Question #8b406

#lim_(x->1)sin(pix)/sin(3pix)#

This limit is in the form #0/0#, so we can use l'Hôpital's rule, which states that, if a limit #lim_(x->c)f(x)/g(x)# is of the indeterminate forms #0/0# or #oo/oo#, then #lim_(x->c)f(x)/g(x)=lim_(x->c)(f'(x))/(g'(x))#.

So, find the derivatives of the numerator and denominator separately:
#d/dx(sin(pix))=d/dx(pix)*d/(d(pix))(sin(pix))=picos(pix)#
#d/dx(sin(3pix))=d/dx(3pix)*d/(d(3pix))(sin(3pix))=3picos(3pix)#

Then,
#lim_(x->1)sin(pix)/sin(3pix)=lim_(x->1)(picos(pix))/(3picos(3pix))=(picos(pi))/(3picos(3pi))=1/3#.

We can verify this limit with a graph:
graph{sin(pix)/sin(3pix) [-5,5,-2.5,2,5]}

#lim_(xrarr1)sin(πx)/sin(3πx)#

We have the #0/0# indeterminate form. Both functions #sin(3πx)#, #sin(3πx)# are differentiable so we can use Rules De L'Hospital

We have,
#lim_(xrarr1)sin(πx)/sin(3πx)=lim_(xrarr1)((sin(πx))')/((sin(3πx))')# #=#

#=# #lim_(xrarr1)(cos(πx)(πx)')/(cos(3πx)(3πx)')# #=#

#lim_(xrarr1)(πcos(πx))/(3πcos(3πx))# #=#

#1/3lim_(xrarr1)(cos(πx))/(cos(3πx))# #=#

#1/3*1=1/3#

because #cosπ=-1# , #cos(3π)=-1#

NOTE: #cos3pi=cos(2i+pi)=cos2pi*cospi-sin2pi*sinpi= 1*(-1)-0*0=-1#

As #xrarr1#, we know #x-1rarr0#, so let #u = x-1#.
(This makes #x = u+1#)

We seek #lim_(urarr0) sin(pi(u+1))/sin(3pi(u+1)) = lim_(urarr0)sin(pi u + pi)/sin(3pi u +3pi)#

Use the sum formula for sine to get

#sin(piu + pi) = -sin(pi u)# #" "# and #" "# #sin(3piu + 3pi) = -sin(3pi u)#

Our limit becomes:

#lim_(urarr0) sin(piu)/sin(3piu) = lim_(urarr0) 1/3 (sin(piu)/(piu)) ((3piu)/sin(3piu))#

# = 1/3(1)(1)#

(using the trigonometric limit #lim_(trarr0)sint/t=1#)

Here, we seek for the Solution without using the L'Hospital's Rule.

Prerequisite : #lim_(theta to 0)sintheta/theta=1#.

Let, #x=1+theta :. x to 1, theta to 0#.

#:." The Reqd. Lim.="lim_(theta to 0)sin(pi(theta+1))/sin(3pi(theta+1))#,

#=limsin(pitheta+pi)/sin(3pitheta+3pi)#,

#=lim(-sinpitheta)/(-sin3pitheta)#,

#=lim{sin(pitheta)/(pitheta)*pitheta}/{sin(3pitheta)/(3pitheta)*3pitheta}#,

#=1/3{lim_(pitheta to 0)sin(pitheta)/(pitheta)}/{lim_(3pitheta to 0)sin(3pitheta)/(3pitheta)}#,

#=1/3*(1/1)#.

#rArr " The Reqd. Lim.="1/3#.