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?

hint
Let $ x\in[0,1] $ fixed.
If $ x=0 $, then $ f_n(0)=1=f(0)$.
If $ x>0 $, then for $ n $ large enough ,
$$0<\frac 1n <x \implies f_n(x)=1=f(x)$$
