Is this a pointwise convergent function?

Question

For the series of functions $f_{k}(x)$ defined by
$$f_{k}(x)=\begin{cases}k \quad 0<x\leq\frac{1}{k},\\0 \quad \text{otherwise}.\end{cases}$$ Is this pointwise converging to the zero function? I would say so (graphically), but I'm not sure how I would show formally. This is what I did:
We want
$$\forall x \in \mathbb{R}, \forall \epsilon >0, \exists K>0: k\geq K \implies |f_k(x) - 0| < \epsilon.$$

So for $0<x\leq \frac{1}{k}$, $|f_k(x) - 0| = k.$ I'm not sure if it actually does pointwise converge now, seeing as if I pick $K= k = 10$ and $\epsilon = 1$, then the inequality obviously does not hold for all $\epsilon$... but my intuition keeps telling me it does, when I look at it graphically.

Answer 1

Yes, this converges pointwise to $f=0$, and the line in the middle of your question is a proof for this statement.

You must not pick some $k$ and then some $\varepsilon$. For pointwise convergence you first pick some (arbitrary) $x$ and some $\varepsilon >0$ and then you have to show there is some $K$ such that for $k\ge K$ the the necessary inequality holds.