Question

Let h(x) be the function that rounds down numbers, e.g. h(1.5)=1, h(2)=2, h(╥)=3.

a. Compute lim x-->10- h(x) and lim x-->10+ h(x).

b. show that h(x) is continuous at x=10/3. (hint: think of a simple formula for h(x) when x is in (3,4).)

THANK YOU SO MUCH FOR LOOKING...if you can help in any way, it would be greatly, EXTREMELY appreciated!

Answer 1

as x approaches 10 from below...h(x) approaches 9.5

as x approaches 10 from above...h(x) approaches 10...so x = 10 is a discontinuous point

b) as x approaches 10/3 (roughly 0.33) from above, h approaches 0, as x approaches 10/3 from below, h approaches 0. so it's continuous.

Answer 2

a) 10-h(x) = 9
10+ h(x) = 10
h(x) = integral factor of x ie removing the fractional part
b) since for all values between 3&4 we will get a straight line parallel to -axis
hence it is continuous at 10/3

Answer 3

THANK YOU jamesm!

Answer 4

THANK YOU Nitin608!

Answer 5

i am so careless...10/3 = 0.333?? in what universe is that true???

sorry. 10/3 is roughly 3.33, so h approaches 3 from below and above

Answer 6

LOL its ok. Your help is greatly appreciated. I wish I had a fraction of your intelligence in this field.

Answer 7

jamesm, what method did you use to get the answers? so that i may try and figure it out myself...

Answer 8

i just used some intuition i guess. there was no systematic method.

for x between 3 and 4, you can write h like so:

h(x) = 3.5, 3.5 < x < 4
h(x) = 3, 3 < x < 3.5

so draw the graph of h against x. as you follow h in the interval [3.3.5] you can see i approaches 3 from the left as x gets closer to 3.5. and similar for the other interval.

you c

Answer 9

thank you again jamesm