Question

Evaluate the limit `lim_(x rarr 1) sin(pix)/(1-x)`?

Answer 1

`lim_(xrarr-1)(sin(pi*x)/(1-x))=pi`

Explanation:

The original expression was in indeterminate form. Since you get `0/0` if you evaluate directly with `x=1`.
Apply the L'Hospital's Rule to solve this problem. How do you use L'hospital's rule to find the limit?
Here are my steps:
`lim_(xrarr1)(sin(pi*x)/(1-x))`
`=lim_(xrarr1)((d/dx*sin(pi*x))/(d/dx*(1-x)))`
`=lim_(xrarr1)((pi*cos(pi*x))/(-1))`
`=lim_(xrarr1)((pi*cos(pi))/(-1))`
`=pi*((-1)/-1)`
`=pi`

Answer 2

Answer: `pi`

Explanation:

Find the limit of
`lim_(x->1)(sin(pix))/(x-1)`

Note that this is an indeterminate form: `0//0`. We can use L'Hopital's Rule to solve this equation.

Applying L'Hopital's Rule, we get that
`lim_(x->1)(sin(pix))/(x-1)=lim_(x->1)(d/dx(sin(pix)))/(d/dx(x-1))`
`=lim_(x->1)(picos(pix))/(-1)`
`=lim_(x->1)-picos(pix)`

The limit of a continuous function at a point is just it's value there.

`=lim_(x->1)-picos(pix)=-picos(pi)=pi`

Answer 3

`lim_(x rarr 1) sin(pix)/(1-x) = pi`

Explanation:

We want to find the limit:

`L = lim_(x rarr 1) sin(pix)/(1-x)`

Let us make a simple substitution:

Let `\ \ u =pi(1-x)=pi-pix`

Then `pix = pi - u`, and `1-x=u/pi`
As `x rarr 1 => u rarr 0`

Then the limit becomes:

`L = lim_(u rarr 0) sin(pi-u)/(u/pi)`
`\ \ = lim_(u rarr 0) pi \ {sinpicosu-cospisinu}/u`
`\ \ = pi \ lim_(u rarr 0) {sinpicosu-cospisinu}/u`

Using `sinpi=0` and `cospi=-1` we have:

`L = pi \ lim_(u rarr 0) (sinu)/u`

And an elementary trigonometry calculus limit is that

`lim_(theta rarr 0) (sintheta)/theta = 1`

Leading to:

`L = pi`

If we look at the graph of the function:
graph{sin(pix)/(1-x) [-4.123, 5.74, -0.818, 4.11]}

We note that at `x=1` the function is continuous (so the limit exists), and has a value slightly above `3` consistence with our analysis.

Answer 4

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

Explanation:

The limit:

`lim_(x->1) sin(pix)/(1-x)`

is in the indeterminate form `0/0` so we can solve it using l'Hospital's rule:

`lim_(x->1) sin(pix)/(1-x) = lim_(x->1) (d/dx sin(pix))/(d/dx (1-x)) = lim_(x->1) (picos(pix))/(-1)=pi`

Alternatively it can be resolved algebraically by substituting `t= pi(1-x)`, so that `x=1-t/pi`:

`lim_(x->1) sin(pix)/(1-x) = lim_(t->0) sin(pi(1-t/pi))/(1-(1-t/pi)) = pi lim_(t->0) sin(pi-t)/t`

Using the formula for the sine of the sum of two angles:

`sin(pi-t) = sinpi cos(t) -cospi sin(t) = sin(t)`

then:

`lim_(x->1) sin(pix)/(1-x) = pi lim_(t->0) sint/t = pi`