it is known that $f(0,0) = 0$ and $f$ is differentiable at $(0,0)$. It is also known that for every $t>0$ we have $f(\cos(t)/t , \sin(t)/t) > 0$.
Show that necessarily $\operatorname{Gradient}(0,0) = (0,0)$.
I have no clue how to solve this. I have tried to show through limits, I have tried to get something from the definition ( that with $\varepsilon, \:(x/y) \to 0$ at the limit)
I am sorry for the bad latex language really.
Suppose that the gradient of $f$ at $(0,0)$ exists and is not $(0,0).$ That would imply that if you look at a small enough neighborhood of $(0,0),$ within that neighborhood $f$ would look like a linear function (informally speaking), and that linear function would be increasing in some directions from $(0,0)$ and decreasing in other directions from $(0,0).$
Now ask how $f$ can look like a decreasing function in some directions from $(0,0)$ and still have $f\left(\frac{\cos t}{t},\frac{\sin t}{t}\right) > 0$ for every positive real number $t.$ If the answer is that it cannot, then $f$ cannot have a non-zero gradient at $(0,0).$
The problem of writing the proof is mainly how to relate the given facts to the formal definition of the gradient and the theorems about it.