clara1223
  • clara1223
find the limit as x approaches 9pi/4 of (cos(x)-1)/6x
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.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
clara1223
  • clara1223
\[\lim_{x \rightarrow \frac{ 9\pi }{ 4 }}\frac{ \cos(x)-1 }{ 6x }\]
anonymous
  • anonymous
substitute
thomas5267
  • thomas5267
\[ \lim_{x\to0}\frac{\cos(x)-1}{x}=0 \]

Looking for something else?

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

More answers

zzr0ck3r
  • zzr0ck3r
it is defined on a continuous function :)
freckles
  • freckles
what is that one commercial that says plug it in plug it in
clara1223
  • clara1223
@freckles @satellite73 Alright plugging it in I get \[\frac{ \cos(\frac{ 9\pi }{ 4 })-1 }{ 6\frac{ 9\pi }{ 4 } }\] \[\frac{ \frac{ \sqrt{2} }{ 2 }-\frac{ 2 }{ 2 } }{ \frac{ 54\pi }{ 4 } }\] \[\frac{ \sqrt{2}-2 }{ 2 }\times \frac{ 4 }{ 54\pi }\] \[\frac{ 4(\sqrt{2}-2) }{ 108\pi }\] \[\frac{ \sqrt{2}-2 }{ 27\pi }\] Is there any way to simplify it further?
clara1223
  • clara1223
is this the final answer?
myininaya
  • myininaya
I don't see anything wrong with your answer
myininaya
  • myininaya
looks totally awesome!
clara1223
  • clara1223
Great! thanks so much!

Looking for something else?

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