Can These Mathematical Proofs Determine the Limit of a Function?

[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 don't 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 don't 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?

how about this one:

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?

 

[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.

 

[M98Ranger]

To prove the statement, we need to show that for any given epsilon > 0, there exists a delta > 0 such that if 0 < |x-c| < delta, then |f(x) - L| < epsilon. This would prove that the limit as x approaches c of f(x) is equal to L.

1. First, let's choose an arbitrary epsilon > 0. We want to find a delta that satisfies the condition above. Since we are given that |f(x) - L| <= K|x-c|, we can rewrite this as |f(x) - L|/|x-c| <= K. This means that for any given epsilon, we can choose delta < epsilon/K.

2. Next, we want to show that if 0 < |x-c| < delta, then |f(x) - L| < epsilon. We can use the condition given to us to rewrite this as |f(x) - L| <= K|x-c|. Since we have already chosen delta < epsilon/K, we can substitute this in and get |f(x) - L| < epsilon/K * |x-c|.

3. Now, we can use the fact that |x-c| < delta to simplify the expression further. We get |f(x) - L| < epsilon/K * delta. Since delta < epsilon/K, we can substitute this in and get |f(x) - L| < epsilon. This completes the proof that the limit as x approaches c of f(x) is equal to L.

For the second question, we can follow a similar approach.

1. Let epsilon > 0 be given. We want to find a delta > 0 such that if 0 < |x-c| < delta, then |√(x) - √(c)| < epsilon.

2. We can rewrite this as |√(x) - √(c)|/|x-c| < K, where K is some constant. We can see that K = 1/√(c) satisfies this condition.

3. Now, we can choose delta < epsilon*K = epsilon/√(c). This satisfies the condition that if 0 < |x-c| < delta, then |√(x) - √(c)| < epsilon.

This proves that the limit as x approaches c of √(x)