# Math Help - proving a limit?

1. ## proving a limit?

I'm working through limit questions and i think I've got them down.....but how do i go about giving a formal proof of a limit.

lim x->1 $\sqrt{x}$ = 1

I know it does...but how do you go about providing a formal proof?

Thanks

2. ## idea

Originally Posted by SimonK
I'm working through limit questions and i think I've got them down.....but how do i go about giving a formal proof of a limit.

lim x->1 $\sqrt{x}$ = 1

I know it does...but how do you go about providing a formal proof?

Thanks
your question's answer involves the basic concept of limits. limits are used when a function has a discontinuity at some point and it cannot be evaluated at that point by directly plugging in any value in the domain. take for example lim x->2(x^2-4)/(x-2).here if we simply put the limit we get an undefined expression. instead if we cancel out the factors we get a definite value. another important point is that when a function is said to have a limit it means that at the neighbourhood of that point (however close be the neighbourhood points doesnot matter because by definition of limit the fuction only approaches that point and not gets it)the function is defined. it would be good if we consult a textbook once again.

3. I'm sorry.....what are you saying?
I understand what a limit is....but i'm not certain how to state this as a formal proof as it relates to the question.
I know if we graph the function it is very easy to see that it approaches 1 as x->1.....but how do you prove it?
I'm also not sure how the example you have given applies to this question?

4. Are you saying that I should try to show that a value on the domain either side of X=1 will be very close to 1?

5. ## suggestion

Originally Posted by SimonK
Are you saying that I should try to show that a value on the domain either side of X=1 will be very close to 1?
i wouldnot be able to give you a formal proof. but we could get a broad idea from the term :neighbourhood of the point in question. ok.let me try this in your case:
let us try the proof by contradiction way.
let us suppose that no limit exists for sqrt(x) ie. the expression doesnot converge to any value ie. it doesnot have any bounds. this is false. so perhaps something should come out. you could also go the way you suggested yourself. but using convergence and series willl be a better way because the concept of limit has come from it.

6. thats a good idea. I'll start with that

Thanks