Let $f_n(x)=\frac{1}{nx+1}$ and $g_n(x)=\frac{x}{nx+1}$ for $x\in (0,1)$. Show that $f_n$ doesn't converge uniformly on $(0,1)$ but that $g_n$ does. I know that it is not something difficult, but I still would like to know if my proofs are correct, please.
First of all, we remark that $\lim_{n\to\infty}f_n(x)=1$ and $\lim_{n\to\infty}g_n(x)=x$. We prove now that $f_n$ doesn't converge uniformly to $1$:
$\bullet$ First approach: By definition, to show that $f_n$ doesn't converge uniformly to $1$, we have to satisfy the following: $ \exists \epsilon>0 \ \forall N \ \exists n\ge N \ \exists x\in (0,1)$: $|\frac{1}{nx+1}-1|>\epsilon$. If we take $x=1/n$ and $\epsilon=1/3$, we satisfy the definition. Therefore, $f_n(x)$ doesn't converge uniformly to $1$ on $(0,1)$
$\bullet$ Second approach: Suppose by absurd that $f_n$ converges uniformly to $1$. Then, we have the following: $\forall \epsilon>0 \ \exists N \ \forall n\ge N \ \forall x\in(0,1)$:
$|\frac{1}{nx+1}-1|<\epsilon$
But, $|\frac{1}{nx+1}-1|=|\frac{-nx}{nx+1}|\le|\frac{nx}{nx}|=1$. Taking $\epsilon=2$ we got the contradiction. So, $f_n$ doesn't converge uniformly to $1$.
Now, we show that $g_n$ converges uniofrmly to $x$ on $(0,1)$. By definition we have to satisfy the following: $\forall \epsilon>0 \ \exists N \ \forall n\ge N \ \forall x\in(0,1)$:
$|\frac{x}{nx+1}-x|<\epsilon$.
But,
$|\frac{x}{nx+1}-x|\le|\frac{x}{nx+1}|\le|\frac{x}{nx}|\le|\frac{1}{n}|$.
If we pick $N$ such that $\frac{1}{N}< \epsilon$ we obtain that $f_n$ converges uniforomly to $x$ on $(0,1)$
Best Answer
Both your pointwise limits for $f_n$ and $g_n$ are wrong.
Hint.
For every $x\in (0,1)$, $$ \lim_{n\to\infty}\frac{1}{nx+1}=0,\quad \lim_{n\to\infty}\frac{x}{nx+1}=x\lim_{n\to\infty}\frac{1}{nx+1}=0 $$
So both $f_n$ and $g_n$ converges to $0$ pointwise on $(0,1)$.
To show that $f_n$ does not converge uniformly to zero, consider the sequence $x_n=\frac1n$ and estimate $$ |f_n(x_n)-f(x_n)| $$ It is easy to see that this quantity does not go to zero.
For $g_n$, consider the estimate $$ |\frac{x}{nx+1}|=|\frac{1}{n+\frac{1}{x}}|\le\frac1n $$