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Limits

Name: Anonymous 2008-10-06 3:10

Can anyone help me. I suck at limits

here's the problem

Lim  (x-5)/(x^2-25)
x -> 5

Name: 4tran 2008-10-06 6:20

(x-5)/(x^2-25) =
(x-5)/(x-5)(x+5) =
1/(x+5)
(if |x| =/= 5)

HERP DERP

Name: Anonymous 2008-10-06 10:54

You just need to admit you have a problem and are ignoring it unconsciously

Name: Anonymous 2008-10-06 15:19

>>2
The existence condition isn't relevant if you're just looking for the limit, so mentioning it without explanation just serves to confuse.
The rest is true, though, and the limit for x going to 5 is 1/10.

Name: Anonymous 2008-10-06 15:25

i could have solved that with my eyes closed. holy shit

Name: Anonymous 2008-10-06 21:46

>>5
Welcome to /8thgradehomework/.

Name: Anonymous 2008-10-10 6:17

dont u have to test both sides and see if the limit exists on both sides of x=5 to say that the limit is true?

Name: Anonymous 2008-10-10 6:19

>>7
Yes but not in high school.

Name: Anonymous 2008-10-11 2:23

>>7
Not particularly in this case (shit should be obvious) because for a small neighborhood about x=5, 1/(x+5) always has the same signum value (in this case, sgn(f(x))=1.

Name: Anonymous 2008-10-12 5:08

>>9


I don't understand why that implies that limit must be the same on both sides.

Surely the function

f(x) = 1/((x+5) 0<x<5
f(x) = 1 + 1/(x+5) 5<=x<inf

this has the property that for a small neighbourhood around x=5 sgn(f(x))=1 but the limit as x->5 doesn't exist.


I don't get what point you're trying to make by bringing the sign function into this.

Name: 4tran 2008-10-12 18:45

>>10
I might as well also point out that f(x) = x-5 is also continuous at x=5, but does not have the same signum value in a small neighborhood.

The signum test is only useful if there's a (non removable) singularity involved at the limit.

Name: Anonymous 2008-10-13 21:03

\lim_{x -> 5} (x-5)/(x^2-25) =
\lim_{x -> 5} (x-5)/[(x-5)(x+5)] =
\lim_{x -> 5} 1/[(x+5)] = 1/10

Don't change these.
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