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Question
Evaluate the following :
`lim_(x -> "a") [(sin(sqrt(x)) - sin(sqrt("a")))/(x - "a")]`
Solution
`lim_(x -> "a") [(sin(sqrt(x)) - sin(sqrt("a")))/(x - "a")]`
Put `sqrt(x)` = y, `sqrt("a")` = b and y = b + h.
As x → a, y → b and h → 0.
∴ `lim_(x -> "a") (sinsqrt(x) - sinsqrt"a")/(x - "a")`
= `lim_(y -> "b") (siny - sin"b")/(y^2 - "b"^2)`
= `lim_(y -> "b") (siny - sin"b")/((y - "b")(y + "b"))`
= `lim_("h" -> 0) (sin("b" + "h") - sin"b")/("h"("b" + "h" + "b"))`
= `lim_("h" -> 0) (2cos(("b" + "h" + "b")/2)sin(("b" + "h" - "b")/2))/("h"(2"b" + "h")`
= `lim_("h" -> 0) (2cos("b" + "h"/2)*sin("h"/2))/("h"(2"b" + "h"))`
= `lim_("h" -> 0) (cos["b" + ("h"/2)])/(2"b"+"h")* (sin ("h"/2))/("h"/2)`
= `lim_("h" -> 0) (cos["b" + ("h"/2)])/(2"b" + "h")* [lim_("h" -> 0) (sin("h"/2))/("h"/2)]`
= `(cos("b" + 0))/(2"b" + 0)*1 ...[because "h" -> 0"," "h"/2 -> 0 "and" lim_(theta -> 0) sintheta/theta = 1]`
= `cos"b"/(2"b")`
= `cossqrt("a")/(2sqrt("a"))`
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