anonymous
  • anonymous
\[\LARGE \lim_{x\to0}\frac{1-\cos 2x}{x\cdot \sin x }\] A) 1 B) 2 C) 3 D) 4 I need a hint please ...
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
TuringTest
  • TuringTest
\[\cos(2x)=1-2\sin^2x\]should help I think
anonymous
  • anonymous
Thank you... I'll see what I can do :D
anonymous
  • anonymous
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 ? :(

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

TuringTest
  • TuringTest
you forgot the squared
anonymous
  • anonymous
I just made a mistake... it is \[\lim_{x\to0}\frac{1-(1-2 \sin^2x)}{x\cdot \sin x}\]
TuringTest
  • TuringTest
right, follow that through
anonymous
  • anonymous
but still the answer holds , what's wrong?
TuringTest
  • TuringTest
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
TuringTest
  • TuringTest
where did sin in the denominator go?
TuringTest
  • TuringTest
...also you can't plug in zero where you did, because you still have x in the denominator.
anonymous
  • anonymous
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 ??
anonymous
  • anonymous
I didn't plug nothing .. that's why should be wrong, but I'm confused ^o)
TuringTest
  • TuringTest
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\[ \lim_{x\to0}\frac{\sin x}{x}=1\]?
anonymous
  • anonymous
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
TuringTest
  • TuringTest
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
anonymous
  • anonymous
ahhh... now I get it, anyway , without your first hint I would not be able even to start it... Thanks in advance ;)
TuringTest
  • TuringTest
you're welcome :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.