1 Answer | Add Yours
Roughly speaking, we need to show that we can make the quantity `1/|x|` as large as we want by taking `x` sufficiently close to 0. More precisely, we want to show that for any given `M>0` , there exists some `delta>0` such that
`1/|x|>M` for all `|x|<delta`. Note that `1/|x|>M` is equivalent to `|x|<1/|M|` , which suggests that we choose `delta=1/M` . Then the proof would just go something like:
Let `M>0` be given. Set `delta=1/M` . Then if `|x|<delta=1/M` , we see that `1/|x|>M` , so `lim_(x->0) 1/|x|=oo`.
One more remark. The page I link to uses `epsilon`, where I've used `M` instead. It makes no difference of course, but since ` ` we usually think of `epsilon` as a "small" number, here we intuitively think of `M` as a large number. That's the only reason I used `M`.
We’ve answered 318,988 questions. We can answer yours, too.Ask a question