Question

LIM (2sinx)/x
x-->0

Answer 1

sin 0 = 0
as 2sin approaches 0 0/x approaches 0

Answer 2

why 0/x @ehuman , why not 0/0 which is undefined if we plug zero into f(x)?

Answer 3

0<sin x<1
0<2sin x<2

0=<2sin x/x<2/x
take the limit for all fun as x-->0

lim 0 =0
lim 2/x =0 = lim 0
then by the sandwich roll " i dnt know if its really name "
lim 2sin x/x =0

Answer 4

good point, then it would be undefined once it reached zero, but would approach zero until it reached it and became undefined.

Answer 5

asymptote at x=0

Answer 6

@ehuman u cant take the direct limit since it become 0/0 so undefined limit

Answer 7

ok, so we would describe the trend as it approaches 0 meaning the limit is 0?

Answer 8

or rather limit is undefined?

Answer 9

CHECK THIS OUT GUYS.....0+ X=0.1 f(x) = 0.998334
0.01 0.999983
0.001 0.9999999983
0.0001 0.9999999999983

0- x= -0.1 f(x) = 0.998334
-0.01 0.999983
- 0.001 0.99999983

thus, if you are to compute the values of sinx/x, you will find that as x approaches 0, f(x) is actually 1

Answer 10

snap
same numerator and denominator (virtually)

Answer 11

intersting..

Answer 12

i checked from the num is som how i got a neybourhood of 0.034

Answer 13

did you compute correctly, am not really sure how u got 0.034

Answer 14

http://www.wolframalpha.com/input/?i=lim+2sinx+%2Fx+as+x-%3E0

Answer 15

hummm its just use the rol of
lim sin x /x =1 when x->0

Answer 16

if you're allowed to use L'hospital rule

LIM (2sinx)/x
x-->0

LIM (2cosx)/1
x-->0

take the limit now

2

Answer 17

@ikram002p thank u for introducing me to wolframalpha, thanx. @ganeshie8 , l'hospita rule is awesome, not permissible in my syllabus, but a useful tool to prove right/wrong.

Answer 18

@ganeshie8 i really need to know how you got 2cosx/1

Answer 19

cool :)
L'hospital rule says this :-

if f(x) = g(x) = 0,

\(\lim \frac{f(x)}{g(x)} = \lim \frac{f'(x)}{g'(x)}\)

Answer 20

(2sinx)' = 2cosx

(x)' = 1

Answer 21

oh, ok, thanx, i see, said th blind man....