Question

How do you find the limit of `sinx/(x-pi)` as `x->pi`?

Answer 1

Use L'Hôpital's rule `lim_(x→pi)sin(x)/(x - pi)= lim_(x→pi)cos(x) = -1`

Explanation:

Use L'Hôpital's rule

The derivative of the numerator is `cos(x)`

The derivative of the denominator is 1 so we will not write it.

`lim_(x→pi)sin(x)/(x - pi) = lim_(x→pi)cos(x) = -1`

NOTE
The question was posted in "Determining Limits Algebraically" , so the use of L'Hôpital's rule is NOT a suitable method to solve the problem

Answer 2

I see that substitution gets us the indeterminate form `0/0`.

Explanation:

A limit like this reminds me that `lim_(thetararr0) sintheta/theta = 1`

And that makes me wonder if can use `sin(x-pi)` somehow.

The difference formula for the sine gets us:

`sin(x-pi) = sinx cos pi - cosx sinpi = sinx(-1) - cosx (0) = -sinx`

Therefore we can replace `sinx` by `-sin(x-pi)`

`lim_(xrarrpi)sinx/(x-pi) = lim_(xrarrpi) (-sin(x-pi))/(x-pi)`

`= - lim_(theta rarr0) sintheta/theta` `" "` (with `theta = x-pi`)

`= -1`