Is the limit of this sequence of function $1$ or $0$?

Question

I was given to compute the limit $\lim_{n\to\infty}f^n(x)$ with $x\in(0,1)$, where $$f^n(x)=\begin{cases} 0\,\,& x\leq0\\ x\,\,& 0<x<1/n\\ 1\,\,& x\geq1/n \end{cases}.$$

Would the answer simply be $1$? Because $$\lim_{n\to\infty}f^n(x)=\begin{cases} 0\,\,& x\leq0\\ 1\,\,& x>0 \end{cases}$$ and given that $x\in(0,1)$ then we pick $1$ instead of $0$.

Answer 1

If $x\le 0$, then $f_n(x)=0$ for all $n$.

If $x>0$, then $f_n(x)=1$ for all $n$ such that $n\ge \frac{1}{x}.$

Therefore your answer is correct.

Answer 2

I think you could simplify your work a little. From the first step, since $x\in(0,1),x>0$. Since $\lim_{n\to\infty} \dfrac{1}{n} = 0$, $\exists N$ st $\forall \epsilon >0, \dfrac{1}{n}<\epsilon\Rightarrow$ $x>\dfrac{1}{n}$. Thus $\lim_{n\to\infty}f^n(x)=1.$

Answer 3

Since

$$\lim_{n\to\infty}f^n(x)=\begin{cases} x\,\,& 0<x<1/n\\ 1\,\,& 1/n<x<1 \end{cases}$$

we have that $$\lim_{n\to\infty}f^n(x)=1$$

indeed $\forall \epsilon >0 \quad \exists n_0>0\quad \forall n>n_0$$

$$|f^n(x)-1|<\epsilon$$

indeed it suffices to take $\frac1{n_0}<x \implies n_0\ge \frac1x$.

Answer 4

I depends what you mean by convergence. This is an example that shows that the set of continuous functions is not closed in the $L^2$ sense.