$f_n$ converges pointwise to $g$

Question

For $n \in N$, define $f_n(x) = e^{-3nx} - \frac {e^x}{n}$. The sequence $(f_n)_{n=1}^{\infty}$ converges to a function $g$ pointwise on $[0,1].$ Find $g(x)$ for all $x \in [0,1].$

When $x = 0$, $f_n(x) = 1.$

When $x \in(0,1],$ I claim that $\lim_{n \to \infty}f_n(x) = 0.$ Let $\varepsilon >0.$ Let $M = \cdot.$ For $n \ge M,$

$|e^{-3nx} - \frac {e^x}{n} - 0| =|\frac{1}{e^{3nx}}-\frac{e^x}{n}|=|\frac{n-e^{x+3nx}}{ne^{3nx}}|<\cdot <\varepsilon$.

I don't know how to proceed from here. I appreciate any hint.

Answer 1

Since, for $x>0$ we have $e^{3nx}> 3nx$ we get $e^{-3nx} \frac{1}{3nx}$, hence

$|f_n(x)| \le \frac{1}{3nx}+ \frac{e}{n}$ for $x \in (0,1]$.

This shows that $f_n(x) \to 0$ for $x \in (0,1]$ as $n \to \infty$. Furthermore:

$f_n(0)=1-\frac1n \to 1$ $n \to \infty$.

Answer 2

If you really need an $\epsilon$-proof, you can use

1) $\left|e^{-3nx}-\frac{e^x}{n}\right|\leq \frac1{e^{3nx}}+\frac{e^x}n\leq \frac1{e^{3nx}}+\frac{e}n$ and

2) $ e^y\geq \frac1{k!}y^k\text{ for all }y\geq 0\text{ and all }k\in\mathbb N $.