Question

How do you find the limit using the epsilon delta function?

Answer 1

In this question we will work with the epsilon delta function by solving the example using the same. We must know that the general definition of the epsilon delta function is:
$\displaystyle \lim_{x \to a}f(x)=\text{L if }\forall \varepsilon >0,\exists \delta >0:$
$0<|x-a|<\delta \Rightarrow |f(x)-L|<\varepsilon$

Complete step-by-step answer:
The epsilon delta function which is generally represented as the $(\varepsilon -\delta )$ function in mathematics is the property of limits which is used to simplify linear limit questions. This proof is only limited to linear problems which means that there is no power to any term in the whole expression and there are no fractions present in the expression.
The general definition of the epsilon delta function is:
$\displaystyle \lim_{x \to a}f(x)=\text{L if }\forall \varepsilon >0,\exists \delta >0:$
$0<|x-a|<\delta \Rightarrow |f(x)-L|<\varepsilon$
Which states that when there is a limit tending to $a$ of a function has a value $L$ for all the values of $\varepsilon$ which are greater than $0$, if there exists $\delta$ which is greater than $0$.
And the value of $|x-a|$ is between $0$ and $\delta$ which implies that value of $|f(x)-L|$ is less than $\varepsilon$.
The epsilon delta function is used to avoid most of the rigor in formal mathematics.
The proof shoes that: $0<|x-a|<\delta \Rightarrow |f(x)-L|<\varepsilon$
Consider the polynomial expression $f(x)={{x}^{2}}-3x+2$
As $x \to 0$, the value of $f(x)\to 0$
Now using the epsilon delta function we need to show that $0<|x-a|<\delta \Rightarrow |f(x)-L|<\varepsilon$
Now observe that:
$\Rightarrow f(x)-2={{x}^{2}}+3x+2-2$
Which can be written as:
$\Rightarrow f(x)-2={{x}^{2}}+3x$
Now this expression can be written in the factorized format as:
$f(x)-2=(x-0)\times (x+3)$
Now from the expression if $(x-0)<\delta$, then you have $(x-0)\times (x+3)<\delta \times (x+3)<\varepsilon$
Now as $x \to 0$, you have $\delta \times (0+3)=3\delta <\varepsilon$ therefore, we choose $\delta =\dfrac{\varepsilon }{3}$.

Note: It is to be remembered that the various symbols which are used to write the epsilon delta function. The symbol $\forall$ stands for “for-all” which means that consider all the elements in the set and $\exists$ sign, which means “there-exists' ' which means that the solution exists in the set.