Question

In the following example, given ∈ > 0, find a δ > 0 such that whenever, |x – a| < δ, we must have |f(x) – l| < ∈.

`lim_(x -> 1) (x^2 + x + 1)` = 3

Answer 1

We have to find some δ > 0 such that

`lim_(x -> 1) (x^2 + x + 1)` = 3

Here a = 1, l = 3 and f(x) = x² + x + 1

Consider ∈ > 0 and |f(x) – l| < ∈

∴ |x² + x + 1 – 3| < ∈

∴ |x² + x – 2| < ∈

∴ |(x + 2) (x – 1)| < ∈    ...(i)

We have to get rid of the factor |x + 2|

As |x – 1| < δ

– δ < x – 1 < δ

∴ 1 – δ < x < 1 + δ

Since δ can be assumed as very small, let us choose δ < 1

∴ 0 < x < 2

∴ 2 < x + 2 < 4

∴ |x + 2| < 4

∴ |(x + 2) (x – 1)| < 5| x – 1|   ...(ii)

From (i) and (ii), we get

4|x – 1| < ∈

∴ `|x - 1| < ∈/4`

If δ = `∈/4`, |x – 1| < δ ⇒ |x² + x – 2| < ∈

∴ We choose δ = `min{∈/4, 1}` then

|x – 1| < δ ⇒ |f(x) – 3| < ∈