I've done a proof for the following question. I'm not a mathematician (or a math student) myself, and I don't know anyone who is. Please evaluate my proof and point out any mistake I've done.
Question:
Prove that $\lim_{x\to 0} {1\over x}$ does not exist.
My Solution:
Proof by contradiction: Assume that the limit exists and is some real number L. Then for every $\epsilon > 0$ there is a $\delta > 0$ such that for all x, if $ \lvert x-0\rvert = |x| < \delta$ then $|{1\over x}-L| < \epsilon.$ I will prove that for some such x the latter inequality does not hold. Since $\epsilon$ is any number, take $\epsilon = 1$. Since x is only restricted above, we can take $x < min(\delta,{1\over |L+1|}).$ Then ${1\over x} > |L+1|$.
Since $|{1\over x} - L| > ||L+1|-L|=1,$ then $1 < \epsilon = 1.$, which is a contradiction.
EDIT:
The part where I concluded $|{1\over x} - L| > ||L+1|-L|$ was false. Instead, I've done a proof by cases.
In all cases, $|{1\over x} - L| \ge |{1\over x}| - |L|.$ Therefore $|{1\over x}| - |L| < \epsilon$.
Case 1: L is positive.
We take $x<min(\delta,{1\over |L+1|})$ and $\epsilon = 1$ as did previously. Since ${1\over x} > |L+1|,$ then $ |L+1| - |L| = 1 < |{1\over x}| - |L|<\epsilon=1.$ Hence a contradiction.
Case 2: L is nonpositive.
We take $ x < min(\delta,1).$ Then ${1\over x} >1.$ Since L is nonpositive and ${1\over x} >1,$ we get $|{1\over x} - L| > 1.$ Putting $\epsilon = 1,$ we get a contradiction.
Thus L does not exist in R.