0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Choose $\epsilon=|M|/2$ in definition of limit. $\endgroup$ – Paramanand Singh Jul 16 '17 at 10:18
  • $\begingroup$ Okay but is this wrong? $\endgroup$ – mathnoob123 Jul 16 '17 at 10:22
1
$\begingroup$

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 $|x-a|<\delta$.

So there's nothing wrong in what you wrote, but the fact that there should be a statement preceding it is to be noted carefully.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.