Question

limits question

Answer 1

\[\lim_{x \rightarrow 0}\frac{1-\cos^{2}{x}}{ 3\sin{x} }\]

Answer 2

@ganeshie8

Answer 3

@aum

Answer 4

@ganeshie8 i know i have to simplify because i have to break that indetermination but i just don know how to

Answer 5

you may use `1-cos^2 = sin^2`

Answer 6

ok

Answer 7

\[\large \lim_{x \rightarrow 0}\frac{1-\cos^{2}{x}}{ 3\sin{x} } = \lim_{x \rightarrow 0}\frac{\sin^{2}{x}}{ 3\sin{x} } = \lim_{x \rightarrow 0}\frac{\sin x}{ 3} \]

Answer 8

take the limit ^^

Answer 9

ok

Answer 10

i guess the question is, how do get that 1-cos(x) = sin^2(x)

Answer 11

sorry 1-cos^2(x)

Answer 12

@ganeshie8 ?

Answer 13

thats a good question,
familiar with below trig identity ?
\[\large \sin^2x + \cos^2x = 1\]

Answer 14

is there a place where i could check the trig identities?

Answer 15

below pdf covers everything needed in calculus :
http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf

Answer 16

*trig identities needed for calculus

Answer 17

thanks