In the following examples, given ∈ > 0, find a δ > 0 such that whenever, |x – a| < δ, we must have |f(x) – l| < ∈: lim (x^2+x+1) =3 (x ∈ 1)

Question

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

\(\lim\limits_{x\longrightarrow1} (x^2+x+1)=3\)

lim (x²+x+1) =3 (x ∈ 1)

Answer 1

We have to find some δ > 0 such that

\(\lim\limits_{x\longrightarrow1} (x^2+x+1)=3\)

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)|< 4 |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| < ∈