# Prove that $\lim_{x \to a} f(x) = \infty$ iff $\lim_{x \to a} \frac{1}{f(x)} = 0$

Let $a \in \mathbb{R}$ and prove that, using the $\epsilon-\delta$ definition of the limit, $\displaystyle \lim_{x \to a} f(x) = \infty$ iff $\displaystyle \lim_{x \to a} \frac{1}{f(x)} = 0$ and that there is $\delta > 0$ such that $f(x) > 0$ whenever $0 < |x-a| < \delta$.

For the first question, we are given that $$\forall M>0, \exists \delta \quad 0<|x-a|<\delta \implies \quad f(x) > M$$ and need to show this is equivalent to $$\forall \epsilon, \exists \delta \quad 0<|x-a|<\delta \implies \quad \left |\dfrac{1}{f(x)} \right |< \epsilon.$$ I am not to sure how to show this part. As for the second part of the question, couldn't we just have $f(x) = -5$ and that never be true?

• Actually that's not right @ThomasAndrews. I think your example is wrong this should work for any function since it approaches infinity as $a$ goes to infinity. Mar 7, 2016 at 2:11
• Remember we are talking about $f(x)$ approaching infinity as $a$ goes to infinity, which it does in your example. But the $\infty$ I define here means $+\infty$ so this works for any function. Mar 7, 2016 at 2:16

For the direct side of the theorem as you have stated we have$$\forall M>0, \exists \delta \quad 0<|x-a|<\delta \implies \quad f(x) > M$$define $$M={1\over \epsilon}$$ with $$\epsilon>0$$. Therefore the above logical statement is equivalent to $$\forall \epsilon>0, \exists \delta \quad 0<|x-a|<\delta \implies \quad f(x) > {1\over \epsilon}\to 0<{1\over f(x)}<\epsilon$$which is the direct definition of limit to zero. Here we do not need the second condition to hold as a necessary condition while as a necessary condition for the converse we must have that the function must be positive on at least on neighborhood of $$a$$. The second condition is in fact a formal mathematical expression of this. Sinilarly in the proof of converse, you only need to take $$\epsilon$$ sufficiently small such that $$\delta<\delta_0$$ where $$\delta_0$$ satisfies the second condition.

Here is the first part: Let $\varepsilon>0$. Then, by assumption, $\exists \delta>0$ such that $\forall x: |x-a|<\delta\implies f(x)>\frac{1}{\varepsilon}$. Thus, $\frac{1}{f(x)}<\varepsilon$ and since $\varepsilon$ was arbitrary, $\lim\limits_{x\rightarrow a}\frac{1}{f(x)}=0$.

So let some $\epsilon > 0$. We need to find a $\delta$ such that when $|x-a|<\delta$, we would have $\left|\frac{1}{f(x)}\right| < \epsilon$.

Let $M = 1/\epsilon$. Now there must exist a $\delta$ such that when $|x-a|<\delta$, we have $f(x) > M = 1/\epsilon$.

Can you finish this by inverting the last inequality?

• We have $\frac{1}{f(x)} < \epsilon$. We can't take the absolute value of both sides, can we? Mar 7, 2016 at 1:31
• @Puzzled417 But $M = 1/\epsilon > 0$ and so $f(x) > 0$ too Mar 7, 2016 at 1:37
• What about the second statement in the question? Mar 7, 2016 at 1:41
• What about the $f(x) > 0$ part? Mar 7, 2016 at 2:31
• @Puzzled417 it is talking about the same $f$ in the second part... Mar 9, 2016 at 17:23