anonymous
  • anonymous
what is the limit as x approaches 0 when ((1-cosx)^2)/x
Mathematics
katieb
  • katieb
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

amistre64
  • amistre64
0 i think
amistre64
  • amistre64
2(1-cosx) . sin(x) = 0
anonymous
  • anonymous
Yes L'pthals

Looking for something else?

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

More answers

amistre64
  • amistre64
yep :)
amistre64
  • amistre64
otherwise you expand the top and see what you can do to it..
anonymous
  • anonymous
((1-cosx)(1-cosx))/x = 0?
amistre64
  • amistre64
keep going....
amistre64
  • amistre64
1 -2cos(X) + cos^2(x) -------------------- .... x
anonymous
  • anonymous
doesn't (1-cosx)/x = o
amistre64
  • amistre64
dunno; its been so long since I tried it the hard way :)
amistre64
  • amistre64
you could try squeeze this on a graph
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=lim{x+to+0+}%28%281-cosx%29^2%29%2Fx
anonymous
  • anonymous
But, (1-cosx)/x times (1-cosx)/x = (1-cosx)^2 / x^2.......
anonymous
  • anonymous
Why would you do that?
amistre64
  • amistre64
i dunno why it cuts it short; but copy paste that link itno your address bar
anonymous
  • anonymous
@amogh - I'm just pointing out that I'm doing it wrong...
amistre64
  • amistre64
1 Attachment
anonymous
  • anonymous
Yes, x will go with either of them, not both of them
anonymous
  • anonymous
@amistre64: do you like ff4?
anonymous
  • anonymous
Sh@t! I did the problem wrong on my test then.......
anonymous
  • anonymous
What did you answer?
anonymous
  • anonymous
Thank you for the Wolfram alpha link amistre64. How much is that?
anonymous
  • anonymous
I answered "0" but I didn't do my work correctly (analytically), so I get the problem wrong

Looking for something else?

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