# 3 questions prove them

1. Mar 8, 2005

### semidevil

let I be an interval in R, let f: I --> R and let c belong to I. suppose there exists a constant K and L such that |f(x) - L <= K|x-c| for x in I. show that the limit as x --> c of f(x) = L.

I"m looking at this, but I dont know how to start. From what I know, this is very very familiar. the sentence ...."suppose there exists a constant K and L such that |f(x) - L| <= K|x-c|," doesn't this pretty much imply the statement?

while I am doing practice problems, I always see this form and as soon as I find a delta and K, it pretty much assumes the answer. I feel that I am close, but I dont know how to go about it.

maybe, let epsilon > 0, then we find a bound for |x - c|, and once we find K, this means |f(x) - L <= K|x-c|. then, we choose delta := inf(delta, 1/k*epsilon),

then, if 0 < | x - c | < delta, it is proved..

close? way off?

lim as x to c, of root(x) = root(c) for c >0.

how would I do that?

can I square both sides? and have | x - c| < e?

2. Mar 9, 2005

### matt grime

YOu can't "find K", it is given in the question.

Given e>0, let d=e/K, then, if |x-c|<d it follows that |f(x)-L| < e.

for the last one sqrt(x) - sqrt(c) = (x-c)/(sqrt(x)+sqrt(c))

so, if |x-c| <d, then c-d < x< c+d.

assume d sufficiently small such that c-d > c/2, so that sqrt(x)+sqrt(c) > sqrt(c)(1+sqrt(3/2)) define this to be K.

then |x-c|<d implies |sqrt(x)-sqrt(c)| < d/K

so, given e>0, pick d sufficiently small such that d/K is less than e, which can be done.