proof involving limits

**myoplex11** · July 26th 2009, 09:48 PM

Need help with the following proof:
prove that if lim x->c 1/f(x)= 0 then
lim x->c f(x) does not exist.I think i need to use the delta epsilon definition i am not sure how to set it up.

**red_dog** · July 27th 2009, 08:00 AM

I think it's not true.

If $f(x)=\frac{1}{(x-c)^2}$ then $\lim_{x\to c}f(x)=\infty$ and $\lim_{x\to c}\frac{1}{f(x)}=0$

**Enrique2** · July 27th 2009, 08:26 AM

Originally Posted by myoplex11

Need help with the following proof:
prove that if lim x->c 1/f(x)= 0 then
lim x->c f(x) does not exist.I think i need to use the delta epsilon definition i am not sure how to set it up.

Actually, not only the red dog example works. Your condition is in fact equivalent to $\lim_{x\to c}|f(x)|=\infty$ . So I suppose you have to show the limit is not a real number, isn't it?

**Opalg** · July 27th 2009, 08:54 AM

Originally Posted by myoplex11

Need help with the following proof:
prove that if lim x->c 1/f(x)= 0 then
lim x->c f(x) does not exist.I think i need to use the delta epsilon definition i am not sure how to set it up.

Proof by contradiction. Suppose that the limit exists, and that $\lim_{x\to c}f(x) = L$ .

We know that $\lim_{x\to c}1/f(x) = 0$ . So for each $\varepsilon>0$ there exists $\delta>0$ such that $|1/f(x)|<\varepsilon$ whenever $|x-c|<\delta$ .

Choose $\varepsilon< 1/(|L|+1)$ . Then $|x-c|<\delta\ \Rightarrow\ |1/f(x)|< 1/(|L|+1)\ \Rightarrow\ |f(x)|> |L|+1$ .

But that means that whenever x is sufficiently close to c, f(x) differs from L by at least 1. That contradicts the assumption that $f(x)\to L$ as $x\to c$ . (I'll leave it to you to make that last sentence rigorous in terms of epsilons and deltas.)

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