Question

`lim_(x→0^+)(int_0^(x^2)(sinsqrt("t"))"dt")/x^3` is equal to ______.

Options

• `2/3`

• 0

• `1/15`

• `3/2`

Answer 1

`lim_(x→0^+)(int_0^(x^2)(sinsqrt("t"))"dt")/x^3` is equal to `underlinebb(2/3)`.

Explanation:

Given: `lim_(x→0^+)(int_0^(x^2)(sinsqrt("t"))"dt")/x^3`

Apply Leibniz's rule:

`"d"/("d"x)int_("f"(x))^("g"(x)) "u"("t")"dt" = "u"("g"(x))."g"^'(x)-"u"("f"(x))."f"^'(x)`

∴ `"d"/("d"x)int_0^(x^2) sinsqrt("t")"dt" = sin(sqrt(x^2)).2x - sin(0).0 = sin(|x|).2x`

Apply t Hospital's rule in

`lim_(x→0^+)(int_0^(x^2)(sinsqrt("t"))"dt")/x^3`

= `lim_(x→0^+) ("d"/("d"x)int_0^(x^2)(sinsqrt("t"))"dt")/("d"/("d"x)(x^3))`

= `lim_(x→0^+)(2x.sin(|x|))/(3x^2)`

= `2/3lim_(x→0^+)(sin|x|)/x`

= `2/3 xx 1 = 2/3`  ...`(∵ lim_(x→0) sinx/x = 1)`