#lim_(xto1) (sinpix)/(x-1)# ?

2 Answers
Aug 20, 2017

#lim_(xto1) (sinpix)/(x-1) = -pi#

Explanation:

The value of #sin(pi(1))/((1)-1)# is the indeterminate form #0/0#, therefore, the use of L'Hôpital's rule is warrented.

Given: #lim_(xto1) (sinpix)/(x-1)#

In accordance with L'Hôpital's rule, differentiate both the numerator and the denominator:

#lim_(xto1) ((d(sinpix))/dx)/((d(x-1))/dx)#

#lim_(xto1) (pi(cospix))/1 = picos(pi(1)) = -pi#

The rule states that the limit of the original expression goes to the same value:

#lim_(xto1) (sinpix)/(x-1) = -pi#

Aug 20, 2017

#lim_(x -> 1) sin(pix)/(x-1) = -pi#.

Explanation:

Let #u=pi(x-1)#. As #x -> 1#, #u -> 0#.

Then,

#lim_(x -> 1) sin(pix)/(x-1) = lim_(u -> 0) sin(u+pi)/(u/pi)#

By the sine addition rule, #sin(u+pi) = sin(u)cos(pi) + cos(u)sin(pi)#, #sin(u+pi)=-sin(u)#.

Then,

#lim_(x -> 1) sin(pix)/(x-1) = -pi lim_(u -> 0) sin(u)/u#.

The limit of #sin(theta)/theta# as #theta -> 0# is a standard result and is #1#.

Then,

#lim_(x -> 1) sin(pix)/(x-1) = -pi#.