Reading guide · Proof index

Limits of functions

Jiří Lebl, Basic Analysis I–II, version 6.3. Free author edition of this section. Selection and attribution · Notation.

L3.1.7: Full function/sequential-limit equivalence.

Proof.

Suppose f(x)→Lf(x) \to L as x→c,x \to c\text{,} and {xn}n=1∞\{ x_n \}_{n=1}^\infty is a sequence such that xn∈S∖{c}x_n \in S \setminus \{c\} and lim⁡n→∞xn=c.\lim_{n\to\infty} x_n = c\text{.} We wish to show that {f(xn)}n=1∞\bigl\{ f(x_n) \bigr\}_{n=1}^\infty converges to L.L\text{.} Let ϵ>0\epsilon > 0 be given. Find a δ>0\delta > 0 such that if x∈S∖{c}x \in S \setminus \{c\} and ∣x−c∣<δ,\sabs{x-c} < \delta\text{,} then ∣f(x)−L∣<ϵ.\babs{f(x) - L} < \epsilon\text{.} As {xn}n=1∞\{ x_n \}_{n=1}^\infty converges to c,c\text{,} find an MM such that for n≥M,n \geq M\text{,} we have that ∣xn−c∣<δ.\sabs{x_n - c} < \delta\text{.} Therefore, for n≥M,n \geq M\text{,}
∣f(xn)−L∣<ϵ.\begin{equation*} \babs{f(x_n) - L} < \epsilon . \end{equation*}
Thus {f(xn)}n=1∞\bigl\{ f(x_n) \bigr\}_{n=1}^\infty converges to L.L\text{.}
For the other direction, we use proof by contrapositive. Suppose it is not true that f(x)→Lf(x) \to L as x→c.x \to c\text{.} The negation of the definition is that there exists an ϵ>0\epsilon > 0 such that for every δ>0\delta > 0 there exists an x∈S∖{c},x \in S \setminus \{c\}\text{,} where ∣x−c∣<δ\sabs{x-c} < \delta and ∣f(x)−L∣≥ϵ.\babs{f(x)-L} \geq \epsilon\text{.}
Let us use 1 ⁣/ ⁣n\nicefrac{1}{n} for δ\delta in the statement above to construct a sequence {xn}n=1∞.\{ x_n \}_{n=1}^\infty\text{.} We have that there exists an ϵ>0\epsilon > 0 such that for every n,n\text{,} there exists a point xn∈S∖{c},x_n \in S \setminus \{c\}\text{,} where ∣xn−c∣<1 ⁣/ ⁣n\sabs{x_n-c} < \nicefrac{1}{n} and ∣f(xn)−L∣≥ϵ.\babs{f(x_n)-L} \geq \epsilon\text{.} The sequence {xn}n=1∞\{ x_n \}_{n=1}^\infty just constructed converges to c,c\text{,} but the sequence {f(xn)}n=1∞\bigl\{ f(x_n) \bigr\}_{n=1}^\infty does not converge to L.L\text{.} And we are done.