check uniform convergence of $f_n(x)=\begin{cases} 1-nx &\text{if}\;x \in [0,1/n]\\\\0 &\text{if}\;x \in [1/n,1] \end{cases}$ Let $$f_n(x)=\begin{cases} 1-nx &amp;\text{if}\;x \in [0,1/n]\\\\0 &amp;\text{if}\;x \in [1/n,1] \end{cases}$$  Then  


$ \lim_{n \to \infty}f_n(x)$ defines a continuous function on $[0,1]$.
$\{f_n\}$ converges uniformly on $[0,1]$.
$ \lim_{n \to \infty}f_n(x)=0$ for all $x \in [0,1]$.  
$ \lim_{n \to \infty}f_n(x)$ exists for  all $x \in [0,1]$.  



  In this question I am not able to evaluate the limit function $f(x)=\lim_{n \to \infty}f_n(x)$.
  as I tried to form the sequence
   $f_1(x)=1-x, x\in[0,1]$ ,
  $f_2(x)=1-2x,x \in [0,1/2]$ and $f_2(x)=0 ,x \in [1/2,1]$
  $f_3(x)=1-3x,x\in [0,1/3]$ and $f_3(x)=0,x\in[1/3,1]$......
  now $f_n(1/2)=\{1/2,0,0...0\}$ so $ \lim_{n \to \infty}f_n(1/2)=0$
  $f_n(1/3)=\{2/3,1/3,0,0...0\}$ so $ \lim_{n \to \infty}f_n(1/3)=0$
  and by archimedean property for any $a \in (0,1]$ there exists $n\in \Bbb N$ such that $1/n\lt a$ so $ \lim_{n \to \infty}f_n(a)=0$ $a \in (0,1]$ but $f_n(0)=\{1,1,1...1\}$ so  $\lim_{n \to \infty}f_n(0)=1$.
  so $\lim_{n \to \infty}f_n(x)=1,x=1$ and $\lim_{n \to \infty}f_n(x)=0,x\in(0,1]$.
  Is This limit function right?
  If yes then option $4$ is right?
Yes, the fourth option is right and you computed correctly the limit of $\bigl(f_n(x)\bigr)_{n\in\mathbb{N}}$, for each $x$. One could add that:


The function is not continuous.
The convergence is not uniform, since otherwise the limit would be continuous.
Let $x &gt; 0$. There exists $N$ such that $x &gt; \frac{1}{N}$. So for all $n \geq N$, $f_n(x)=0$. So $\lim_{n \rightarrow +\infty} f_n(x)=0$.

For $x=0$ now, you have $f_n(x)=1$ for all $n$ so $\lim_{n \rightarrow +\infty} f_n(x)=1$.

That means that the simple limit of $(f_n)$ is the function $f$ such that $f(0)=1$ and $f(x)=0$ if $x &gt; 0$.

In particular, $f$ is not continuous meanwhile all the $f_n$ are. This implies that the convergence cannot be uniform, otherwise the simple limit would also be continuous.