Does $\displaystyle \frac {sinx}{k} $ converges pointwise or uniformly as k approaches infinity?
Isn't the limit 0?
converges uniformly on what set? $\displaystyle \mathbb{R}$? yes it does (to 0). you can use either of the two main definitions to show this. there is an $\displaystyle \epsilon$ definition and a limit definition involving a supremum
there is also a theorem that has to do with the derivative being bounded implies uniform convergence (or maybe it's uniform convergence implies the derivative is bounded. i can never remember which. i should look it up)
Not sure if this argument is good or not. I think it converges to the zero function. It's been a while since I studied calculus.
If we define the domain as the real line, then we know $\displaystyle \sin x$ is bounded by $\displaystyle 1$ and $\displaystyle \frac {1}{k} \rightarrow 0$ for all $\displaystyle k \geq N.$ Let $\displaystyle x$ be in the domain. Given $\displaystyle \epsilon > 0$, there is some $\displaystyle N \in \mathbb{N}$ such that $\displaystyle \left | \frac{1}{k} \right | < \epsilon$ for all $\displaystyle k \geq N$. Then $\displaystyle |f_k(x)-0| = \left |\frac {\sin x}{k}\right | = \left|\frac {1}{k} \right ||\sin x | < \epsilon \cdot 1 = \epsilon.$