Question

Question #8b406

Answer 1

`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]}

Answer 2

`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#

Answer 3

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`)

Answer 4

`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`.