Question

This limit is driving me crazy

Lim x^2COSX
x->0 ---------
2Sin^2(x/2)

Answer 1

\[\lim_{x \rightarrow 0}\frac{ x ^{2}cosx}{ 2\sin^2(x/2) } \]

Answer 2

does your question is like this????

Answer 3

yes

Answer 4

what always gets to me are the COS and SIn

Answer 5

Im supposed to calculate it but im having trouble with the COS and the SIN

Answer 6

ok wait let me solve it then i will explain it to you

Answer 7

if you substitute 0 for x you will get 0/0 form which is undefined case right

Answer 8

yes i understand

Answer 9

so you have to look for the denominator which is 2(sin(x/2))^2
now use half angle trigonometric identity

Answer 10

[sin(x/2)]^2 = 1/2[1-cosx]

Answer 11

putting this denomenator you will get

Answer 12

\[\lim_{x \rightarrow 0}\frac{ x^2cosx }{ 2.1/2.(1-cosx) } \]

Answer 13

2 . (1/2)(1-COSX) right?

Answer 14

ya now cancel 2 with 1/2

Answer 15

And it becomes one leaving me with

x^2COSX
---------
(1-COSX)

Answer 16

at this point i m stuck a little plz wait

Answer 17

use\[\lim_{x\to0}{\sin x\over x}=1\]

Answer 18

\[\lim_{x\to0}{x^2\cos x\over\sin^2(x/2)}=\lim_{x\to0}\cos x\cdot\lim_{x\to0}\left({1\over\frac{\sin (x/2)}x}\right)^2\]\[=4\lim_{x\to0}\cos x\cdot\lim_{x\to0}\left({1\over\frac{\sin (x/2)}{x/2}}\right)^2\]

Answer 19

@Chef86 does this make sense to you?

Answer 20

@TuringTest you missed a 2 in denominator in frist step, rest is fine

Answer 21

true the 2 is missing

Answer 22

ah you are right :)

Answer 23

\[\lim_{x\to0}{x^2\cos x\over2\sin^2(x/2)}=2\lim_{x\to0}\cos x\cdot\lim_{x\to0}\left({1\over\frac{\sin (x/2)}{x/2}}\right)^2\]

Answer 24

is there part of that you are having trouble with?

Answer 25

The Sin(x/2) after dividing would be Sin1 right

Answer 26

@TuringTest thanks for the info

Answer 27

@Chef86 I don't think I understand what you mean by "after dividing"
@03453660 welcome, please help Chef understand if you could, I have to leave
Bye y'all

Answer 28

leme solve it for u

Answer 29

from where to start?

Answer 30

Where Turing left off.

Answer 31

now apply limits to get result

Answer 32

is it fine now?