Question

find the limit (attached)

Answer 1

\[\lim_{x \rightarrow 0} \frac{ \sin2x }{ 4x }\]

Answer 2

im just not sure how to simplify sin2x

Answer 3

do you know the lim x->0 sinx/x = 1

Answer 4

\[1/2\lim_{x \rightarrow 0} \sin2x/2x\]

Answer 5

yes i did know that

Answer 6

\[1/2 * 1 = ?????\]

Answer 7

you can also open sin2x as 2 sin x cos x

Answer 8

yeah, but that more work.. mathmaticians are lazy

Answer 9

she was asking she doesnt know how to simplify sin2x that is why i wrote so.........

Answer 10

\[\frac{\sin(2x)}{4x}=\frac{\sin(2x)}{2x}\times \frac{1}{2}\]

Answer 11

@harsh314 i know that sin2x = 2sinxcosx but i dont know how i can use that in this

Answer 12

oh sorry, what @baldymcgee6 said

Answer 13

\[\frac{ 2 sinx cosx }{ 1 } * \frac{ 1 }{ 4x }\]
is that right?

Answer 14

hope it is clear that saying
\[\lim_{x\to 0}\frac{\sin(x)}{x}=1\] is not different from saying
\[\lim_{x\to 0}\frac{\sin(2x)}{2x}=1\]

Answer 15

lim\[\lim_{x \rightarrow 0} f(x)g(x)=\lim_{x \rightarrow 0}f(x) \times \lim_{x \rightarrow 0}g(x)\]

Answer 16

@harsh314 you're making this more complicated than it has to be :)

Answer 17

haha i understand. thanksss guys.

Answer 18

i am not making it complicated its always better to know more than 1 method
and of course i understand that if x is an angle that approaches 0 which means that the length of the arc it projects diminshes to zero then of course double of that angle would also be tending to 0.....but still in that method some physical errors are there as 2x is not as close as x towards zero...........

Answer 19

@baldymcgee6 i hope you dont mind that

Answer 20

@harsh314 I also think it is a very good idea to have more than one approach to any given problem, but I also think it is important to not over complicate things when they can be written faster and more efficiently.