# MATH1151 HELP (1 Viewer)

#### liamkk112

##### Well-Known Member
The sum of all probabilities is 1, so alpha*(1/4+1/4^2+1/4^3....) = 1. Using the formula for a limiting sum of a geometric series, the LHS is
alpha*(1/4)/(1-1/4) = 1/3*alpha. So alpha = 3
oh shoot i thought it was continuous variables yup this is right

For part (a) use the definition of a derivative at the point x = 9, so find an expression for f'(9) = lim_{x->9} ((f(x)-f(9)/(x-9)). You'll end up getting the expression lim_{x->9} ((x-9)^(n-1)*sin(1/(x-9)^2)+x+9). This limit only exists if (x-9)^(n-1) approaches 0 since sin(1/(x-9(^2)) will oscillate rapidly as x approaches 9. This means that n > 1 since (x-9)^0 = 1 as x->9, so the smallest integer is 2. For part (b), f'(9) using the limit found in part (a) is 18, since lim_{x->9} ((x-9)^(n-1)*sin(1/(x-9)^2)+x+9) = 0 + 9 + 9 = 18. For part (c) repeat the process used in part (a) and you should get n = 5
i think this is also right and my response was wrong

#### scaryshark09

##### âˆžâˆ† who let 'em cook dis long âˆ†âˆž

does anyone know how to do part (c)? thanks