Proving If f(x)>0 for All x: A Guide for Dogma

[dogma]

Hi all.

I'm doing some self studying on limits, and...I have the following problem with this problem...

Prove: If $$f(x)>0$$ for all $$x$$, then $$\lim_{x\rightarrow x_o} f(x)\geq 0$$ for any $$x_o$$

I'm assuming the best way to prove this is through contradiction:

Assume $$\lim_{x\rightarrow x_o} f(x) = A < 0$$

This as far I get before vapor lock sets in. I guess I need to find an appropriate $$\epsilon$$ and then try to show/not show that $$f(x) < 0$$ for at least one $$x$$.

Can someone please point me on the right direction?

Thanks,
dogma

 

[Timbuqtu]

$$\lim_{x\rightarrow x_o} f(x) = - A$$ means: for all $$\epsilon > 0$$ there exist a $$\delta > 0$$ such that $$f( \left]x_0 - \delta, x_0 + \delta \right[) \subset \left]-A-\epsilon, -A+\epsilon \right[$$. So in particular what do we get for $$\epsilon = A$$?

 

[tacman]

Hi dogma,

To prove this statement, we can use the definition of a limit. According to the definition of a limit, if \lim_{x\rightarrow x_o} f(x) = A, then for any \epsilon > 0, there exists a \delta > 0 such that for all x satisfying 0 < |x - x_o| < \delta, we have |f(x) - A| < \epsilon.

Now, since we are assuming that \lim_{x\rightarrow x_o} f(x) = A < 0, we can choose \epsilon = -A. Then, by the definition of a limit, there exists a \delta > 0 such that for all x satisfying 0 < |x - x_o| < \delta, we have |f(x) - A| < -A. This can be rewritten as -A < f(x) - A < A, which simplifies to 0 < f(x) < 2A.

But since we know that f(x) > 0 for all x, this implies that 0 < f(x) < 2A < 0, which is a contradiction. Thus, our assumption that \lim_{x\rightarrow x_o} f(x) = A < 0 must be false, and therefore \lim_{x\rightarrow x_o} f(x) \geq 0.

I hope this helps guide you in the right direction. Keep up the good work with your self-studying on limits!