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प्रश्न
Evaluate the following limit :
`lim_(x -> "a")[1/(x^2 - 3"a"x + 2"a"^2) + 1/(2x^2 - 3"a"x + "a"^2)]`
उत्तर
`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`
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