# Proving limit of a function is greater than half its limit

If $$\lim_{x\rightarrow a}g(x)=M$$ where $M\neq0$, show there exists a number $\delta>0$ such that $$0<|x-a|<\delta \implies |g(x)|>|M|/2$$

I tried $$\epsilon>|g(x)-M|=|M-g(x)|\geq|M|-|g(x)|$$ $$\epsilon+|g(x)|>|M|>|M|/2$$ if $\epsilon=0$ since $\epsilon$ is arbitrary.

Is this working correct?

• Choose $\epsilon=|M|/2$ in definition of limit. – Paramanand Singh Jul 16 '17 at 10:18
• Okay but is this wrong? – mathnoob123 Jul 16 '17 at 10:22

Whatever you have written is correct, but in the wrong order. The inequality $\epsilon > |M| - |g(x)|$ is correct only for certain $x$, and we must specify which $x$ these are.
In our case, by definition of $\lim_{x \to a} g(x) = M$, we can write down this statement:
For all $\epsilon > 0$ there is $\delta > 0$ (depending on $\epsilon$) such that if $|x-a| < \delta$ then $|g(x) - M| < \epsilon$.
Now, let $\epsilon > 0$. Whenever $|x-a| < \delta$, only then can we say that $|g(x) - M | < \epsilon$ , or that $\epsilon > |M| - |g(x)|$. Hence, this statement is not true for all $x$, so it can't be the first statement of a proper proof.
Now, to finish the argument, all you need is to choose $\epsilon$ properly. In our case, since $\epsilon > 0$, we can choose $\epsilon = \frac{|M|}2$, which gives that for some $\delta > 0$, (depending upon $\epsilon = \frac{|M|}2$), we have that if $|x-a| < \delta$ then $\epsilon > |M| - |g(x)|$. Knowing that $\epsilon = \frac{|M|}2$, and substituting above gives us that $|g(x)| > \frac{|M|}2$, whenever $|x-a|<\delta$.