limit x->0, tan2x/x find the limit, help!!

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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

limit x->0, tan2x/x find the limit, help!!

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

l'Hopital's rule top and bottom. (2sec(x)^2)/1, limx->0 of 2sec(x)^2
we havent learned that rule, yet, so u wouldnt have tan=cos/sin?
sin/cos*

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

is it\[\tan(2x)\]or\[\tan^2(x)\]?
tan2x not ^2
I wonder, I don't think it's possible to solve tan(2x)/x without l'Hopital's. Unless I'm derping. Maybe some weird half angle identity?
my calc class just started last week n my professor has mentioned that rule but has yet to teach it, so wa does that rule state?
http://mathworld.wolfram.com/LHospitalsRule.html Apparently I spelled it wrong, but w/e. The proof is lovely.
you could try \[\lim_{x\to 0}\frac{\sin(2x)}{x\cos(2x)}\]
thats as far as i got what do i after that satel.?
break into to parts \[\lim_{x\to 0}\frac{1}{\cos(2x)}= 1\] and \[\lim_{x\to 0}\frac{\sin(2x)}{x}=2\]
How about\[\frac{\tan(2x)}x=\frac{\sin(2x)}{x\cos(2x)}=2\frac{\sin x}{x}\cdot\frac{\cos x}{2\cos^2x-1}=2\]sat beat me as usual...
yeah but you are quicker because i did not explain my second limit
Man, algebra. I forget that if I can't solve a problem, it's probably just me doing algebra badly.
I would write \[\frac{\sin(2x)}{x}=2\frac{\sin(2x)}{2x}\]
dang...lol
or replace x by x/2
turning test, so 2cos^2x-1 cancels the top, n ur left wit 2?
I didn't even need to do that identity, don't know why I did but you can just plug in x=0 to that second part and get 1
the answer is 2 tho
that's what you get in total\[\frac{\tan(2x)}x=\frac{\sin(2x)}{x\cos(2x)}=2\frac{\sin x}{x}\cdot\frac{\cos x}{\cos(2x)}=2\cdot1\cdot1=2\]
I didn't write the limits in ...lazy I know
got it!!

Not the answer you are looking for?

Search for more explanations.

Ask your own question