# Calculate limit x tend to 0 1/x integral sign from x+3 to 2x+3 of tsquare root(t^3+9)dt?

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### 1 Answer

You should remember how to evaluate the definite integral of the function `int_a^b f(x) dx = F(b) - F(a).`

Hence, `lim_(x-gt0) (1/x)int^(x+3)^(2x+3) tsqrt(t^3+9)dt = lim_(x-gt0) (F(2x+3) - F(x+3))/x`

Substituting 0 for x in the limit yields:

`lim_(x-gt0) (F(2x+3) - F(x+3))/x = (F(2x+3) - F(x+3))/x = 0/0`

You should use l'Hospital's theorem such that:

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

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

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

`lim_(x-gt0) (2f(2x+3)-f(x+3)) = 2f(3) - f(3) = 3sqrt(27+9)`

`lim_(x-gt0) (2f(2x+3)-f(x+3)) = 3*6 = 18`

**Hence, evaluating the given limit yields`lim_(x-gt0) (1/x)int^(x+3)^(2x+3) tsqrt(t^3+9)dt = 18` .**

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