Question

limx-->0 of (a-cosbx)/x^2 =8 find a and b
without lhopital

Answer 1

we know that sin x / x as x->0 =1 and

Answer 2

and lim (1 - cos x ) / x = 0 as x->0

Answer 3

I would use Lhopitals

lim (-(-sin bx ) *b / (2x ) = 8

Answer 4

i know using lhopital,but without it how can i start it

Answer 5

let b = 4

Answer 6

if b = 4 then
limx-->0 of (a-cosbx)/x^2
=lim (-(-sin bx ) *b / (2x )
= lim sin(4x)*4 / (2x)
= lim sin(4x)*4*2 / (2*2x)
= lim sin(4x)/(4x) * 8
= 8 * lim sin(4x)/4x)
= 8 *1

note that a = any constant . pick 0 if you want

Answer 7

thank you

Answer 8

see if you can generalize this problem

Answer 9

limx-->0 of (a-cosbx)/x^2 =k

find a,b

Answer 10

I thought he wanted it without using L'Hospitals

Answer 11

right

Answer 12

so the question is still open

Answer 13

clearly a=1 otherwise the limit would not exist

then multiply by \(1+\cos(bx)\) to the top and bottom and simplify

Answer 14

good point, if a equals zero that the limit diverges

Answer 15

i was thinking about the squeeze thm, however the =8 is throwing me off. I wanted to
say -1/x^2 <_ a-cosbx/x^2 <_ 1/x^2, but that does nothing

Answer 16

n121 i would use zarkon's suggestion

Answer 17

squeeze theorem probably won't be of use here

Answer 18

Zarkon's suggestion makes a nice tidy simplification

Answer 19

i will try

Answer 20

also use the suggestion that a = 1, otherwise the limit diverges (do you see why?)

Answer 21

yes