it. What comes first, $n$ or $x$ ? If it is $n$, then the result would be $0$ but if it is $x$, the result would be $1$ (as $x$ tends to $0$). And if I should take both into account, $nx$ would be indeterminate.

Any help? Thanks.

Answer 1

Hint :

In your last calculation , at $x=\frac{1}{n}\in (0,1)$ , $\displaystyle\left|\frac{1}{nx+1}\right|=\frac{1}{2}$. So , $\displaystyle \lim_n \sup_{x\in (0,1)}\left|\frac{1}{nx+1}\right|\ge\frac{1}{2}$.

So , $\{f_n\}$ is NOT uniformly convergent in $(0,1)$.

Answer 2

You are on the right track. Note that the function $\dfrac 1 {nx + 1}$ is decreasing on $]0,1[$ (because the denominator is obviously increasing). Then its $\sup$ on this interval will be its limit towards $0$, which is $1$, which does not converge to $0$, so the convergence is not uniform.

On the other hand, had the functions been considered on $[q,1[$ with $0<q<1$, the convergence would have been uniform, because the $\sup$ on this interval would have been $\dfrac 1 {nq + 1}$, which does tend to $0$.