Question

I need help with this limit. NO L'H'S !

Answer 1

\[\lim_{x \rightarrow 0}~\frac{1-\cos(x)}{x^2}\]

Answer 2

multiple ways to solve this

multiply and divide by 1+ cos x

Answer 3

alternatively , you can use double angle formula for cos x
1-cos 2x = 2 sin^2 x

Answer 4

Okay, then I get ( 1-cos^2x) / x(1+cos(x) )

Answer 5

x^2 in the denominator, right??

and write 1-cos^2 x as sin^2 x

Answer 6

( 1-cos^2x) / x^2(1+cos(x) )

Answer 7

use the standard limit
lim x->0 sin x/x =1

Answer 8

after multiplying times 1+cos(x)\[\lim_{x \rightarrow 0} \frac{1-\cos^2x}{x^2(1+\cos(x)}\]then re-write as sin^2x,
\[\lim_{x \rightarrow 0} \frac{\sin^2x}{x^2(1+\cos(x)}\]
\[1+\lim_{x \rightarrow 0} \frac{1}{(1+\cos(x)}\]

Answer 9

this is what I am up to.

Answer 10

Wait no

Answer 11

\[1\times \lim_{x \rightarrow 0} \frac{1}{(1+\cos(x)}\]

Answer 12

misused the rule

Answer 13

now you can simply plug in x=0 :)

Answer 14

1+(1+1) =1/2

Answer 15

tnx man

Answer 16

\(\huge \checkmark\)
welcome ^_^

Answer 17

if curious, you can try the alternative method too :)

Answer 18

well, this looks the most simple.

Answer 19

tnx for the help ! Didn't think of a conjugate :)

Answer 20

you won't appreciate the simplicity of an easy method unless you try the not-so-easy one :)

Answer 21

yeah? which one ?

Answer 22

No, I really DO appreciate the easiness of this approach, because it is quite simple, even though I didn't come up with it on my own.