Looking For Help On Calc Limit Graphs - Topic

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Sep 8 2013 06:52pm

Quote (Azrad @ Sep 9 2013 12:49am)

heh a tricky one.... see at you approach x =1 from both sides... the y value gets closer and closer to 2... so the limit does exist!

so the limit would be x=1 or 2?

Motown

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#12

Sep 8 2013 06:57pm

and one last one!

would it be 0 or 12? would it be 1(6)+6 or -6+6

Azrad

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#13

Sep 8 2013 07:01pm

Quote (FiestaBeast @ Sep 8 2013 05:52pm)

so the limit would be x=1 or 2?

in English we would say "The limit as x approaches 1 of f(x) is 2"
we would write like this

This post was edited by Azrad on Sep 8 2013 07:02pm

brmv

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#14

Sep 8 2013 07:04pm

Quote (FiestaBeast @ 9 Sep 2013 00:52)

so the limit would be x=1 or 2?

with the expectation that lim progresses continuously towards the limit
my answer would be 2

Quote (FiestaBeast @ 9 Sep 2013 00:57)

and one last one!
http://i44.tinypic.com/2cymo11.png
would it be 0 or 12? would it be 1(6)+6 or -6+6

12

now let's see what 'Azrad' says

Motown

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#15

Sep 8 2013 07:06pm

Haha thanks guys, i'm pretty bad at math as you can tell...
but I knew the answer couldn't be 0, seemed fishy..

Azrad

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#16

Sep 8 2013 07:06pm

Quote (FiestaBeast @ Sep 8 2013 05:57pm)

and one last one!
http://i44.tinypic.com/2cymo11.png

For it to be continuous, there can't be any funny holes. Which means f(1) had better equal the lim as x -> 1 of f(x)

a=12 seems to work cuz with that we get:

f(1) = 6

lim as x -> 1 of f(x) = 6

Azrad

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#17

Sep 8 2013 07:09pm

Quote (FiestaBeast @ Sep 8 2013 06:06pm)

Haha thanks guys, i'm pretty bad at math as you can tell...
but I knew the answer couldn't be 0, seemed fishy..

yeah a=0 won't work... if you were to plot it on a graph, you would see why, there would be a "jump" at x=1 (it would jump from -6 to 6 which is certainly not continuous)

This post was edited by Azrad on Sep 8 2013 07:11pm

Motown

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#18

Sep 8 2013 07:11pm

alright I have them all mainly answered
just trying to do that green graph one which is pretty lame cause she threw this curve ball of adding pi

Azrad

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#19

Sep 8 2013 07:11pm

post of your solutions to those 2 and we'll check em before you submit them

Motown

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#20

Sep 8 2013 07:18pm

Quote (Azrad @ Sep 9 2013 12:21am)

1)
Just calculate the area of each green rectangle, then sum them up.

all of the width's for the first problem are pi/4, the heights are gotten by evaluating the function at each x values (x = pi/4, x=pi/2, x=3pi/4)

so the height of the first rectangle is sin(pi/4), and its width is pi/4, so its area is:
A= L * W = Sin(pi/4) *pi/4 = sqrt(2)*(1/2) * pi/4 = sqrt(2)*pi*(1/8)

Do this for all 3 and sum them up and you will get a decent approximation for the area under the curve.

still trying to figure this out
so the work you did there was the area for the first rectangle?

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