$||.||_{\infty}$ in sobolev space

Question

Let $u_n ∈ W^{1,1} (I), I=(0,1)$ defined by:

$u'_n (x) = n$ if $x < 1/n$

$u'_n (x) = 0$ if $x > 1/n$

$u_n (0) = 0$.

Find $||u_n − 1||_∞$.

My attempt:

We have $u_n(x)=u_n(0)+\int_0^{1/n}ndt=1$ a.e. hence $||u_n − 1||_∞=0$ but there is a case I can't really grasp is when $n=\infty$ therefore $||u_n − 1||_∞=1$ so what is the correct answer?

Edit

Correcting my dumb mistake thanks to comments.

if $x<1/n$ , we have $u_n(x)=u_n(0)+\int_0^{x}n\ dt$ hence $||u_n − 1||_∞=1$ as $x\to 0$ when $n\to \infty$.

if $x>1/n$ we have $u_n(x)=u_n(0)+\int_0^{1/n}n\ dt=1$ hence $||u_n − 1||_∞=||1-1||_∞=0$.

We deduce finally that $||u_n − 1||_∞=1$

is That correct?

Thank you!