Question

\[\LARGE \lim_{x\to0}\frac{1-\cos 2x}{x\cdot \sin x }\]

A) 1
B) 2
C) 3
D) 4

I need a hint please ...

Answer 1

\[\cos(2x)=1-2\sin^2x\]should help I think

Answer 2

Thank you... I'll see what I can do :D

Answer 3

I think I made a mistake somewhere...
attempted:

\[ \lim_{x\to0}\frac{1-(1-2\sin x)}{x\cdot \sin x}=\lim_{x\to0}2\cdot \frac{\sin x}{x}\cdot \sin x=\]

\[=2\cdot \sin 0= 2\cdot 0=0 \]

What's wrong ? :(

Answer 4

you forgot the squared

Answer 5

I just made a mistake... it is \[\lim_{x\to0}\frac{1-(1-2 \sin^2x)}{x\cdot \sin x}\]

Answer 6

right, follow that through

Answer 7

but still the answer holds , what's wrong?

Answer 8

no you made a mistake
how did you wind up with sin^2 in the numerator and no sin in the denom?
you made an algebra mistake

Answer 9

where did sin in the denominator go?

Answer 10

...also you can't plug in zero where you did, because you still have x in the denominator.

Answer 11

so...

\[\lim_{x\to0}\frac{1-(1-2 \sin^2x)}{x\cdot \sin x}= \lim_{x\to0}\frac{2\sin x}{x}=2 !!\]

Dumb answer !! how it is ??

Answer 12

I didn't plug nothing .. that's why should be wrong, but I'm confused ^o)

Answer 13

you are supposed to recognize that\[\lim_{x\to0}\frac{1-(1-2 \sin^2x)}{x\cdot \sin x}= \lim_{x\to0}\frac{2\sin x}{x}=2 \lim_{x\to0}\frac{\sin x}{x}=2(1)=2\]did you not know that\[ \lim_{x\to0}\frac{\sin x}{x}=1\]?

Answer 14

Absolutely .... I don't need to lie check out my work that I did a few minutes ago, I USED that formula... but here's a question : We didn't plug nothing is it supposed x to attend to zero ?? x->0 we didn't plug anything in

Answer 15

yes, x tends to 0
it doesn't have to "become" zero in the limit
it is a removable discontinuity at x=0 because\[ \lim_{x\to0^-}\frac{\sin x}{x}=\lim_{x\to0^+}\frac{\sin x}{x}=1\]even though the function is not defined at x=0
that is the definition of a removable discontinuity: no plug-ins allowed

Answer 16

ahhh... now I get it, anyway , without your first hint I would not be able even to start it... Thanks in advance ;)

Answer 17

you're welcome :)