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**kalagota** Q: when you say $\displaystyle a$ is a non negative number and $\displaystyle a<x$ for **any** positive real number, what can you conclude about $\displaystyle a$, i mean, what is a(the) possible value for $\displaystyle a$? (give me an answer..)

**Definition: **Let $\displaystyle a\in X\subseteq\mathbb{R}$ and let $\displaystyle f: X \rightarrow \mathbb{R}$ be a function. We say that *limit of* $\displaystyle f(x)$ *as* $\displaystyle x$ *approaches* $\displaystyle a$ is $\displaystyle L$, denoted by $\displaystyle \lim_{x\rightarrow a} f(x) = L$ if and only if for every **(epsilon)** $\displaystyle \varepsilon > 0$, there is a **(delta)** $\displaystyle \delta>0$ such that $\displaystyle |f(x) - L| < \varepsilon$ whenever $\displaystyle 0<|x-a|<\delta$.

Remark: We shall assume that $\displaystyle L<\infty$, that is, $\displaystyle L$ exists.

Remark: The $\displaystyle \delta$ is usually in terms of $\displaystyle \varepsilon$, that is $\displaystyle \delta$ is a function of $\displaystyle \varepsilon$. Hence, $\displaystyle \delta$ depends on the value of $\displaystyle \varepsilon$.

So how do we use this?

Let us take the first example:

$\displaystyle f(x)=2x+1$

By our first method, we concluded that the *limit of* $\displaystyle f(x)$ is 3 as $\displaystyle x$ goes to 1, and we had that conclusion by observing the behavior of $\displaystyle f(x)$. The formal way to prove it is using the epsilon-delta definition.

Proof:

So, we let $\displaystyle \varepsilon >0$ be an arbitrary positive real number. We want to find a necessary $\displaystyle \delta>0$ (in terms of epsilon) such that if $\displaystyle 0<|x-1|<\delta$, then $\displaystyle |f(x) - 3| < \varepsilon$.

Let us take a look on $\displaystyle |f(x) - L|$.

$\displaystyle |f(x) - L| = |2x+1 - 3| = |2x - 2| = |2||x - 1| =2|x - 1| $ and we want it to be less than epsilon, that is $\displaystyle 2|x - 1| < \varepsilon$.

So let us take $\displaystyle \delta = \frac{\varepsilon}{2}$. Then whenever $\displaystyle 0<|x-1|<\delta=\frac{\varepsilon}{2}$, we have $\displaystyle |f(x) - L| = 2|x - 1| < 2\delta = 2 \cdot \frac{\varepsilon}{2} = \varepsilon$ . Therefore $\displaystyle \lim_{x\rightarrow 1} 2x+1 = 3$.

Can you still follow? Just raise your questions if you have any.

And also, prove that $\displaystyle \lim_{x\rightarrow2}\frac{x^2-4}{x-2} = 4$.