Math Help - Prove that lim (x -> 0) (1/x^2) does not exist

1. Prove that lim (x -> 0) (1/x^2) does not exist

Hello everyone,

I am to use the delta-epsilon definition of a limit to show that the following limit does not exist. Unfortuantely, neither my instructor nor my textbook offer any worked examples or guides as to do this so I have been quite disillusioned so far. I have posted my attempt at a proof below.

Thank you very much for your help.

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Prove that $\lim_{x \rightarrow 0} \frac{1}{x^2}$ does not exist.

Proof #1: I came up with this but I don't think that it works because x depends solely on L. I think that delta should be in my $x$ somewhere since I have to show that for any $\delta > 0$, the definition of a limit does not hold.

2. Originally Posted by scherz0
Hello everyone,

I am to use the delta-epsilon definition of a limit to show that the following limit does not exist. Unfortuantely, neither my instructor nor my textbook offer any worked examples or guides as to do this so I have been quite disillusioned so far. I have posted two attempts at proof below.

Thank you very much for your help.

---

Prove that $\lim_{x \rightarrow 0} \frac{1}{x^2}$ does not exist.

Proof #1: I came up with this but I don't think that it works because x depends solely on L. I think that delta should be in my $x$ somewhere since I have to show that for any $\delta > 0$, the definition of a limit does not hold.

Proof #2: I was suggested to use this value of x by my instructor. However, I don't understand how this version would generate a contradiction.

Are you trying to prove that $\lim_{x\to0}\frac{1}{x^2}$ DNE or $\lim_{x\to0}\frac{1}{x}$ DNE? You use one in one proof and the other in the second.

If it's the former, you want to show that $\lim_{x\to0}\frac{1}{x^2}=\infty$, not that it does not exist. There is a difference between the two. If it's the latter, all you have to show is that $\lim_{x\to0^-}\frac{1}{x}\neq\lim_{x\to 0^+}\frac{1}{x}$.

3. Originally Posted by redsoxfan325
Are you trying to prove that $\lim_{x\to0}\frac{1}{x^2}$ DNE or $\lim_{x\to0}\frac{1}{x}$ DNE? You use one in one proof and the other in the second.

If it's the former, you want to show that $\lim_{x\to0}\frac{1}{x^2}=\infty$, not that it does not exist. There is a difference between the two. If it's the latter, all you have to show is that $\lim_{x\to0^-}\frac{1}{x}\neq\lim_{x\to 0^+}\frac{1}{x}$.
Thank you for your reply, redsoxfan325. I apologise for the confusion above. I am trying to prove that $\lim_{x\to0}\frac{1}{x^2}$ DNE and so have removed Proof #2.

My course assumes that a limit which goes to $\pm \infty$ is the same as a limit that does not exist. Although I can show that $\lim_{x\to0}\frac{1}{x^2} = \infty \Rightarrow$ DNE by evaluating the right- and left-hand limits, I have to show that it does not exist using the delta-epsilon definition of a limit and my attempt is in my first post.

Therefore, could my proof above be reviewed? I really am uncertain as to whether it is valid.