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प्रश्न
Evaluate the following limits: `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)/((3"a" + x) - 4x) xx (sqrt(3"a" + x) + 2sqrt(x))/(sqrt("a" + 2x) + sqrt(3x))]`
= `lim_(x -> "a")[("a" - x)/(3"a" - 3x) xx(sqrt(3"a" + x) + 2sqrt(x))/(sqrt("a" + 2x) + sqrt(3x))]`
= `lim_(x -> "a")[(-(x - "a"))/(-3(x - "a")) xx (sqrt(3"a" + x) + 2sqrt(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 x - "a" ≠0)]`
= `(sqrt(3"a" + "a") + 2sqrt("a"))/(3(sqrt("a" + "2a") + sqrt(3"a"))]`
= `(sqrt(4"a") + 2sqrt("a"))/(3(sqrt(3"a") + sqrt(3"a"))`
= `(2sqrt("a") + 2sqrt("a"))/(3(2sqrt(3"a"))`
= `(4sqrt("a"))/(6sqrt(3) sqrt("a")`
= `2/(3sqrt(3)`
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