Question

Evaluate the following limit :

`lim_(x -> "a")[1/(x^2 - 3"a"x + 2"a"^2) + 1/(2x^2 - 3"a"x + "a"^2)]`

Answer 1

`lim_(x -> "a")[1/(x^2 - 3"a"x + 2"a"^2) + 1/(2x^2 - 3"a"x + "a"^2)]`

Consider,

x² – 3ax + 2a² = x² – 2ax – ax + 2a²

= x(x – 2a) – a(x – 2a)

= (x – 2a) (x – a)

2x² – 3ax + a² = 2x² – 2ax – ax + a² 

= 2x(x – a) – a(x – a)

= (x – a) (2x – a)

∴ `lim_(x -> "a") [1/(x^2 - 3"a"x + 2"a"^2) + 1/(2x^2 - 3"a"x + "a"^2)]`

= `lim_(x -> "a") [1/((x - 2"a")(x - "a")) + 1/((x - "a")(2x - "a"))]`

= `lim_(x -> "a") ((2x - "a") + (x - 2"a"))/((x - 2"a")(x - "a")(2x - "a"))`

= `lim_(x -> "a") (3x - 3"a")/((x - 2"a")(x - "a")(2x - "a"))`

= `lim_(x -> "a") (3(x - "a"))/((x - 2"a")(x - "a")(2x - "a"))`

= `lim_(x -> "a") 3/((x - 2"a")(2x - "a"))  ...[(because  x -> "a""," therefore x ≠ "a"),(therefore x - "a" ≠ 0)]`

= `3/(("a" - 2"a")(2"a" - "a"))`

= `3/((-"a")("a"))`

= `(-3)/"a"^2`