Chapter 7 Limits Ex 7.1 Solutions

We have to find some δ so that Here a = 2, l = 1 and f(x) = 2x + 3 Consider ∈ > 0 and |f(x) – l| < ∈ ∴ |(2x + 3) – 7| < ∈ ∴ |2x + 4| < ∈ ∴ 2(x – 2)|< ∈ ∴ |x – 2| < ∴ δ ≤ such that |2x + 4| < δ ⇒ |f(x) – 7| < ∈

We have to find some δ so that Here a = -3, l = -7 and f(x) = 3x + 2 Consider ∈ > 0 and |f(x) – l| < ∈ ∴ |3x + 2 – (-7)| < ∈ ∴ |3x + 9| < ∈ ∴ |3(x + 3)| < ∈ ∴ |x + 3| < ∴ δ < such that |x + 3| ≤ δ ⇒ |f(x) + 7| < ∈

We have to find some δ > 0 such that Here, a = 2, l = 3 and f(x) = x2 – 1 Consider ∈ > 0 and |f(x) – l| < ∈ ∴ |(x2 – 1) – 3| < ∈ ∴ |x2 – 4| < ∈ ∴ |(x + 2)(x – 2)| < ∈ …..(i) We have to get rid of the factor |x + 2| As |x – 2| < δ -δ < x – 2 < δ ∴ 2 – δ < x < 2 + δ Since δ can be assumed as very small, let us choose δ < 1 ∴ 1 < x < 3 ∴ 3 < x + 2 < 5 …..(Adding 2 throughout) ∴ |x + 2| < 5 ∴ |(x + 2)(x – 2)| < 5|x – 2| ……(ii) From (i) and (ii), we get 5|x – 2|< ∈ ∴ x – 2 < If δ = , |x – 2| < δ ⇒ |x2 – 4| < ∈ ∴ We choose δ = min{, 1} then |x – 2| < δ ⇒ |f(x) – 3| < ∈

We have to find some δ > 0 such that 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| < If δ = , |x – 1| < δ ⇒ x2 + x – 2 < ∈ ∴ We choose δ = min{, 1} then |x – 1| < δ ⇒ |f(x) – 3| < ∈