How can I analytically do an epsilon-delta proof without using a graph?

Question

I'm taking Calculus 1 - Analytic Geometry. We are learning how to analytically prove the limit of a function. For most, I will simply find the limit using an ordered pairs table.

Then use |f(x) - L| < E (epsilon) and through typical algebraic operations the inequality will end up being 0 < |x - c| < d in order to find d (delta). Is this a sure fire way to prove limits? It seems that most problems I've been doing, this works rather well. However I've been stuck on one for days and it makes absolutely no sense to me because through regular algebra, the x values will cancel out, I've tried several ways of doing this and perhaps I'm just making some algebraic errors but I really can't tell. My professor is one of those extremely intelligent math gurus but his explanations are often far too complicated. I understand methodology behind direct substitution and others for finding limits, but proving this one is difficult for me. Please help! f(x) = 2 - (1/x); x->1; E = .1 and L = 1. Best explanation will get best answer. Thanks!

Answer 1

First, note that f(1)=1 is well defined, and is equal to L.

You know that c = 1 (as c is the value x is converging to), let us find d.

Problem formulation: find d such that, for all x satisfying

0&lt;|x-c|&lt;d, we have |f(x)-1|&lt;1/10

First, note that if x&lt;0, f(x)&gt;2, and |f(x)-1|&gt;1&gt;E. Thus we conclude that x&gt;0

|f(x)-1|&lt;1/10 is equivalent to

-1/10&lt;2-1/x-1&lt;1/10. Let us first concentrate on the first inequality.

-1/10&lt;1-1/x or 1/x&lt;11/10. Since x&gt;0, we conclude

10/11&lt;x

From the second inequality, we have

1-1/x&lt;1/10 or 9/10&lt;1/x. Since x&gt;0, we conclude

x&lt;10/9

Thus we conclude 10/11&lt;x&lt;10/9, which yields

-1/11&lt;x-1&lt;1/9 (and also 1/11&gt;1-x&gt;-1/9). Since you want a unique value for d, you need to satisfy the stricter of the two inequalities. Since 1/11&lt;1/9, it follows that

|x-1|&lt;1/11, and thus d = 1/11