Limit of Function by Convergent Sequences

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[edit] Theorem

Let M_1 = \left({A_1, d_1}\right) and M_2 = \left({A_2, d_2}\right) be metric spaces.

Let S \subseteq A_1 be an open set of M_1.

Let f be a mapping defined on S, except possibly at the point c \in S.


Then \lim_{x \to c} f \left({x}\right) = l iff, for each sequence \left \langle {x_n} \right \rangle of points of S such that \forall n \in \N^*: x_n \ne c and \lim_{n \to \infty} x_n = c, it is true that \lim_{n \to \infty} f \left({x_n}\right) = l.


[edit] Proof

  • Suppose that \lim_{x \to c} f \left({x}\right) = l.

Let \epsilon > 0.

Then by the definition of the limit of a function, \exists \delta > 0: d_2 \left({f \left({x}\right), l}\right) < \epsilon provided 0 < d_1 \left({x, c}\right) < \delta.

Now suppose that \left \langle {x_n} \right \rangle is a sequence of points of S such that such that \forall n \in \N^*: x_n \ne c and \lim_{n \to \infty} x_n = c.

Since \delta > 0, from the definition of the limit of a sequence, \exists N: \forall n > N: d_1 \left({x_n, c}\right) < \delta.

But \forall n \in \N^*: x_n \ne c.

That means 0 < d_1 \left({x_n, c}\right) < \delta.

But that implies that d_2 \left({f \left({x_n}\right), l}\right) < \epsilon.

That is, given a value of \epsilon > 0, we have found a value of N such that \forall n > N: d_2 \left({f \left({x_n}\right), l}\right) < \epsilon.

Thus \lim_{n \to \infty} f \left({x_n}\right) = l.


  • Now suppose that for each sequence \left \langle {x_n} \right \rangle of points of S such that \forall n \in \N^*: x_n \ne c and \lim_{n \to \infty} x_n = c, it is true that \lim_{n \to \infty} f \left({x_n}\right) = l.

What we will try to do is assume that it is not true that \lim_{x \to c} f \left({x}\right) = l, and try to find a contradiction.

So, if it not true that \lim_{x \to c} f \left({x}\right) = l, then:

\exists \epsilon > 0: \forall \delta > 0: \exists x: 0 < d_1 \left({x, c}\right) < \delta: d_2 \left({f \left({x_n}\right), l}\right) \ge \epsilon

In particular, if \delta = \frac 1 n, we can find an x_n where 0 < d_2 \left({x, c}\right) < \frac 1 n such that d_2 \left({f \left({x_n}\right), l}\right) \ge \epsilon.

But then \left \langle {x_n} \right \rangle is a sequence of points of S such that \forall n \in \N^*: x_n \ne c and \lim_{n \to \infty} x_n = c, but for which it is not true that \lim_{n \to \infty} f \left({x_n}\right) = l.

So there is our contradiction, and so the result follows.


\blacksquare


[edit] Corollary

Let f: \R \to \R be a real function.

The above result holds for f tending to a limit both from the right and from the left:

  • \lim_{x \to b^-} f \left({x}\right) = l \iff \forall \left \langle {x_n} \right \rangle: \lim_{n \to \infty} f \left({x_n}\right) = l;
  • \lim_{x \to a^+} f \left({x}\right) = l \iff \forall \left \langle {x_n} \right \rangle: \lim_{n \to \infty} f \left({x_n}\right) = l

where f is defined on the open interval \left({a \, . \, . \, b}\right).


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