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प्रश्न
Evaluate the following limit :
`lim_(x -> "a") [(sqrt("a" + 2x) - sqrt(3x))/(sqrt(3"a" + x) - 2sqrt(x))]`
उत्तर
`lim_(x -> "a") (sqrt("a" + 2x) - sqrt(3x))/(sqrt(3"a" + x) - 2sqrt(x))`
= `lim_(x -> "a") (sqrt("a" + 2x) - sqrt(3x))/(sqrt(3"a" + x) - 2sqrt(x)) xx (sqrt("a" + 2x) + sqrt(3x))/(sqrt("a" + 2x) + sqrt(3x)) xx (sqrt(3"a" + x) + 2sqrt(x))/(sqrt(3"a" + x) + 2sqrt(x))`
= `lim_(x -> "a") ([("a" + 2x) - 3x](sqrt(3"a" + x) + 2sqrt(x)))/([(3"a" + x) - 4x] (sqrt("a" + 2x) + sqrt(3x))`
= `lim_(x -> "a") (("a" - x)(sqrt(3"a" + x) + 2sqrt(x)))/(3("a" - x)(sqrt("a" + 2x) + sqrt(3x))`
= `lim_(x -> "a") (sqrt(3"a" + x) + 2sqrt(x))/(3(sqrt("a" + 2x) + sqrt(3x))) ...[(because x -> "a" "," x ≠ "a"),(therefore "a" - x ≠ 0)]`
= `(lim_(x -> "a") (sqrt(3"a" + x) + 2sqrt(x)))/(lim_(x -> "a") [3(sqrt("a" + 2x) + sqrt(3x))]`
= `(sqrt(3"a" + "a") + 2sqrt("a"))/(3(sqrt("a" + 2"a") + sqrt(3"a"))`
= `(4sqrt("a"))/(3(2sqrt(3"a"))`
= `2/(3sqrt(3))`
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