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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?

Theorem:

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

Proof:

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

(ii)$$0<\frac{1}{f(x)}<\frac{1}{M}$$

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 $\frac{1}{M}$.

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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 $x \to c$.

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

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  • $\begingroup$ 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
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    $\begingroup$ @Red It's a fix to your imperfect reasoning. $\endgroup$ – egreg Aug 13 '15 at 21:36

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