Expert: eigensteve - 7/9/2009

Question
QUESTION: I am working on a proof that should be fairly simple but
keep getting stuck.

the problem is prove that

lim (1/x-1) does not exist
x-->1

Now what I did was assume the limit does exist and tried to
find a contradiction.

So for any number e>0 we can find a number d>0 such that

|f(x)-L|<e

whenever 0<|x-a|<d

now x=(d/2)+1 and x=(-d/2)+1 both satisfy the above
equation.

so I plug in (d/2)+1 and (-d/2)+1 into the equation for
|f(x)-L|<e

and I get |(1/((d/2)+1-1) - L|<e

and |(1/((-d/2)+1-1) - L|<e

which simplify to

|2/d - L|<e
and
|-2/d - L|<e --> which also equals |2/d + L|<e

picking an arbitrary value for e (like 1 or 1) and solving
for L does not seem to present me with a contradiction.

Can you help?

ANSWER: Hi Richard,

The issue is in the condition "whenever 0<|x-a|<d".  Given some specific e>0, we should be able to find a d so that /for all/ 0<|x-a|<d, the inequality |f(x)-L|<e holds.

However, in the example you chose with x=(d/2)+1, if d is chosen small enough, then clearly 2/d blows up.

Hope this helps,

Steve

---------- FOLLOW-UP ----------

QUESTION: Thanks, but I don't quite see how that brings me to a proof.

It is obvious geometrically, since as x-->1 from the left
the equation 1(x-1) approaches -infinity, while as x-->1 the
equation approaches +infinity, but the book wants me to
prove it using delta and epsilon. Can you give me some
further pointers?

Answer
Hi,

I think that this does constitute a proof by contradiction:

Assume that the limit exits (and =L).

Then for all e>0 there exist d>0 so that

|f(x)-L|<e for all |x-a|<d

However, we have shown that this cannot possibly be true since f(x) becomes unbounded as x->a.  Therefore, there can be no such d-e proof and hence no limit.

I am pretty sure that this is how you would use d-e for the problem.

-Steve