Verification of Epsilon-Delta proof.

Question

$$\lim _{ x\rightarrow 3 }{ { -(x-3 })^{-2 }=-∞ } $$

$\ -(x-3 )^{-2 }< M<0$

$\ (x-3 )^{-2 }> |M|$

$\ (x-3 )^{2 }< 1/|M| $

$\ |x-3|<1/{\sqrt{|M|}} $

$\ 0<|x-3|<$

$\ <1/{\sqrt{|M|}}$

Since a delta can be obtained for all negative values of M, the limit is correct

Is the proof logically correct?

Answer 1

The idea is correct, but not your way of expressing it. It should be:\begin{align}-(x-3 )^{-2 }< M&\iff(x-3 )^{-2 }> |M|\text{ (assuming $M<0$)}\\&\iff\ (x-3 )^{2 }< \frac1{|M|}\\&\iff\ |x-3|<\frac1{\sqrt{|M|}}.\end{align}Therefore, if we take $\delta=\frac1{\sqrt{|M|}}$, we have$$|x-3|<\delta\implies-(x-3 )^{-2 }< M$$