In previous article ** here** we have introduced the basic idea of limit. If a function

**is not defined at a point**

*y=f(x)***, means at**

*a***,**

*a***become $\frac{0}{0}$, or**

*f(a)***. Then we approach closer and closer to the point**

*indeterminate form***that the function approaches closer and closer to some value**

*a***. Where L is the limit of the function for**

*L***. However, In this article we will understand more fundamental definition, known as epsilon delta definition of limit**

*x→a***. According to**

*L***A function**

__Limit definition:__**is defined when**

*f(x)***is near the number**

*x***, but not**

*a***. Also,**

*a***has a value on some open interval that contains**

*f***, except possibly at**

*a***itself. Then we write:**

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

So the above definition means we can get ** f(x)** “

*” to*

__as close as we want__**L**by making

**sufficiently close to**

*x***. Let say we want**

*a***to get close to L within some positive number epsilon (**

*f(x)***). Then we need to find**

*ε***within another number delta (**

*x***) of**

*δ***that f(x) will be within**

*a***of**

*ε***.**

*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

**then**

*|x-a|<δ***. where x is any point closer to a within δ, and f(x) is any value of the function near its limit L.**

*|f(x)-L|<ε**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

**, if**

*ε-δ definition***is any point close to**

*x***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**

*5**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,

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