Question

lim as x approaches 4 of (2x^2+32)

Answer 1

replace x by 4

Answer 2

lol satellite this wasn't for you, come on..

Answer 3

lgba go ahead and give me the answer.. then i'll show you something else..

Answer 4

maybe you meant lim x-> 4 of (2x^2 - 32)?

Answer 5

no i meant what i posted, just give me the answer..

Answer 6

64

Answer 7

All right, that's what i was getting at. Now prove that the lim as x->4 (x^2+32) = 64 lol

Answer 8

*sorry meant 2x^2+32

Answer 9

i already proved this, just wanna see your approach

Answer 10

prove that 2x^2 + 32 = 64?

Answer 11

lol satellite's answer was kinda embarrassing for me... :/ a junior in college and a math major and can't solve limits

Answer 12

No, prove that the lim as x->4 of 2x^2 +32 = 64

Answer 13

U might not be familiar with the proof if u didn't take advanced calc.. i actually liked how the proof works

Answer 14

With epsilons and deltas? The formal way of doing limits is so boring.

Answer 15

my proof is that polynomials are continuous functions, cue ee dee

Answer 16

by the way.. hint:
delta = min {1, epsilon/18}

Answer 17

yea lol rogue.. haha satellite

Answer 18

my proof is the definition of limits

Answer 19

that's for derivatives lol

Answer 20

meh.. i guess we can stop here then. haha just factor |2x^2+32-64|

Answer 21

Then u'll end up with:
2|x-4||x+4|<epsilon
|x-4||x+4|<epsilon /2

Answer 22

Suppose |x-4| < 1 (let delta = 1)

Answer 23

too much math terms @_@ error error error

Answer 24

that means:
-1 < x-4 < 1
adding 8 to everything gives:
7 < x+4 < 9
So you get:
|x-4||x+4|<|x-4|*9 (since |x+4|<9)