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प्रश्न
Evaluate the following limits:
`lim_(x -> 0) (sqrt(x^2 + "a"^2) - "a")/(sqrt(x^2 + "b"^2) - "b")`
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उत्तर
We know `lim_(x -> "a") (x^"n" - "a"^"a")/(x -"a") = "na"^("n" - 1)`
`lim_(x -> 0)((sqrt((x^2 + "a"^2)) - "a")/(sqrt((x^2 + "b"^2)) - "b")) = lim_(x -> 0)(((x^2 + "a"^2)^(1/2) - ("a"^2)^(1/2))/((x^2 + "b"^2)^(1/2) - ("b"^2)^(1/2)))`
= `lim_(x -> 0) ((x^2 + "a"^2)^(1/2) - ("a"^2)^(1/2))/x^2 xx x^2/((x^2 + "b"^2)^(1/2) - ("b"^2)^(1/2))`
= `lim_(x -> 0) ((x^2 + "a"^2)^(1/2) - ("a"^2)^(1/2))/((x^2 + "a"^2) - "a"^2) xx ((x^2 + "b"^2) - "b"^2)/((x^2 + "b"^2)^(1/2) - ("b"^2)^(1/2))`
= `lim_(x -> 0) ((x^2 + "a"^2)^(1/2) - ("a"^2)^(1/2))/((x^2 + "a"^2) - "a"^2) xx lim_(x -> 0) 1/(((x^2 + "b"^2)^(1/2) - ("b"^2)^(1/2))/((x^2 + "b"^2) - "b"^2)`
Put x2 + a2 = y
Put x2 + b2 = z
When x = 0
⇒ y = a2
When x = 0
⇒ z = b2
= `lim_(y -> "a"^2)((y^(1/2) - ("a"^2)^(1/))/(y - "a"^2)) xx 1/(lim_(x -> "b"^2)(("z"^(1/2) - ("b"^2)^(1/2))/("z" - "b"^2))`
= `1/2 ("a"^2)^(1/2 - 1) xx 1/(1/2 ("b"^2)^(1/2 - 1))`
= `(("a"^2)^(- 1/2))/(("b"^2)^(- 1/2))`
= `("a"^(- 1))/("b"^(- 1))`
= `"b"/"a"`
`lim_(x -> 0)((sqrt((x^2 + "a"^2)) - "a")/(sqrt((x^2 + "b"^2)) - "b")) = "b"/"a"`
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