Question

may you solve it?
limx→02x2/1−cosx

Answer 1

Ok ok this will try to write the bottom with one term by multiplying by bottom's conjugate
We still need to use those two limits (or at least of the two) for this one also

Answer 2

\[\lim_{x \rightarrow 0}\frac{2x^2}{1-\cos(x)}\]
So we multiply (1+cos(x)) on top and bottom

Answer 3

ok freind

Answer 4

well...

Answer 5

not a pal of l'hopital i guess

Answer 6

lol thats a good guy to use here but i like the manipulations ;)

Answer 7

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

Answer 8

that is the first step I was asking you to do
so you can write the bottom as one term

Answer 9

but if you prefer sat's way I will let him by himself with you

Answer 10

fine. while you struggle i will write
\[\lim_{x\to 0}\frac{2x^2}{1-\cos(x)}=\lim_{x\to 0}\frac{4x}{\sin(x)}=\lim_{x\to 0}\frac{4}{\cos(x)}\]

Answer 11

and then wait for the algebra

Answer 12

\[\lim_{x \rightarrow 0}\frac{2x^2 (1+\cos(x))}{\sin^2(x)}=2 \lim_{x \rightarrow 0}\frac{x^2}{\sin^2(x)}(1+\cos(x))\]

Answer 13

What does x/sin(x)->? as x->0

Answer 14

x/2

Answer 15

\[\lim_{x \rightarrow 0}\frac{x}{\sin(x)}=1 \text{ by squeeze thm}\]

Answer 16

actually this is pretty snappy isn't it?

Answer 17

\[\lim_{x \rightarrow 0}(\frac{x}{\sin(x)})^2=1^2=1\]

Answer 18

For next part (1+cos(x))
This part is continuous at x=0 and se therefore you can just plug in x=0

Answer 19

so then you have 2*1*( ?)

Answer 20

satellite73 You must have sense

Answer 21

What does that mean?

Answer 22

I wasn't with you

Answer 23

I was with her

Answer 24

well...
You say..