Question

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

Answer 1

`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`

Answer 2

`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`.