Limit (Real Analysis)

The simplified definition of a limit states that f(x) approaches L as x approaches a. But what exactly does approach mean? How close does x get to a?

Definition §

Limit ^definition

Let $E\subseteq \mathbb{R}$ and let $f$ be a real function $f:E\to \mathbb{R}$. Let $a$ be a limit point of $E$. Then we say that $f(x)$ converges to $a$, or equivalently:

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

Iff:

$$\forall ϵ>0,(\exists \delta >0,(\forall x\in E,[0<|x-a|<\delta \to |f(x)-L|<ϵ]))$$

• It’s a bit hard to grasp initially, but the key point is the second bound, $0<|x-a|<\delta$. Since $|x-a|$ is an absolute value, the only time it is equal to zero is when $x=a$. Therefore, for this inequality to hold, $x\ne a$ (which is what is defined casually anyway).

  • This helps avoiding cases where there is discontinuity at the limit point itself, $a$
  • And also it means we do not need to define $f(a)$

• When we get ‘closed enough’ to $a$ (a distance of $\delta$), then the function value should be ‘close enough’ to the limit (a tolerance of $ϵ$)

• In formal proof writing, it helps to not take the left inequality $0<|x-\alpha |$.

Examples §

Prove: For the function $f:\mathbb{R}\to \mathbb{R},f(x)=x^4$

$$\underset{x\to 2}{\lim }{x}^4=16$$

Let $ϵ>0$. We need to show that:

$$|x^4-16|<ϵ$$

We can use polynomial long division to get:

$$|x^4-16|=|x-2||x^3+2x^2+4x+8|$$

Since we use the assumption that $0<|x-2|<\delta$, we know $|x-2|<\delta$.

Let $\delta \le 1$. Then, the greatest value that satisfies the first inequality is $x=3$, meaning we have an upper bound of:

$$|x-2||x^3+2x^2+4x+8|\le |x-2|(3^3+2(3{)}^2+4(3)+8)=65|x-2|$$

To ensure this is always less than $ϵ$, we can just take:

$$65|x-2|<65\delta$$

To show that $65\delta <ϵ$, we just take $\delta <\frac{ϵ}{65}$.

Our actual proof is:

Let $ϵ>0$, and take $\delta =\min (1,\frac{ϵ}{65})$. If $|x-a|<\delta$ then we have:

$$\begin{array}{rl}& \\ |x^4-16|& =|x-2||x^3+2x^2+4x+8|\\ & <65|x-2|\\ & <\frac{65ϵ}{65}<ϵ\end{array}\text{\hspace{1em}(∣x−2∣<δ<1)}$$

Hence proven.

1: Use in Proof

Using the formal definition of a limit, prove that ${\lim }_{x\to 1}2x=2$:

To prove this, we need to show that $|f(x)-L|<ϵ\leftrightarrow 0<|x-a|<\delta$ for $2x$ $|f(x)-L|=|2x-2|=2|x-1|<ϵ$ $2|x-1|=2\times |x-a|<ϵ$ $|x-a|<\frac{ϵ}{2}$ If we choose $\delta =\frac{ϵ}{2}$ then $|2x-2|<ϵ\leftrightarrow 0<|x-1|<\delta$ I.e. $|f(x)-L|<ϵ\leftrightarrow 0<|x-a|<\delta$. Therefore, ${\lim }_{x\to 1}2x=2$