# Limit Of A Function From Definition

by Juggler123
Tags: definition, function, limit
 P: 82 I have to prove from the definiton of the limit of a function that the limit of f(x) as x tends to 2 equlas 3, given that; f(x)=(2x-1) I know that I have to find an epsilon such that |f(x)-l| $$\leq$$ $$\epsilon$$ and delta such that 0 $$\leq$$ |x-a| $$\leq$$ $$\delta$$ Nowing putting in the conditions for this f(x); |2x-4| $$\leq$$ $$\epsilon$$ and 0 $$\leq$$ |x-2| $$\leq$$ $$\delta$$ But I don't where to go from here. Any help would be great!
 Sci Advisor HW Helper PF Gold P: 12,016 Well, you have: |2x-4|=2|x-2| Can this be made arbitrarily small (i.e, below some fixed "epsilon"), as long as we pick a small enough "delta"? In particular, IF we want $2|x-2|\leq\epsilon$, what is the largest value we can pick for "delta" in order to ENSURE the validity of the above inequality?
 P: 49 Often, we need to find $$\delta = \delta(\epsilon)$$ as a function of the $$\epsilon$$ we were given. That is, write $$|f(x)-l|$$ as a function $$|x-a|$$ and use your delta to show that it is less than epsilon
 P: 82 Limit Of A Function From Definition So I know that |f(x)-l|=2|(x-2)|=2|x-a| Would it then be right to say that since |x-a|=|x-2| $$\leq$$ $$\delta$$ then |f(x)-l|=2|x-2| $$\leq$$ 2$$\delta$$ hence taking 2$$\delta$$ $$\leq$$ $$\frac{\epsilon}{2}$$ would satisfy the problem??
 Quote by Juggler123 So I know that |f(x)-l|=2|(x-2)|=2|x-a| Would it then be right to say that since |x-a|=|x-2| $$\leq$$ $$\delta$$ then |f(x)-l|=2|x-2| $$\leq$$ 2$$\delta$$ hence taking 2$$\delta$$ $$\leq$$ $$\frac{\epsilon}{2}$$ would satisfy the problem??
 P: 82 Yeah that is what I meant! Woops. So is the following proof correct then; Definition of the Limit of a Function at a Point. Suppose that f(x) is defined on (a-R,a)U(a,a+R), for some R $$\succ$$ 0. Then f(x) tends to l as x tends to a if, given any $$\epsilon$$ in R$$^{+}$$ , there exists $$\delta$$ in R$$^{+}$$ such that, |f(x)-l| $$\prec$$ $$\epsilon$$ whenever 0 $$\prec$$ |x-a| $$\prec$$ $$\delta$$ Now in this case f(x)=2x-1, l=3 and a=2 Hence |2x-4| $$\prec$$ $$\epsilon$$ Now 2|x-2| $$\prec$$ $$\epsilon$$ = |x-a|$$\prec$$ $$\delta$$ Hence taking $$\delta$$ $$\leq$$ $$\frac{\epsilon}{2}$$ Then |f(x)-l| $$\prec$$ $$\epsilon$$ whenever 0 $$\prec$$ |x-a| $$\prec$$ $$\delta$$ As required