Question

lim [abs(x-7)] / [x-7]
x -> 7 (-)

Answer 1

how do you deal with the (-)? i think that means from the left? right?

how do you find the limit form only one direction?

Answer 2

and how does that compare to:

lim [abs(x-7)] / [x-7]
x -> 7

where you find the limit form both directions?

Answer 3

(-) means take its limit from the left

Answer 4

and this means that you should multiply the absolute value by -1

Answer 5

7-x/x-7=-1

Answer 6

so, is the limit from the left -1?

Answer 7

but the limit from both sides is positive 1? right?

Answer 8

yes, it would be +1 from the right which means limit does not exist at all, for limit from the left is not equal to the limit from the right

Answer 9

'cause you mulitply it by 1?

Answer 10

yes

Answer 11

aha!

Answer 12

in fact the point is not to multiply it by 1 but to understand the absoulte value will be the same if the limit is taken from the right side

Answer 13

I have one more question...

Answer 14

it's about the intermediate value theorem

Answer 15

f(x) = x^3*cos(x) - x*sin(x) + 1

in the interval [-2,2]

they want me to prove that there are at least two zeros within the interval... what does that mean? like, where they cross the axis? do you plug in zeros for the x? how do you find more than one zero? there's only one solution if you plu in the zero?

Answer 16

or, should i post this as a question?

Answer 17

please let me check for a second

Answer 18

plug in the x_values -pi/2, 0, pi/2

f(-pi/2) = 0 - (-pi/2)(-1) +1 = -.57
f(0) = 0 -0+1 = 1
f(pi/2) = 0 -(pi/2)(1) +1 = -.57

f(x) goes from - to + to -, since it is a continuous function it proves f(x) = 0 at least twice in that interval