Ask your own question, for FREE!
Mathematics 8 Online
OpenStudy (anonymous):

Lim cotg2x/cotg5x X-->0

OpenStudy (gorv):

cotg means cot???

OpenStudy (anonymous):

ah yes

OpenStudy (gorv):

convert it into tan and then use (tanx)/x=1 as x-->0

OpenStudy (anonymous):

that would be tan5x/tan2x.now what? i mean there is not x.

OpenStudy (anonymous):

Expand the denominator using double angle results.

OpenStudy (anonymous):

Wait, actually don't do that. Try rationalising using tan2x/tan2x.

OpenStudy (anonymous):

Bah, that doesn't work either. Where did you get this question from?

OpenStudy (anonymous):

Wait, just change tan2x into (sin2x)(cos2x)

OpenStudy (anonymous):

and then you get y=0

OpenStudy (anonymous):

Try it out and tell me whether you get that.

OpenStudy (anonymous):

(sin2x)/(cos2x)*

OpenStudy (anonymous):

a book of mine.i was thinking about deviding both with an x.and then it would be like (tan2x/x and thhis devided by tan5x/x .and we know that tan2x/x is equal 2 and tan5x devided by x is 5 so it all equals 2/5 .what do u think?

OpenStudy (anonymous):

Read what I wrote above.

OpenStudy (anonymous):

Dang sorry, that doesn't work.

OpenStudy (anonymous):

Sorry tan2x/x does not equal 2.

OpenStudy (anonymous):

It's getting really late here, so I have to leave this with you. I'm sure someone else here will be glad to help you more than I did. Sorry about the inconvenience.

OpenStudy (anonymous):

its ok,thanks anyway]

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Latest Questions
Arriyanalol: i need help
16 hours ago 7 Replies 1 Medal
Arriyanalol: CONGRATS to me
19 hours ago 8 Replies 2 Medals
Legomaster346: What colleges are good that are near Arab Alabama?
1 week ago 5 Replies 0 Medals
Arriyanalol: i need help
1 week ago 6 Replies 2 Medals
Arizona: does anyone know the percentage rate of 30,000 and 40,000
1 week ago 0 Replies 0 Medals
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!