$\lim_{n\to 1}(\frac{1}{1-n})=\infty$

Question

$\lim_{n\to 1}(\frac{1}{1-n})=\infty$

When $n=1$, the equation, of course, becomes undefined, as it becomes $\frac10$

I know this can be proven since $\frac{1}{1-n}=\sum_{k=0}^{\infty}n^k$, which would make $\frac1{1-1}=1^1+1^2+1^3+1^4...=1+1+1+1...=\infty$

But my question is, how can you prove this using the epsilon-delta definition of a limit?

Thanks!

Answer 1

The statement should be $\lim_{n\to1^\pm}\frac{1}{1-n}=\mp\infty$. By definition, the $\pm=+$ case means $\forall N>0\exists\epsilon>0\left(1<n<1+\epsilon\implies\frac{1}{1-n}<-N\right)$ (just choose $\epsilon<\frac1N$), and you can handle the other case similarly.