What is the pointwise limit of the following function: What is the pointwise limit of:
$$f_n(x) = \begin{cases} n-n^2x & if &  0\leq x \leq\frac{1}{n} 
           \\0 & if & \frac{1}{n} \leq x \leq 1
     \end{cases} $$
 A: The pointwise limit is:
$$f(x) = \begin{cases} \infty & if & x=0 
           \\0 & if &  0<x\leq1 
     \end{cases} $$
To see this, notice that on $[0,1/n]$, $f_n$ is a line with $y$-intercept at $n$ and slope of $-n^2$ and an $x$-intercept at $1/n$. As $n$ approaches $\infty$ think about what happens to these lines.
A: Let $x_0 \in [0,1]$. Note that if $x_0 > 0$ we have $\frac{1}{n} \leq x_0 \leq 1$ for all large enough $n \in \mathbb{N}$ and so $f_n(x_0) = 0$ for all large enough $n \in \mathbb{N}$. This implies that
$$ \lim_{n \to \infty} f(x_0) = \lim_{n \to \infty} \begin{cases} n - n^2x_0 & 0 \leq x_0 \leq \frac{1}{n} \\ 0 & \frac{1}{n} \leq x_0 \leq 1 \end{cases} = \begin{cases} \lim_{n \to \infty} n & x_0 = 0 \\ \lim_{n \to \infty} 0 & 0 < x_0 \end{cases} = \begin{cases} +\infty & x_0 = 0 \\ 0 & 0 < x_0 \leq 1 \end{cases}. $$
A: The pointwise limit is the always vanishing function on $(0,1]$ and the sequence of functions does not converge (or diverges to $+\infty$) at $0$.
A: The domain of the first part shrinks to $0$. It follows that the limit is $0$ for all $x>0$. For the unique case $x=0$ it is enough to check by hand that it is $\infty$.
