Thank you for your time!
I would like to check if my proof is ok here:
Prove: Let $\lim_{x\to\infty} f(x)= 4$. Prove that for every $M>0, x>M$, $15.5 < [f(x)]^2 < 16.5$.
So, I said that: $-\sqrt{15.5} < f(x) < \sqrt{16.5}$
$ |f(x)-L| < \epsilon$,
I chose $\epsilon = 0.05$, and than $-0.05 < f(x)-4 < 0.05$
From here, $3.95 < f(x) < 4.05$ and $15.6025 < [f(x)]^2 < 16.4025$
So $15.5 < [f(x)]^2 < 16.5$
Is that legit?
Thanks a lot!
I suspect the question should read, prove that if $\lim_{x\to\infty} f(x) = 4$ then there exists a number $M$ such that for all $x > M$, $$15.5 < f(x)^2 < 16.5.$$
With this understanding, by definition of the concept of limit, for every $\varepsilon > 0$ there exists $M$ (dependent on $\varepsilon$) such that when $x > M$, $|f(x) - 4| < \varepsilon$. That means, $$ 4-\varepsilon < f(x) < 4 + \varepsilon.$$ Now you can square the inequality provided all terms are positive; that will be the case if $\varepsilon$ is small enough, and so, $$16 - 8 \varepsilon + \varepsilon^2 < f(x)^2 < 16 + 8 \varepsilon + \varepsilon^2.$$ Now just choose $\varepsilon$ so small that the left extreme is greater than $15.5$ and the right extreme is less than $16.5$. For example $\varepsilon = 0.01$. Then with the corresponding $M$ the required statement is true.
Of course you can get many different values for $M$ but your are only asked to show one exists. Once you have on, any larger $M$ would also work. Some smaller $M$ might also work depending on how finely you chose the first one.