/sci/ - Prove — World4ch

Name: Anonymous 2009-01-07 5:41

lim as x->1 of 1/(x+3) is 1/4.

Name: Anonymous 2009-01-07 6:35

wat

Name: Anonymous 2009-01-07 7:18

I can.

But I'm sure it'd help both of us more if you tried to do it
yourself first.

Take a good look at your definitions.

Perhaps it's easier if you substitute x with something such
as 1 - (1/n), with n -> inf, and n being natural.

Name: Anonymous 2009-01-07 9:14

>>1

Well 1/(x+3) is continuous at x=1, so the limit is the value of the function.

Everything in that sentence is just as easy to prove as the shitty isolated case you're proving, so why even bother?

Name: Anonymous 2009-01-07 11:01

The proof has to use the epsilon delta definition of a limit

http://en.wikipedia.org/wiki/(%CE%B5,_%CE%B4)-definition_of_limit

Name: Anonymous 2009-01-07 17:59

LOL PLUG IN 1.

Name: Anonymous 2009-01-07 19:04

>>6
Do you know what an epsilon delta proof means, zombified high-achiever?

Name: Anonymous 2009-01-07 19:21

Let \epsilon > 0 and let \delta = 16\epsilon. Then

\left|\frac{1}{(1+\delta)+3} - \frac{1}{4}\right| = \left|\frac{\delta}{4(4+\delta)}\right| < \frac{\delta}{16} = \epsilon

Name: Anonymous 2009-01-07 21:22

no i do not. explain.

Name: Anonymous 2009-01-07 23:28

do not explain epsilon delta. it was a painful day when I learned of those

Name: Anonymous 2009-01-08 11:50

>>8

How can you replace x with 1 + \delta?

Name: Anonymous 2009-01-08 20:10

>>11
for any \epsilon>0, there exists a \delta such that, x \in (1-\delta,1+\delta) \implies f(x) \in (\frac{1}{4}-\epsilon,\frac{1}{4}+\epsilon)

find a \delta for an arbitrary \epsilon.

Prove

1 Name: Anonymous 2009-01-07 5:41

2 Name: Anonymous 2009-01-07 6:35

3 Name: Anonymous 2009-01-07 7:18

4 Name: Anonymous 2009-01-07 9:14

5 Name: Anonymous 2009-01-07 11:01

6 Name: Anonymous 2009-01-07 17:59

7 Name: Anonymous 2009-01-07 19:04

8 Name: Anonymous 2009-01-07 19:21

9 Name: Anonymous 2009-01-07 21:22

10 Name: Anonymous 2009-01-07 23:28

11 Name: Anonymous 2009-01-08 11:50

12 Name: Anonymous 2009-01-08 20:10