Consider the function $g:[1,\infty)\to\mathbb{R}$ given by $g(x) = x^2$. For each $a\geq 1$ find the largest possible value of $\delta > 0$ such that the following holds: $$x\geq 1\space \text{with}\space|x-a|<\delta \implies|x^2-a^2|<1.$$
I started by playing with the right hand side:
$$|x^2-a^2|=|(x-a)(x+a)|$$ $$=|x-a||x+a|$$ $$<\delta|x+a| = 1$$
We have the largest possible value of delta when $|x+a| $ is minimal (this is obvious since $\delta=1/|x+a|$). $|x+a|$ is minimal when both $x$ and $a$ are minimal, which by definition is when they both equal $1$. So: $$\delta_{max}|2|=1$$ $$\delta_{max} = \frac{1}{|2|}=\frac{1}{2}$$
Is this correct? I have a memory of doing this type of question previously and it being more difficult to find delta. I can't find a reason why this isn't correct, though.
It is not correct. If $|x-a| < \delta$ we have $$|x^2-a^2| = |x-a||x+a| \le |x-a|(|x-a|+2a) < \delta(\delta + 2a)$$
where both inequalities are sharp. Solving the equation $\delta(\delta+2a) = 1$ we get the positive solution $$\delta_{\max} := -a + \sqrt{a^2+1}$$
which is the desired largest possible delta.
Indeed, for $\delta > \delta_{\max}$ consider $x = \sqrt{a^2+1}$. Then $$|x-a| = -a+\sqrt{a^2+1} < \delta$$ but $|x^2-a^2| = 1$.