Question

pls help

(see attachment)

Answer 1

1. lim x->3 (x^2 -2x -3) / (x-3)
If you put x = 3, you get: (3^2 - (2)(3) - 3) / (3 - 3) = 0/0
It is an indeterminate form of the type 0/0 and so we can use L'Hopital's rule.
Differentiate numerator and denominator separately:
lim x->3 (2x - 2) / 1 = 4

Similar procedure for the rest of the problems.

Answer 2

@ranga pls help me in numbers 7,8,and 9 in the 1st attachment :(

Answer 3

For 7, let y = (ln(x))^(x-1)
Take ln on both sides:
ln(y) = (x-1)ln(ln(x)) = ln(ln(x)) / 1 / (x-1)
put x = 1+, you have -infinity / infinity
Apply L'H and find the limit.
ln(y) = the limit that you find.
y = e^(the limit that you find)

Answer 4

@ranga how to answer the calcprobset2 attachment?

Answer 5

I have to log off now but integrate each function and put the upper and lower limits and compute the difference.
If you integrate 3/x^5 you get 3(x^(-5+1)/(-5+1) = -3/4 * 1/x^4
At the upper limit of infinity, it is 0
At the lower limit of 1, it is -3/4.
The difference is 3/4.

Answer 6

@zaisav: Learning how to do these problems yourself is the end goal of all this practice. The more you yourself are involved in finding solutions, the better you'll understand the material. Instead of asking "how to answer the calcprobset2 attachment," why not post one question at a time, do as much work on that question as you can, and then explain what kind of help you think you need? Note that OpenStudy rules state that participants here may neither ask for answers nor provide them.