Problem. Let $U \subset \mathbb{R}^{m}$ be an open. Show that if $f: U \to \mathbb{R}$ has bounded partial derivatives in $U$, then $f$ is continuous on $U$.
Idea. $$\lim_{\lambda \to 0}|f(x + \lambda x_{i}) - f(x)| = \lim_{\lambda \to 0}\left|\frac{f(x + \lambda x_{i}) - f(x)}{\lambda}\right||\lambda| \leq M\lim_{\lambda \to 0}|\lambda| = 0,$$ so $$\lim_{\lambda \to 0}f(x+\lambda x_{i}) = f(x).$$
This is almost $$\lim_{x \to a}f(x) = f(a),$$ but the problem is that I have this result for each $x_ {i}$ "separately". I don't think that is enough. Can someone help me?
Let first do it for dimension $m=2$ and then try to generalize it(it is almost the same with a little more writing):
Look at $f(x+\lambda_1e_1+\lambda_2e_2)-f(x)$ where $e_1=(1,0),e_2=(0,1),\lambda_{1,2}\in\Bbb R$. We can add and subtract $f(x+\lambda_1e_1)$ to get $$f(x+\lambda_1e_1+\lambda_2e_2)-f(x)=(f(x+\lambda_1e_1+\lambda_2e_2)-f(x+\lambda_1e_1))+(f(x+\lambda_1e_1)-f(x))$$Let $y=x+\lambda_1e_1$ to get $$f(x+\lambda_1e_1+\lambda_2e_2)-f(x)=(f(y+\lambda_2e_2)-f(y))+(f(x+\lambda_1e_1)-f(x))$$Now continue like you thought:$$f(x+\lambda_1e_1+\lambda_2e_2)-f(x)=\frac{f(y+\lambda_2e_2)-f(y)}{\lambda_2}\lambda_2+\frac{f(x+\lambda_1e_1)-f(x)}{\lambda_1}\lambda_1$$
Can you see to where is it going?