Question

who can help me in limit theorem of trigonometric functions?

Answer 1

do u have some examples ? then post them i'l see them and try work them out

Answer 2

lim sin3(teta) over teta where teta approaches 0

Answer 3

I don't think this is right. It is probably \[sin^3(\theta)\] not \[sin(3\theta)\]

Answer 4

\[\lim{x\to0}\frac{sin^3(x)}{x}\]

Answer 5

And you need to then apply l'hopitals rule

Answer 6

The limit is 0

Answer 7

Yes but your answer is incorrect, sorry.

Answer 8

The limit is 0

Answer 9

One second

Answer 10

Yes but you said it was \[\theta^3\]

Answer 11

The limit is 0 and you need to apply l'hopitals rule.

Answer 12

wit 1 sec ei need to cinfirm

Answer 13

oh yes, i'm sorry it would be zero.....

Answer 14

for theta^3 but wt abt sin(Ntheta)

Answer 15

it should be zero too

Answer 16

i did it all wrong .... lol

Answer 17

thanks alchemista

Answer 18

this is the equation:

Evaluate:
Lim sin3Θ over Θ
limΘ−>0

Answer 19

sine cubed or sine 3*theta?

Answer 20

\[sin^3(\theta)\]?

Answer 21

sine3theta

Answer 22

\[\sin3\theta \over \theta\]

Answer 23

ah ok

Answer 24

how is it?

Answer 25

Its similar to the limit of sin(x)/x as x approaches 0 but instead of 1 its 3

Answer 26

then its 3

Answer 27

Yeah it turns out it wasnt \[sin^3\]

Answer 28

:)

Answer 29

while in my reference book 3 multiply in lim\[\lim_{\Theta \rightarrow 0}\]

Answer 30

then \[\sin 3\theta \over 3\theta\]

Answer 31

so the answer will be 3..

Answer 32

then its 1

Answer 33

told u the concept bfore ...it works on that

Answer 34

\[\lim_{\Phi \rightarrow 0}\sin \phi \div \phi = 1\]