Show the sequence converges uniformly Here is another uniform convergence problem I have been struggling with. Show that $e^{\frac{x}{k}}$ converges uniformly on $[0,1]$.
Attempt: I see that the function sequence converges pointwise to $1$.Again, I am having difficulty finding a bound for $|e^\frac{x}{k}-1|$ for all $x \in [0,1]$, so that for any $\epsilon$ I can choose an $N$ for all $x \in[0,1]$ such that $k \geq N \implies |e^\frac{x}{k}-1|<\epsilon$ . What am I supposed to do here?
Edit: So I looked at one of the answers and I decided that this would be sufficient.
Let $\epsilon>0$. Choose $N>\frac{1}{\text{ln}(1+\epsilon)}$ then $k \geq N \implies |e^\frac{x}{k}-1|<|e^\frac{1}{k}-1|<\epsilon$ for all $x \in [0,1]$, can I get a confirmation that this is ok?
 A: The functions $f_n(x)=e^{x/n}$ are positive in $\mathbb R$ and $f_n\in\mathcal C^1(\mathbb R)$.
For $x_0\in\mathbb R$ fixed, $\lim_{n\to+\infty}f_n(x_0)=1:=f(x)$, $\forall x\in\mathbb R$. So the convergence is punctual in $\mathbb R.$
Since $|f_n(n)-1|=|e-1|\ne 0$, the convergence is not uniform in $\mathbb R$.
The functions $f_n(x)-1$ have derivative $\dfrac{1}{n}e^{x/n}$, which is always postive. So, for a fixed  positive real number $M>0$, we have
$$\sup_{x\in(-\infty,M]}|f_n(x)-f(x)|\le|f_n(M)-1|\underset{n\to+\infty}\longrightarrow 0$$
and this shows that the convergence is uniform $\forall x\in D_M:=(-\infty,M]$, $M>0.$
A: hint
For any $ k>0$, the function $$g_k: x\mapsto e^{\frac xk}-1$$
is positive and strictly increasing at $ [0,1]$.
$$M_k=\max_{x\in[0,1]}|g_k(x)|=$$
$$g_k(1)=e^{\frac 1k}-1$$
$$\lim_{k\to +\infty}M_k=1-1=0$$
So, the convergence is uniform at $ [0,1]$.
A: The definition of uniform convergence may help: you need to verify that
$$sup_{x\in[0,1]}\|e^{\frac{x}{k}}-1\| \to 0 $$ and it may worth to notice where do you get that supermum (which is maximum in this case). from there you can complete the proof
