Jump to content
Mental Support Community
  • entries
  • comments
  • views

gosh i feel dumb...



not feeling too much like myself today, got some math homework covering things that we worked on last year and have just totally blanked on all of it.

it looks like a foreign language to me.

its not just math homework either. its every subject. i am looking at my homework as if it might as well be written in chinese.

i don't know what is up today, but i literally can not think at all.

i think im done with this homework, i'll just take the F and move on...


Recommended Comments

The same thing happens to me every once in a while. In my experience the best way to fix it is to get out of the house for a while and clear your head. Even just going to the store usually helps.

Although using pseud as a calculator works too :P

Link to comment

sorry pseud, i don't think that that'll do me much good at the moment, i just can't figure out anything from any of my subjects, i must be preoccupied with something, but i just can't figure out what is gnawing at me right now.

im just feeling really down today, nothing bad has happened, but all my thoughts are like clouded over and concentrating is something of a wonder.

Link to comment

ok its given as lim x->2 (3x+2)

and another problem of lim x->2 (x^2 - 3)


find limit L. then find "upside down g symbol" > 0 such that |f(x) - L| < 0.01 whenever 0 < |x-c| < "upside down g symbol"

i can find the limit, but that's all i can do :o

Link to comment

Yeah, yeah--c is 2, the thing x was approaching.

Ok, so the first one goes something like this:

You know already that lim x->2 of f(x) = 3x+2 = 8

so |f(x) - L| = |3x + 2 - 8| = |3x - 6|

So you need to find out what x has to be to satisfy:

|3x - 6| < 0.01

the absolute value is less than 0.01 as long as what's inside the absolute value bars lies between -0.01 and 0.01:


add 6 all the way through, then divide by 3 all the way through:

5.99/3 < x <6.01/3

x-c = x-2 so 5.99/3 - 2 < x - 2 < 6.01/3 -2

hence 0< |x -c| < delta-------> the delta is in this case .00333333333333..... (because that's what 6.01/3 - 2 is and what |5.99/3 - 2| is. Note that if those two yeilded different numbers, you'd take the smaller one.)

Does that make sense?

Link to comment

lim x->2 of f(x) = x^2 - 3 = 1

so |f(x) - L| = |x^2 - 3 - 1| = |x^2 - 4|

|x^2 - 4|<0.01 => -0.01< x^2 - 4 < 0.01

you need to solve for x. Add 4 all the way through:


sqrt all the way through, but note that x could be negative:

sqrt(3.99)< x < sqrt(4.01) or -sqrt(3.99)>x>-sqrt(4.01). I don't think the negative solutions actually matter.

now for |x-2| < delta, we need sqrt(3.99) - 2 < x - 2 < sqrt(4.01) - 2

which is something like -.00250.... < x-2 < .002498......

so you want to pick delta to be the smaller (in absolute value) of these. And you probably want an exact number, so leave it as delta = sqrt(4.01) - 2. Though any delta smaller than this also works...

Again, basically these questions are asking you how small you have to squish the x range down so that the f(x) stays within .01 of the limit as x approaches 2.

Link to comment

Join the conversation

You are posting as a guest. If you have an account, sign in now to post with your account.
Note: Your post will require moderator approval before it will be visible.

Add a comment...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

  • Create New...