Hi guys I have this question here from my assignment
and here's my attempt at the above question
(apologies for the handwriting) I wonder is it possible if someone could tell me if I have the right approach or maybe to try something different. Any help is much appreciated.Thanks guys.
It's good (apart notational slips) until you start putting in $\delta$.
You have to show that the inequality $|x|^2+|x|<\varepsilon^2$ is satisfied for $|x|<\delta$ for some $\delta>0$ to determine.
You needn't find the better $\delta$, just one that does the job. So you can restrict yourself to $|x|<1$, which implies $|x|^2<|x|$. So, if you satisfy $|x|+|x|<\varepsilon^2$, you're done; but this is $$ |x|<\frac{\varepsilon^2}{2} $$ so you can take $$ \delta=\min\left\{\frac{\varepsilon^2}{2},\frac{1}{2}\right\} $$ (because, remember, you want to ensure $|x|<1$).
A more complicated approach would be to solve the inequality $$ |x|^2+|x|-\varepsilon^2<0 $$ that is satisfied for $$ \frac{-1-\sqrt{1+4\varepsilon^2}}{2}<|x|<\frac{-1+\sqrt{1+4\varepsilon^2}}{2} $$ which is the same as $$ |x|<\frac{-1+\sqrt{1+4\varepsilon^2}}{2} $$ because the least root is negative.
Therefore you could take $$ \delta=\frac{-1+\sqrt{1+4\varepsilon^2}}{2} $$
