maths
posted by khela .
find the lim (cos x1)/ (sin 2x)
x>0

maths 
Reiny
can you use L'Hopital's rule?
if so
lim (cos x1)/ (sin 2x)
x>0
= lim (sinx)/(2cos 2x) as x > 0
= 0/2 = 0 
maths 
khela
no u cant

maths 
Reiny
some common trig limits you should have come across:
lim( (sinx)/x ) as x > 0 = 1
lim( x/sinx ) as x > 0 = 1
lim( (cosx  1)/x) as x  0 = 0
so lets rewrite the question as
lim[ ((cos x1)/x) * x(2sinxcosx) ]
= lim( (cosx  1)/x) * lim (x/sinx) * lim (1/2cosx)
= 0(1)(1/2) = 0
Respond to this Question
Similar Questions

tigonometry
expres the following as sums and differences of sines or cosines cos8t * sin2t sin(a+b) = sin(a)cos(b) + cos(a)sin(b) replacing by by b and using that cos(b)= cos(b) sin(b)= sin(b) gives: sin(ab) = sin(a)cos(b)  cos(a)sin(b) … 
algebra
Can someone please help me do this problem? 
Calculus
yes! tnk u ok? It's actually (x>0.) Find the limit of cot(x)csc(x) as x approached 0? 
Mathematics  Trigonometric Identities
Let y represent theta Prove: 1 + 1/tan^2y = 1/sin^2y My Answer: LS: = 1 + 1/tan^2y = (sin^2y + cos^2y) + 1 /(sin^2y/cos^2y) = (sin^2y + cos^2y) + 1 x (cos^2y/sin^2y) = (sin^2y + cos^2y) + (sin^2y + cos^2y) (cos^2y/sin^2y) = (sin^2y … 
TRIG!
Posted by hayden on Monday, February 23, 2009 at 4:05pm. sin^6 x + cos^6 x=1  (3/4)sin^2 2x work on one side only! Responses Trig please help!  Reiny, Monday, February 23, 2009 at 4:27pm LS looks like the sum of cubes sin^6 x + cos^6 … 
Calculus 12th grade (double check my work please)
1.)Find dy/dx when y= Ln (sinh 2x) my answer >> 2coth 2x. 2.)Find dy/dx when sinh 3y=cos 2x A.2 sin 2x B.2 sin 2x / sinh 3y C.2/3tan (2x/3y) D.2sin2x / 3 cosh 3yz...>> my answer. 2).Find the derivative of y=cos(x^2) … 
Limit Calculas
Evaluate lim>4 sin(2y)/tan(5y) Here is what I have so far. I am not sure the next steps. Can someone help me? 
calculus
Evaluate lim>4 sin(2y)/tan(5y) Here is what I have so far. I am not sure the next steps. Can someone help me? 
Calculus L'Hopital Rule
lim x>0 (x)sin(x)/1cos(x). They both go to 0 so 0/0. now take L'H and product rule of top? 
calculus
using the squeeze theorem, find the limit as x>0 of x*e^[8sin(1/x)] what i did was: 1<=sin(1/x)<=1 8<=8*sin(1/x)<=8 e^(8)<=e^[8*sin(1/x)]<=e^(8) x*e^(8)<=x*e^[8*sin(1/x)]<=x*e^(8) lim x>0 [x*e^(8)] …