Question #2f5d1

1 Answer
Jan 25, 2018

#lim_(xrarrpi)(cosx-sin((3x)/2))/(cosx + (sinx)/2)#

#lim_(xrarrpi)(cosx-sin((3x)/2))/(cosx + (sinx)/2) =(cospi-sin((3pi)/2))/(cospi + (sinpi)/2) #

#= (-1 - sin (66/14))/(-1 + 1)#

#= (-1+1)/(-1+1)#

#= 0/0#

Apply L'Hopital's rule

#lim_(xrarra)f(x)/g(x) = lim_(xrarra)(f'(x))/(g'(x))#

#f(x) = cosx-sin((3x)/2)#
#g(x) = cosx + (sinx)/2#

Taking derivative of numerator

#d/dx(cosx-sin((3x)/2)) = d/dx(cos(x)) - d/dx(sin((3x)/2))#

#d/dx(cos(x))= -sin(x) #

Apply the chain rule

#d/dx(sin((3x)/2)) #

#f= sin(u),u = (3x)/2#

#= \frac{d}{du}(\sin \ (u\ )\ )\frac{cancel(d)}{cancel(d)cancel(x)}\ (\frac{3cancel(x)}{2}\ )#

#= cos(u)3/2 = (cos((3x)/2)(3)/2)#

And bring back first derivation

#= -sin(x) - (cos((3x)/2)(3)/2)#

Taking derivative of denominator

#d/dx(cos(x) + sin((x)/2))#

#= d/dx(cos(x)) + d/dx(sin((x)/2))#

# d/dx(cos(x)) = -sin(x)#

Apply the chain rule

#f = u,u = x/2#

#d/dx(sin((x)/2)) = d/(du)(sin(u))cancel(d)/cancel(dx)(cancel(x)/2)#

#= cos(u)1/2#

#= cos((x)/2)1/2#

Bring back 1st derivation

#= -sin(x)+ cos((x)/2)1/2#

Looking both the numerator and denominator i can guess its again a zero , so L'Hopital's rule again.

Numerator

#d/dx(-sin(x) - cos((3x)/2)) = -d/dxsin(x) - d/dxcos((3x)/2)#

#-d/dxsin(x) = -cos(x)#

#d/dx (cos((3x)/2)(3)/2)) = 3/2(d)/(du)cos(u)d/dx((3x)/2) #

#= 3/2(-sin(u))3/2#
#= 3/2(-sin((3x)/2))3/2#
#= -9/4 sin((3x)/2)#

Numerator = #-cos(x)+9/4 sin((3x)/2)#

Denominator = #-sin(x)+ cos((x)/2)1/2#
#d/dx(-sin(x)+ cos((x)/2)1/2) = d/dx(-sin(x)) + d/dx(cos((x)/2)1/2)#

#= -cos(x) + 1/2d/(du)cos(u)d/dxx/2#

#= -cos(x) + 1/2(-sin(x/2))1/2#

#sin(x) = 1/(csc(x))#

#= -cos(x) + 1/(csc(x/2) *2 *2)#

#= -cos(x) + 1/(4csc(x/2))#

#= -cos(x) + (-sin(x/2))/4#

#= -cos(x) - (sin(x/2))/4#

Now plug in #pi# instead of #x#

Numerator
#-cos(x)+9/4 sin((3x)/2) = -cos(pi)+9/4sin((3pi)/2)#

#sin(x) = cos(pi/2-3pi/2)#

#= -cos(pi)+9/4 cos(pi/2-(3pi)/2)#

#= -cos(pi)+9/4 cos((cancel(-2)^(-1)pi)/cancel(2)) #

# -cos(pi)+9/4 cos(-pi)#

Remember cos(-x) = cos(x)

# = -cos(pi)+9/4 cos(pi)#

#= cos(pi)(-1 + 9/4)#

#= cos(pi)((-4 + 9)/4)#

#= -1 * 5/4#

#= -5/4#

Denominator

#-cos(x) - (sin(x/2))/4#

#= -cos(pi) - (sin(pi/2))/4#

# -1(-1)-1/4#

#(1*4-1)/4#

#= 3/4#

So #lim_(xrarrpi)(cosx-sin((3x)/2))/(cosx + (sinx)/2) = -5/4 *4/3#

#=- 5/3#