The limit of sequence of functions

Let $(f_n)(x)=2n$, $0 < x ≤ 1/n$ and $ f_n(x) = 1$, otherwise. Find the pointwise limit function $f$ on $[0, 1]$.

I tried observing $f_n(x)$ for $n= 1,2,..$. We see that the value of $f_n$ is $2n$ on $(0,1/n]$, and $0$ elsewhere. My question is, as the interval in which the function takes non zero value is decreasing in size, and its size tends to $0$ as $n\rightarrow \infty$, will the point wise limit be $1$?

Edit: If N is any natural number, for all n greater than N, $1/n \leq 1/N$. So, $f_n(x)=2n$ for all $n\geq N$. Does this imply that there is no point wise limit?

Solution 1:

Note that $$f_n(x)=\begin{cases} 2n & 0<x\leq\frac{1}{n}\\ 1 & \text{else} \end{cases} $$ now I claim that the pointwise limit of $f_n(x)=1$ on $[0,1]$. First observe that by definition $f_n(0)=1$ for all $n$. Now let $s\in(0,1]$ to show that the pointwise limit is $1$, it suffice to show that for this arbitrary $s$, we have that $$\lim_{n\rightarrow\infty}f_n(s)=1$$ Indeed since there is some $N$ such that for $n>N$ we have that $\frac{1}{n}<s$, so for $n>N$ we will have that $f_n(s)=1$, so $$\lim_{n\rightarrow\infty}f_n(s)=1$$ and we conclude that the pointwise limit is $1$.