I'm learning about sequences of functions and need some help with this problem:
Investigate pointwise and uniform convergence of the sequence of functions $$f_n(x) = \frac{nx^2}{1 + nx}, x \in [0, 1].$$
My work and thoughts:
If $x = 0$ then $\lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} f_n(0) = \frac{0}{1 + 0} = 0$.
If $x \in (0, 1]$ (i.e. $x \neq 0)$ then $\lim_{n \to \infty} f_n(x) = \frac{\infty}{\infty}$ (undetermined).
Applying L'Hôpital's rule we get $\lim_{n \to \infty} \frac{(nx^2)'}{(1 + nx)'} = \lim_{n \to \infty} \frac{2nx}{n} = 2x$.
However, if instead of using L'Hôpital's rule, I divide numerator and denominator by $n$ I get $$\lim_{n \to \infty} \frac{\frac{nx^2}{n}}{\frac{1}{n} + \frac{nx}{n}} = x.$$
What am I doing wrong here and what about uniform convergence of $f_n(x)$ on the given interval.
The problem is that you're using L'hospitals rule incorrectly. The limit is in terms of $n$ not $x$ so you'd have to differentiate in terms of $n$ not $x$.
$$\lim_{n\to \infty}f_n(x)=\lim_{n\to \infty}\frac{d}{dn}f_n(x)=\frac{x^2}{x}=x$$
Let us try prove uniform convergence using the $\epsilon$ definition. Fix an $\epsilon >0$
$$|f_n(x)-f(x)|=\frac{|nx^2-x(1+nx)|}{|1+nx|}=\frac{x}{1+nx}\leq \frac{1}{1+n} (for \ all \ x \ in\ the\ interval)$$
Now pick $K(\epsilon)=\frac{1-\epsilon}{\epsilon}$. Then for all $n>K(\epsilon)$ $$|f_n(x)-f(x)|<\frac{1}{1+\frac{1-\epsilon}{\epsilon}}=\epsilon$$
So by definition $\{f_n\}$ is uniformly convergent on the interval.