# Formal Definition of a Limit at a Point

## Key Questions

See below

#### Explanation:

The definition of limit of a sequence is:

Given $\left\{{a}_{n}\right\}$ a sequence of real numbers, we say that $\left\{{a}_{n}\right\}$ has limit $l$ if and only if

AA epsilon>0, exists n_0 in NN // AAn>n_0 rArr abs(a_n-l))< epsilon

• Before writing a proof, I would do some scratch work in order to find the expression for $\delta$ in terms of $\epsilon$.

According to the epsilon delta definition, we want to say:

For all $\epsilon > 0$, there exists $\delta > 0$ such that
$0 < | x - 1 | < \delta R i g h t a r r o w | \left(x + 2\right) - 3 | < \epsilon$.

$| \left(x + 2\right) - 3 | < \epsilon \Leftrightarrow | x - 1 | < \epsilon$

So, it seems that we can set $\delta = \epsilon$.

(Note: The above observation is just for finding the expression for $\delta$, so you do not have to include it as a part of the proof.)

Here is the actual proof:

Proof

For all $\epsilon > 0$, there exists $\delta = \epsilon > 0$ such that
0<|x-1| < delta Rightarrow |x-1|< epsilon Rightarrow |(x+2)-3| < epsilon

• Precise Definitions

Finite Limit
${\lim}_{x \to a} f \left(x\right) = L$ if
for all $\epsilon > 0$, there exists $\delta > 0$ such that
$0 < | x - a | < \delta R i g h t a r r o w | f \left(x\right) - L | < \epsilon$

Infinite Limits
${\lim}_{x \to a} f \left(x\right) = + \infty$ if
for all $M > 0$, there exists $\delta > 0$ such that
$0 < | x - a | < \delta R i g h t a r r o w f \left(x\right) > M$

${\lim}_{x \to a} f \left(x\right) = - \infty$ if
for all $N < 0$, there exists $\delta > 0$ such that
$0 < | x - a | < \delta R i g h t a r r o w f \left(x\right) < N$