Prove that $\lim_{x \to \infty} x^2 \neq10^{10}$ using a proof by contradiction style epsilon delta. Prove that $\lim_{x \to \infty} x^2 \neq10^{10}$ using a proof by contradiction style epsilon delta.
My Attempt:
We can show pretty straightforwardly that $\lim_{x \to \infty} x^2 = \infty$, hence we need to show a contradiction in this function having two limits. By definition
$(1)$ for all $ M_1 > 0$ there exists an $N_1 > 0 $ s.t. $x > N_1 \rightarrow x^2 > M_1$
$(2)$ for all $ \epsilon_2 > 0$ there exists an $N_2 > 0 $ s.t. $x > N_2 \rightarrow |x^2 - 10^{10}| < \epsilon_2$
Let $N = max(N_1, N_2)$. Then for $x > N \rightarrow x^2 > M$. However $M > L + \epsilon_2$ which would make $|x^2 - 10^{10}| > \epsilon_2$. Hence by contridiction, the limit does not hold.
Is this proof correct?
 A: Yes your proof is correct, although you need to take $N=\max(N_1,N_2)$ so that you can use both (1) and (2).
Also I would use $\Rightarrow$ in place of $\to$, since that is usually meant to denote a limit in these kinds of proofs.
I think you can shorten it just by saying that if $\lim x^2=10^{10}$, given $\epsilon > 0$, we should be able to find $N$ such that $|x^2-10^{10}|<\epsilon$ for all $x>N$, but when $x>10^5+\epsilon$, it is clear that this is false.
A: Maybe to be more clear we could proceed as follows, suppose
$$\lim_{x \to \infty} x^2 =10^{10}$$
that is
$$\forall  \epsilon> 0 \quad \exists N > 0 \quad \forall x > N \quad |x^2 - 10^{10}| < \epsilon$$
but for $x^2 > 10^{10}+\epsilon$ we obtain a contradiction.
A: There is a problem when you write “However $M>L+\varepsilon_2$”. Why? I had not mentioned $M$ nor $L$ before. Why is $M$ greater than $L+\varepsilon_2$?
Besides, it is not clear to me that you can use the fact that $\lim_{x\to\infty}x^2=\infty$.
You can do it as follows. In order to prove that we don't have $\lim_{x\to\infty}x^2=10^{10}$, I will take $\varepsilon=1$. Take $M>0$. Now, take some $x$ which is both greater than $M$ and greater than $10^6$. Then$$x>10^6\implies x^2>10^{12}\implies|x^2-10^{10}|\geqslant1.$$So, I have proved that there is a number $\varepsilon>0$ such that, for every $M>0$, there is some $x>M$ such that $|x^2-10^{10}|\geqslant\varepsilon$, or$$(\exists\varepsilon>0)(\forall M>0)(\exists x>M):|x^2-10^{10}|\geqslant\varepsilon,$$which the same thing as asserting that we don't have $\lim_{\to\infty}x^2=10^{10}$.
