Check my argument for this proof for Quotient Rule (limits)

Question

I have seen that calculus books have other ways to prove this theorem (involving triangle inequality). Just wanted to know if this reasoning is okay or I have done some circular reasoning in this proof.

Given : $\lim_{x\to a}$ g(x) = M

To prove : $\lim_{x\to a} {1 \over g(x)} = {1 \over M}$

Proof :

For every $\epsilon$>0, we have to find a $\delta$>0 such that, whenever 0<|x-a|<$\delta$

it implies: $|\frac{1}{g(x)}-\frac{1}{M}|<\epsilon$

or $|\frac{M-g(x)}{g(x).M}|<\epsilon$

or $\frac{|g(x)-M|}{|g(x)|.|M|}<\epsilon$

or $|g(x)-M|<\epsilon$.|g(x)|.|M|

As $\lim_{x\to a}$ g(x) = M, therefore we can find a $\delta_1$>0 such that, whenever 0<|x-a|<$\delta_1$

it implies: $|g(x)-M|<\epsilon$.|g(x)|.|M| (because $\epsilon$.|g(x)|.|M|>0)

finally, choose $\delta$ < $\delta_1$

so the condition, whenever 0<|x-a|<$\delta$

it implies: $|\frac{1}{g(x)}-\frac{1}{M}|<\epsilon$, is satisfied

Therefore, limit is true.

Answer 1

No, this is not correct. The fact that $\lim_{x\to a}g(x)=M$ means that for each constant $\varepsilon>0$, there is $\delta>0$ such that $\lvert x-a\rvert<\delta\implies\bigl\lvert g(x)-M\bigr\rvert<\varepsilon$. You applied this to $\varepsilon'=\varepsilon.\bigl\lvert g(x)\bigr\rvert.M$, which is not a constant, since it depends upon $x$.

Answer 2

It's not correct. First, you must have $M\neq 0$. Now, the fact that $g(x)\to M$ when $x\to a$ gives you that $|g(x)|\geq \frac{|M|}{2}$ when $|x-a|<\delta $ for some $\delta >0$. Therefore, $$\frac{|g(x)-M|}{|g(x)||M|}\leq \frac{|g(x)-M|}{2M^2}.$$

I let you finish the proof.