Question

How do you find the limit of `(sin(5x)) / (x-π)` as x approaches pi?

Answer 1

I used the fundamental trigonometric limit `lim_(thetararr0)sin theta / theta = 1`. I got a limit of `-5`. (Which I did check using graphing technology.)

Explanation:

To use the fundamental limt, the argument =of the sine function must be the same as the denominator and must have a limit of `0` adn `xrarr" whatever"` (In this case, as `xrarrpi`.)

We can get a `5x` in the denominator by multiplying by `5`, but when we do that we get

`sin(5x)/(x-pi) = 5 sin(5x)/(5x-5pi)`.

We need `sin(5x-5pi)` in the numerator.

Let's try using the difference formula for sine (if that doesn't work we'll try another formula or a different approach.)

`sin(5x-5pi) = sin(5x)cos(5pi)-cos(5x)sin)5pi)`

`= sin(5x)(-1)-cos(5x)(0)`

`= -sin(5x)`.

We can conclude that `sin(5x) = -sin(5x-5pi)`.

So we want

`lim_(xrarrpi)sin(5x)/(x-pi) = lim_(xrarrpi)[-5*sin(5x-5pi)/(5x-5pi)]`

`= -5 lim_(theta rarr0) sintheta/theta`

`= -5(1) = -5`