# Showing Uniform Convergence of $f_n(x) = \frac{nx}{1+nx}$ for $x \geq 0$.

We've been going over Uniform convergence in my class, but I'm very uncertain as to how to effectively show the correct proof.

The $\lim_{n \rightarrow \infty}f(x) = 1$, understood using L'Hopital indeterminate form, which is not rigorously shown. (Help needed)

I'm having the most trouble showing that for $a > 0, \{f_n\}$ converges uniformly to $f$ on $[a, \infty)$.

• "The limn→∞f(x)=1, understood using L'Hopital indeterminate form, which is not rigorously shown. (Help needed)" Is help still needed about this step? I am asking this because the accepted answer does not deal with this part of the question. – Did Dec 1 '13 at 15:55
• @Did Even though the limit is "intuitive" and can be found using different methods, the analysis requires that I find the limit using an argument of the form, "Let epsilon > 0 ......... |f_n(x) - f(x)| < epsilon. Or something along those lines. – Neurax Dec 1 '13 at 17:26
• My question is: are you able to provide such an argument? If you are not, you might want to ask the answerer to add this to their answer. – Did Dec 1 '13 at 20:04

$$\sup_{x \in [a,\infty)} \lvert f_n(x)-f(x) \rvert=\sup_{x \in [a,\infty)} \left\lvert \frac{1}{1+nx} \right\rvert=\frac{1}{1+n a}$$ and the latter tends to $0$ as $n \to \infty$.
• @Neurax I'm sorry. In that case observe that $\frac{nx}{1+nx}=\frac{x}{1/n+x} \to \frac{x}{x}$. – user1337 Dec 1 '13 at 16:33
• @Neurax This is so because if the supremum is taken over $[0,\infty)$ the result is $1$ which doesn't tend to zero as $n \to \infty$. – user1337 Dec 1 '13 at 16:37