Ask your own question, for FREE!
Mathematics 20 Online
OpenStudy (anonymous):

can anyone help me with this ? lim x approach 2 ( (x^2 -4)^3 cos (π/x+2) )

OpenStudy (anonymous):

0

OpenStudy (anonymous):

how ? can you show me how you can get zero? Are you sure ?

OpenStudy (anonymous):

if you put 2 for x, you get (2^2-4)cos(pi/2+2)

OpenStudy (anonymous):

so you have 0*cos(pi/2+2)

OpenStudy (anonymous):

and result is 0

OpenStudy (anonymous):

I'm so sorry but I put wrong question. It should be x approach -2, not 2. if then do you know how to do it ?

OpenStudy (anonymous):

its the same: put -2 for x and you will get 0*(cos/-2+2)

OpenStudy (anonymous):

and its 0 again

OpenStudy (anonymous):

we cannot do that since cos ( π/ 0 ) is not zero. we will get math error for that

OpenStudy (anonymous):

it is cos(pi/(x+2)) or cos(pi/x+2) ?

OpenStudy (anonymous):

|dw:1324129387774:dw|

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!
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!