Proof: Suppose that the limit of a function f : R > R as x approaches a, element of R exists.?

Question

Then the limit is unique. Can anyone help with this one?

Answer 1

Let L = lim (x -->a ) f(x). Then, according to the definition, for every eps >0, there exists d >0, such that, if 0 < |x -a| < L, then |f(x) - L| < eps.

Suppose L' is also limit of f at a. Then, applying again the definition now to L,' with the same eps used for L, we get a d' such that |f(x) - L'| < eps for every x satisfying 0 < |x -a| < d'.

So, for every x satisfying 0 < |x -a| < minimum {d, d'}, we have

|f(x) - L| < eps and |f(x) - L'| < eps.

Applying the triangle inequalty, we get

|L - L'| = |L - f(x) + f(x) - L'| <= |L - f(x)| + |f(x) - L'| < eps + eps = eps, so that

0 <= | L - L'| < eps. Since this holds for every eps >0, we must have |L - L'| = 0, that is L = L'. This proves the limit, when it exists, is uniquely determined.

This conclusion is not restricted to functions from R to R, it is valid in any metric space (actually, in any Hausdorff topological space).

Answer 2

You can do this by assuming the limit has two values, say L1 and L2.

Then using the definition of limit and a triangle inequality trick:

|L1 - L2| = |L1 - f(x) + f(x) - L2| ≤ |L1 - f(x)| + |f(x) - L2|,

you can show that for any ε > 0, |L1 - L2| ≤ ε, which implies that |L1 - L2| ≤ 0, which implies that L1 = L2.