Expert: Paul Klarreich - 9/20/2007

Question
prove ,if limit of a function f(x) as x approaches 'a' is 'L'
than 'L' is unique.

Answer
Questioner:   prem
Category:  Calculus
Private:  No
 
Subject:  limit
Question:  prove ,if limit of a function f(x) as x approaches 'a' is 'L'
than 'L' is unique.
..........................................
Hi, Prem,

I am a bit surprised that, after rating me as 'mediocre' -- 5 out of 10 -- on politeness, you sent me another question.

Ah, well, some questioners don't bother to send a rating at all, and -- would you believe it -- some don't even read the answer.  But you did both and I cannot criticize you.  [That's criticise if you are asking from some country outside the U.S.]

I am not sure exactly what approach your text/instructor uses for limits; some texts use a strictly sequence approach.  That is, a limit would be defined in terms of sequences of  x[i] and of f(x[i]).  But I think your proof should go something like this:

Lim  f(x) = L  means that
x->a

given  e > 0, there exists  d > 0  such that whenever  | x - a | < d,  we have
| f(x) - L | < e.

Suppose we have both:

Lim  f(x) = L1    and
x->a

Lim  f(x) = L2
x->a

where we can assume  L2 > L1

Let  k = L2 - L1.  Then choose  e < k/2.  Since:

Lim  f(x) = L1
x->a

we can find d such that whenever  | x - a | < d,  we have | f(x) - L1 | < e = k/2

But then whenever  | x - a | < d, we have:

- e < f(x) - L1 < e  and so

L1 - e < f(x)  < L1 + e

L1 - k/2 < f(x) < L1 + k/2

L1 - (L2 - L1)/2 < f(x) < L1 + (L2 - L1)/2

L1 - L2/2 + L1/2 < f(x) < L1 + L2/2 - L1/2

3L1/2 - L2/2  < f(x) < L1/2 + L2/2

Now  -f(x) > -L1/2 - L2/2, and so:

L2 - f(x) > L2 - L1/2 - L2/2

L2 - f(x) > L2/2 - L1/2

L2 - f(x) > (L2 - L1)/2

L2 - f(x) > k/2 = e

But we have to have  | f(x) - L2 | < e, so this is not possible.

This is a lot of inequality stuff, which really boils down to this:

If L2 /= L1, then take our epsilon to be half the difference.  Then if we are closer than epsilon to L1, we have to be further than epsilon from L2.  We can't be close to both at the same time.