I wanted to ask if someone can do me the favor pointing out the mistakes I might of made in proving the theorem below. Also is there a way to prove the theorem without using the definition of limits?


If $\lim_{x\to c}$f(x) =$\infty$ then $\lim_{x\to c} \frac{1}{f(x)}$ = $0$


Rewriting the above statement using the definition of limits and definition of infinite limits:

($ 0<|x-c|<\delta \Rightarrow f(x)>M)$ $\Longrightarrow (0<|x-c|<\delta \Rightarrow |\frac{1}{f(x)} -0|< \epsilon $)

where $\epsilon$ and $M$ are both greater than zero and M denotes any real number.

Taking the second conditional statement: ($0<|x-c|<\delta \Rightarrow |\frac{1}{f(x)} -0|< \epsilon $)

we can see that

(i) $$-\epsilon <\frac{1}{f(x)} <\epsilon$$

Taking the first conditional statement: $ 0<|x-c|<\delta \Rightarrow f(x)>M$, we can conclude using the reciprocal of inequalities that $f(x)>M>0$ is equivalent to


From (i) and (ii) we get$$0<\frac{1}{f(x)}<\frac{1}{M}<\epsilon.$$

Meaning for any chosen value of $\epsilon>0$ and $M>0$ we can rewrite the theorem as ($ 0<|x-c|<\delta \Rightarrow \frac{1}{f(x)}<\frac{1}{M}<\epsilon )$ $\Longrightarrow (0<|x-c|<\delta \Rightarrow |\frac{1}{f(x)} -0|< \epsilon $) which shows that for any $\epsilon$ there exists a$\frac{1}{f(x)}$ that is always lower than $\frac{1}{M}$.


I think you are lost in symbolism. It is better to use some amount of language. The starting statement which you used about $\delta, M, \epsilon$ is simply not making any sense.

What we need to show is the following statement:

If "(P) corresponding to any $M > 0$ we can find a $\delta > 0$ such that $f(x) > M$ whenever $0 < |x - c| < \delta$" then "(Q) corresponding to any $\epsilon > 0$ we can find a $\delta' > 0$ such that $|1/f(x)| < \epsilon$ whenever $0 < |x - c| < \delta'$".

This is a logical statement of type $P \Rightarrow Q$ where $P$ are $Q$ are marked clearly in italics in the last paragraph. These statements $P, Q$ can further be written using logical symbols. For example $P$ can be written as $\forall M > 0,\, \exists \delta > 0,\, 0 < |x - c| < \delta \Rightarrow f(x) > M$. Similarly we can write $Q$. But you haven't written in this manner which makes the statement very vague and confusing.

Another fundamental mistake is that you have used the same $\delta$ in both $P$ and $Q$. This is so very wrong. I have explicitly used $\delta'$ to make it clear that both $\delta$ and $\delta'$ have no relation with each other.

Why $P$ implies $Q$? This is obvious. For truth of $Q$ we need to find a $\delta'$ for an $\epsilon$. Let $M = 1/\epsilon$ and from the truth of $P$ we find a $\delta$ based on $M$ and set $\delta' = \delta$. This chosen $\delta'$ will ensure that $0 < |x - c| < \delta' \Rightarrow |1/f(x)| < \epsilon$.

I have used the above logical symbols only to align with your post. A better answer goes like this:

Let $\epsilon > 0$ be given and set $M = 1/\epsilon > 0$. Since $f(x) \to \infty$ as $x \to c$, it is possible to find a $\delta > 0$ such that $f(x) > M$ whenever $0 < |x - c| < \delta$. Thus $0 < 1/f(x) < \epsilon$ whenever $0 < |x - c| < \delta$. This implies that it is possible to find a $\delta > 0$ such that $|1/f(x)| < \epsilon$ whenever $0 < |x - c| < \delta$. Hence $1/f(x) \to 0$ as $x \to c$.


You are saying nowhere what's the relationship between $\varepsilon$ and $M$.

The proof is more straightforward.

Suppose $\varepsilon>0$. Then, by assumption, there exists $\delta>0$ such that, for all $x$ with $0<|x-c|<\delta$, we have $f(x)>1/\varepsilon$. This implies that $$ \text{for all $x$, if $0<|x-c|<\delta$, then } 0<\frac{1}{f(x)}<\varepsilon $$ Since $\varepsilon$ is arbitrary, we have proved that $$ \lim_{x\to c}\frac{1}{f(x)}=0 $$

  • $\begingroup$ Is this in addition to what I did? because I don't see how that is enough by itself, could you please elaborate it a little more. $\endgroup$ – Red Aug 13 '15 at 21:22
  • 1
    $\begingroup$ @Red It's a fix to your imperfect reasoning. $\endgroup$ – egreg Aug 13 '15 at 21:36

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.