Convergence of triangle function to a constant

Consider a real valued function $$f$$ with $$f(0)=0$$ and $$f(1/n)=1$$ which is linear on $$[0,1/n]$$ and $$0$$ everywhere else. Does $$f$$ converge pointwise to $$0$$?

Specifically im trying to understand the behavior at $$0$$. The value at $$0$$ as $$n$$ goes to infinity seems illdefined.

• Yes, $f$ converges pointwise to $0$. The sequence of values at $0$ is, by definition, the constant sequence with constant value $0$, which converges to $0$; what is the problem? Aug 5 at 22:00
• The limit of the location of the peak is 0. So while $f(0)=0$ for all $n$, i expect the limit is undefined at this point.
– dmh
Aug 5 at 22:38
• $f_n(x)$ is pointwise convergent to $f(x)$ if $$lim_{n\to\infty}f_n(x)=f(x)$$ The convergence of $f_n(0)$ is independent of the values of $f_n(x)$ where $x \neq 0$ Aug 6 at 0:44