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

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Answer 1 · 2016-07-08T19:29:19

-2

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

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.

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