Let $I$ be an interval in $\mathbb{R}$, $f$: $I$$\rightarrow$$\mathbb{R}$ and c$\in$$I$. Suppose there exists constants $K$ and $L$ s.t.
$|f(x)-L|$ $\leq$ $K|x-c|$, $\forall$$x$$\in$$I$. Show that $\lim_{x \to c} f(x) = L$. This is taken from Bartle and Sherbert's Introduction to Real Analysis.
My proof goes like this;
Proof:
Let $(x_n)$ be an arbitrary sequence in $I$ such that $lim(x_n)=c$ and $x_n$$\neq$$c$, $\forall$$x$$\in$$I$. Note that $lim(|x_n-c|)=0$ and $x_n-c\neq0, \forall$$x\in$$I$. So, $|f(x_n)-L|$ $\leq$$K|x_n-c|$. Hence, $lim(f(x_n))=L$ by a theorem. Therefore by the Sequential Criterion,, $\lim_{x \to c} f(x) = L$. $\blacksquare$
My only problem is that I can conclude that $lim(f(x_n))=L$ if $K>0$, but I cannot justify it. Can someone help?
If $K<0$, $0\leq |f(x)-L|\leq K|x-c| < 0$ for any $x\neq c$, which is a contradiction. Therefore, we have $K\geq 0$.
But I do agree with FShrike, why does your expression proves that $\lim_{n\rightarrow \infty}f(x_n) = L$? It is the same expression as given by the question for $x\in I$, so you could "do the same step" and just say $\lim_{x\rightarrow c}f(x) = L$. That is why it feels like skipping the point of the exercise. I'd say you need to show more clearly that $|f(x)-L|$ gets arbitrarily close to $0$.