Question

time for another cal problem :)

Answer 1

for what values of a and b is the following true?
\[\lim_{x \rightarrow 0}(\frac{\sin(2x)}{x^3}+a+\frac{b}{x^2})=0\]

Answer 2

i'll give others some time

Answer 3

lol

Answer 4

no, I have no Idea

Answer 5

i'm still looking too

Answer 6

I was trying to remove x from denominator

Answer 7

i have written it one fraction

Answer 8

applied l'hopital (however you spell it)

Answer 9

now i'm working with two equations not sure if it is right yet

Answer 10

obviously ignore the 'a'...save for last

Answer 11

Yeah L'H might be the way to go

Answer 12

i'm not sure yet but b might be 0

Answer 13

nope

Answer 14

lol

Answer 15

oh wait you are right i missed something

Answer 16

b=-2

Answer 17

I know I'm right ;)


that is correct

Answer 18

when i plugged in 0 into cos for some reason i just okay thats 0 instead of 1

Answer 19

you find a?

Answer 20

no don't say it yet

Answer 21

i won't

Answer 22

ok i need to write this over because i just made this look so messy

Answer 23

ok i need to apply l'hospital again

Answer 24

\[\frac{1}{6} (6 a-8 \text{Cos}[2 x])\]

Answer 25

we need -8cos(2x)+6a=0 as x->0

Answer 26

so we have -8+6a=0

Answer 27

so a=8/6=4/3

Answer 28

yep

Answer 29

do you got it imranmeah?

Answer 30

Yeah, L'H three time

Answer 31

I did it 2 times

Answer 32

x^3
3x^2
6x
6

Answer 33

i think i did it 5 times lol

Answer 34

used the fact that

\[\frac{sin(u)}{u}\rightarrow 1\text{ as }u\rightarrow 0\]

Answer 35

nice

Answer 36

for next one ask multivariable calc

Answer 37

you ask one

Answer 38

if you want i mean
i didn't mean to sound bossy lol

Answer 39

no,ones that I find challenging seems like basic to you

Answer 40

i bet thats not true

Answer 41

everything is easy to zarkon

Answer 42

\[\int\limits_{0}^{\infty}\frac{\sin(x)}{x}dx\]

Answer 43

i have to think a little more than he does

Answer 44

myininaya<zarkon

Answer 45

lol

Answer 46

me<myininaya<zarkon

Answer 47

myininaya<<zarkon

Answer 48

the problem i posted above is a multivar calc problem..believe it or not

Answer 49

imranmeah>myininaya

Answer 50

i don't like the one you post it
its too hard

Answer 51

prove that any continuous function with domain [0,1] and range a
subset of [0,1] must have a fixed point

Answer 52

calc 1 problem

Answer 53

But I did figured out the 1=0 one that you posted

Answer 54

can you figure out either of the two I just posted ;)

Answer 55

i;m looking at the second one now

Answer 56

just so we are all on the same page...

A fixed point of a function is a number c in its domain such that f(c) = c

Answer 57

we need to use the intermediate value thm to show that there is a c btw 0 and 1 such that f(c)=c

Answer 58

yes

Answer 59

i dont know how to say it in words...i'll draw a picture <.<

Answer 60

as long as your function is continuous from 0 to 1, f(x) will always cross the line y = x

Answer 61

look at f(x)-x

Answer 62

http://planetmath.org/encyclopedia/BrouwerFixedPointInOneDimension.html

Answer 63

there is some a in [0,1] such that f(a) - a is positive, then we can also find a b such that f(b)-b is negative. So by the IVT, there there exists a point c such that f(c) - c = 0

i still dont feel im saying it correctly...

Answer 64

Let g(x)=f(x)-x
g(0)=f(0)-0
and
g(1)=f(1)-1

but g(1) does not equal g(0)
so by ivt there exist a number between 0 and 1 such f(c)=N

but how do we show N=c?

ok
g(c)=f(c)-c=N-c
0=N-c
c=N

Answer 65

i didnt get to enjoy my orange :( i was too busy thinking about this problem <.< i only recall eating the last piece!

Answer 66

As stated in the article above its a special case of Brouwer's fixed point theorem.

Answer 67

I would start with

if f(0)=0 we are done so assume f(0)>0
if f(1)=1 we are done so assume f(1)<1

let g(x)=f(x)-x

then g(0)>0 and g(1)<0
...

Answer 68

A bigger challenge would be proving the general theorem.

Answer 69

how cute you chose the same letter to represent your function

Answer 70

good question zarkon
how do you the first one you mentioned

Answer 71

\[\int\limits\limits_{0}^{\infty}\frac{\sin(x)}{x}dx \]

Answer 72

rewrite it a double integral

Answer 73

as a double integral

Answer 74

thats above me >.< wah

Answer 75

i don't like calculus 3

Answer 76

i just need to sit down for two months and read a book.

Answer 77

i like Calculus-Calculus 3

Answer 78

I think you should start a new thread. They still haven't fixed the rendering speed for long threads.

Answer 79

\[\sum_{0}^{\infty}(-1)^n(x)^(2n+1)/(2n+1)!*(1/x)\]
\[\sum_{0}^{\infty}(-1)^n(x)^(2n)/(2n+1)!\]
\[\int\limits_{}^{}\sum_{0}^{\infty}(-1)^n(x)^(2n)/(2n+1)!\]
\[[\sum_{1}^{\infty}(-1)^n(x)^(2n)/(2n+1)!(2n)]\]

Answer 80

(-1)^n+1