Question #8b406

4 Answers
Jan 9, 2018

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

Explanation:

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]}

Jan 9, 2018

lim_(xrarr1)sin(πx)/sin(3πx)=1/3

Explanation:

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

Jan 9, 2018

To find the limit without l'Hospital's Rule see below.

Explanation:

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)

Jan 9, 2018

1/3.

Explanation:

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.