Prove that function is not uniformly convergent

Question

I am in a trouble of showing that the sequence of function

$$f_n(x) = \frac{e^{-nx}}{3n^2x^2+1}, \ \ n = 1, 2, \cdots$$

does not converge uniformly on $[0,\infty)$.

I tried to show that $f_n(x)$ does not point wise converge with $f(x)$ or $\sup|f_n(x)-f(x)|$ is not equal to zero as $n$ goes to infinity. But I couldn't end up with a satisfied answer. Please help.

Answer 1

The given sequence is point-wise convergent to the function $f$ defined by $f(x)=0$ for $x>0$ and $f(0)=1$ and since $f$ isn't continuous at $0$ while the functions $f_n$ are continuous then the convergence isn't uniform on $[0,+\infty)$.

Answer 2

Note that $f_n({1 \over n}) = {1 \over 4e}$, $f({1 \over n}) = 0$.

Answer 3

I think you could also prove this with contradiction. If you suppose it is uniformly convergent, then obviously the sequence $(f_n)_{n=1}^\infty$ converges and is hence a Cauchy sequence. So given $\varepsilon>0$ there must be some $N \in \Bbb{N}$ where for all $m,n \geq N$ you know $$\left|\frac{e^{-nx}}{3n^2x^2+1}-\frac{e^{-mx}}{3m^2x^2+1} \right|<\varepsilon$$ With a little algebra and a tasteful choice of $x$ (and/or $m,n$) a contradiction should pop out.