Question

What is `lim x->1` of `sin(pix)/(x-1)`?

Answer 1

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

Explanation:

Here we have an indeterminate case where the function tends to `0/0`, an indeterminate case. We can use L'Hospital's rule to obtain the limit, which states:

if `lim_(x->h) f(x)/g(x) ->0/0` then `lim_(x->h) f(x)/g(x) =lim_(x->h) (f'(x))/(g'(x))`

In other words differentiate the denominator and numerator separately and then evaluate the limit.

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

Thus the final answer is `-pi`. We can use the graph of `f(x)` to check:
graph{sin(pix)/(x-1) [-10, 10, -5, 5]}