# Explain how apply Hospital in calculate limit x->0 (integral between x+3 and 2x+3 of t squareroot(t^3+9))/x?(without calculate integral )

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### 1 Answer

You need to calculate the limit such that:

`lim_(x-gt0) (1/x) int_(x+3)^(2x+3) t*sqrt(t^3+9)dt = lim_(x-gt0) (F(2x+3) - F(x+3))/x`

Notice that integral was substituted by F(2x+3) - F(x+3).

Plugging x = 0 in the limit yields:

`lim_(x-gt0) (F(0+3) - F(0+3))/0 = 0/0`

This result is so called indeterminate form, hence you may use l'Hospital's rule such that:

`lim_(x-gt0) (F(2x+3) - F(x+3))/x = lim_(x-gt0) ((F(2x+3) - F(x+3))')/(x)'`

`lim_(x-gt0) ((F(2x+3) - F(x+3))')/(x)' = lim_(x-gt0) (F'(2x+3)(2x+3)' - F'(x+3)(x+3)')/1`

`int f(x)dx = F(x)+c=> F'(x) = f(x)=> `

`=>lim_(x-gt0) (F'(2x+3)(2x+3)' - F'(x+3)(x+3)') = lim_(x-gt0) (2f(2x+3) - f(x+3))`

`lim_(x-gt0) (2f(2x+3) - f(x+3)) = 2f(3) - f(3) = f(3)`

`f(3) = 3*sqrt(3^3 + 9) = 3sqrt36 = 3*6 = 18`

**Hence, evaluating the limit using l'Hospital's rule yields: `lim_(x-gt0) (1/x) int_(x+3)^(2x+3) t*sqrt(t^3+9)dt = 18.` **