(tests solution)
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==tests solution==
 
==tests solution==
 +
1)
 +
    a)
 +
    b)
 +
2)a) f(x) = int(o,sqrtx)of e^t^2 dt
 +
    chain rule
 +
    where u = squrt x
 +
    using funamental therem of calc
 +
    f(u)= int(0,u) of e^t^2 dt
 +
    = e^u^2(1/2*x^(-1/2))
 +
    = (e^x)/(2sqrt(x))
 +
 +
    lim of x as x goes to inf x/f(x)=
 +
    lim of x as x goes to inf 1/f'(x)=
 +
    lim of x as x goes to inf 1/(e^x/(2sqrt(x))=
 +
    lim of x as x goes to inf (2sqrt(x))/(e^x)=
 +
    lim of x as x goes to inf 1/(sqrt(x)e^x)=
 +
    0
 +
3)
 +
 +
4)

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tests solution

1)

    a)
    b)

2)a) f(x) = int(o,sqrtx)of e^t^2 dt

    chain rule 
    where u = squrt x
    using funamental therem of calc
    f(u)= int(0,u) of e^t^2 dt
    = e^u^2(1/2*x^(-1/2))
    = (e^x)/(2sqrt(x))
    lim of x as x goes to inf x/f(x)=
    lim of x as x goes to inf 1/f'(x)=
    lim of x as x goes to inf 1/(e^x/(2sqrt(x))=
    lim of x as x goes to inf (2sqrt(x))/(e^x)=
    lim of x as x goes to inf 1/(sqrt(x)e^x)=
    0

3)

4)

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