Mathematics Calculus Epsilon delta definition of Limit | Formal Definition of...

# Epsilon delta definition of Limit | Formal Definition of Limit | with Prove

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In previous article here we have introduced the basic idea of limit. If a function y=f(x) is not defined at a point a, means at a, f(a) become $\frac{0}{0}$, or indeterminate form. Then we approach closer and closer to the point a that the function approaches closer and closer to some value L. Where L is the limit of the function for x→a. However, In this article we will understand more fundamental definition, known as  epsilon delta definition of limit L. According to Limit definition: A function f(x) is defined when x is near the number a, but not a. Also,  f has a value on some open interval that contains a, except possibly at a itself. Then we write:

$$\lim_{x\to a}f(x)=L$$

So the above definition means we can get f(x) “as close as we want” to L by making x sufficiently close to a. Let say we want f(x) to get close to L within some positive number epsilon (ε). Then we need to find x within another number delta (δ) of a that f(x) will be within ε of L.

## Epsilon delta definition of limit

According to limit definition, if for every number ε>0 there is number a number δ>0 such that if |x-a|<δ then |f(x)-L|<ε. where x is any point closer to a within δ, and f(x) is any value of the function near its limit L.

So this definition is trying to refer that if any x is getting closer and close to a within the boundary of δ, then f(x) get closer and close to L within ε. learn more.

#### Now lets see how we can prove that limit exists using epsilon delta definition:

Problem: $\lim_{x\to 5}f(x)=3x+2=17$ prove that the limit exists.

Solution: According to ε-δ definition, if x is any point close to 5 within δ, then any value of f(x) closer to limit 17 within ε. Therefore, |x-5|<δ, as well as |(3x+2)-17|<ε. now let’s prove this relations solving for x.

|(3x+2)-17|<ε,
=> -ε < 3x+2-17<ε,
=> -ε<3x-15<ε,
=> -ε+15<3x<ε+15,
=> 5-$\frac{ε}{3}$< x <$\frac{ε}{3}+5,$
This term is now approximately similar to |x-5|. So to match with this term subtracting 5 from all terms,
$-\frac{ε}{3}$<x-5<$\frac{ε}{3}$, therefore,
|x-5|<$\frac{ε}{3}$. Now comparing this with |x-5|<δ, we can determine that $\frac{ε}{3}$ is equal to δ. So we can say that for any given value ε, our delta is, $δ=\frac{ε}{3}$. Now,
|x-5|<$\frac{ε}{3}$,
=> $-\frac{ε}{3}$< x-5 < $\frac{ε}{3}$
=> $-\frac{ε}{3}+5$< x < $\frac{ε}{3}$+5,
=> 15-ε< 3x < 15+ε,
=> -ε< 3x-15 < ε, this gives the same function we have |(3x+2)-17|<ε. So it means that if we know what ε is, then we can chose our δ. The the value of x within δ will force out function to be in epsilon (ε) of 17. So this satisfy the definition of limit, therefore,

$$\lim_{x\to 5}3x+2=17\quad\text{ … limit does exist.}$$

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