Question

Given `(sin(-2x)) / x` how do you find the limit as x approaches 0?

Answer 1

-2

Explanation:

one way

`lim_{x to 0} (sin(-2x)) / x`

let u = -2x

`= lim_{u to 0} (sin u)/( -u/2)`

`= -2 color{blue}{lim_{u to 0} (sin u)/( u)}`

`= -2 * 1 = -2` as the term in blue is a well known limit often proved using squeeze theorem

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OR

`lim_{x to 0} (sin(-2x)) / x`

`= - lim_{x to 0} (sin(2x)) / x`

then using the fact that `sin 2 psi = 2 sin psi cos psi`
`= - lim_{x to 0} (2sinx cos x) / x`

`= -2 lim_{x to 0} (sinx cos x) / x`

then lifting `cos x` out as it is continuous throught the limit and `cos 0 = 1`

`= - 2cos 0 color{green}{lim_{x to 0} (sinx ) / x}`

the term in green is indeterminate but as stated it is also a well known limit

you cannot use LHopital to prove this limit.