Problem: If $f: \mathbb{R}^2 \rightarrow \mathbb{R}$, is of class $C^1$ and verifies:
I) $f( 1 , 2 ) < 0$
II) $(Q_n)_n$ is a sequence of $\mathbb{R}^2$ such that $\lim _{n\to \infty}f(Q_{n})=+\infty$
Prove:
a) There is $ P\in \mathbb {R} ^{2}$ such that $f(P)=0$
b) If there are two points $P,Q\in \mathbb{R}^2$ such that $f(P)=f(Q)=0$, then there is $ R\in \mathbb {R} ^{2}$ such that $\nabla f(R)$ is orthogonal to the vector $P-Q.$
Attempt of solution:
a) If $\lim _{n\to \infty}f(Q_{n})=+\infty$, then it has to be a $n_0 \in \mathbb{N}$ such that $ n_0<n$ and $f(Q_{n_0})>0$, because $f$ is $C^1$, therefore it's continuous in its whole domain (I'm really insecure with this part, I find it very vague).
Define a function $\alpha:[0,1]\rightarrow\mathbb{R}^2$, continuous in $[0,1]$, where $\alpha(0)=(1,2)$ and $\alpha(1)=Q_{n_0}.$ Let's define the function $g(t)=f(\alpha(t))$, which is continuous for being a composition of continuous. Also $g(0)<0$ and $g(1)>0$, so I can apply Bolzano's theorem:
Let's call $S$ the set of the $x\in [0,1]$ where $g(x)<0. \quad S\ne \emptyset$ because at least $g(0)\in S$ and it's superior bounded for at least $g(1)$. Then, it must have a supremum. Let $s$ be the supremum of $S$. We can define the sequence $S_n$ where $g(S_n) \rightarrow s$, for S being bounded and closed. Then we have to see if $g(s)<0$ or $g(s)=0$. If $g(s)<0$ then there is an interval $(s-\epsilon,s+\epsilon)\cap [P,s]$ where $g(x)<0 \quad \forall$ x in that interval. Besides, $s<g(1)$ therefore there is a $\tilde{x}\in(s,s+\epsilon)$ such that $g(\tilde{x})<0$. But then $\tilde{x}\in S$ and $\tilde{x}>s$, and this can't be possible, because s is the supremum of the set. Then $g(s)=0$, and I take a $P=\alpha(s)$. QED.
b) I don't really know where to start here. I think it has something to do with Rolle's Theorem, which I could apply doing something similar to what I have done with Bolzano's in a), so I could prove that there's a point where the derivative is 0 in that point, so I will have something like $0=f'(\alpha (c))\cdot \alpha ' (c)$, but this leads me to nothing.
So I managed to do something with the a) but I'm not sure at all if my attempt is correct, and I can't even start the b). I would really appreciate any help you can provide.
If you write out the definition of a limit as it applies to $\lim _{n\to \infty}f(Q_{n})=+\infty$, it's something equivalent to
I say "equivalent to" because your textbook might put it in a slightly different form. I would recommend to use the textbook's form rather than mine, but try to apply the same ideas.
You don't necessarily need to quote the definition in the proof. You can take some steps to apply it to your particular problem before writing it in the proof, but it helps if you don't skip too many steps. And if you get confused, you can write all the steps on a piece of scrap paper in as much detail as you need to sort them out before writing this part of the proof.
So you might choose to let $M = 0$ for your application. I think I might write the result into the proof before going any farther:
Now you want a particular $n$ as described in the sentence above. An easy way is to let $n_1 = n_0+1,$ and now you know such a number $n_1$ exists (because $n_0$ exists) and you know something about $f(Q_{n_1}).$
Unless your definition of limit has "$n \geq n_0$" where I wrote "$n > n_0$", however, you won't get much use out of $f(Q_{n_0}).$
Also, so far none of this has anything to do with $f$ being continuous. It's simply and purely about the definition of a limit.
Next you set up $\alpha:[0,1]\rightarrow\mathbb{R}$, but I think you want $\alpha:[0,1]\rightarrow\mathbb R^2$, since $(1,2)$ is not in $\mathbb R.$
I think later you meant to say $g(1) > 0$ rather than $g(1) > 1.$ A typo?
If you've already done Bolzano's theorem, you don't need to repeat its proof. Just state the theorem, show you have certain things corresponding to its premises, and then write the corresponding conclusion. When you get more fluent with proof-writing you won't even have to state the theorem literally, just write something like, "Then by Bolzano's theorem, ... ."
For part b), the proof depends on what tools you've already developed. But I would probably look at $(P-Q) \cdot \nabla f(\alpha(t))$ as a function of $t.$
I would like to point out that it can happen in part b) that the only way for $\nabla f$ to be "orthogonal" to $P-Q$ is when $\nabla f(x,y) = 0.$ For example, consider the function $f(x,y) = x^2 - 4$ and the points $P = (2,0)$ and $Q = (-2,0).$