# Thread: proof involving limits

1. ## proof involving limits

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.

2. I think it's not true.

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

3. 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?

4. 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.)