How do you evaluate the limit $sin(3x)/x$ as x approaches $0$ ?

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Answer 1 · 2016-10-22T18:43:01

Use $lim_(theta rarr 0)sin theta /theta = 1$ .

One way to use $lim_(theta rarr 0)sin theta /theta = 1$ is to use $theta = 3x$

But now we need $3x$ in the denominator.

No problem, multiply by $3/3$

$lim_(xrarr0) sin(3x)/x = lim_(xrarr0) 3 * sin(3x)/((3x))$

As $xrarr0$ , s also $3x rarr0$ . We can substitute to get

$lim_(theta rarr0)3*sin theta/theta = 3*1 = 3$

I like the first method (above) Here's a second method.

$sin(3x) = sin(x+2x) = sinx cos(2x)+cosx sin(2x)$

$= sinx(cos^2x-sin^2x) + cosx(2sinxcosx)$

$= sinx(3cos^2x-sin^2x)$

So

$lim_(xrarr0) sin(3x)/x = lim_(xrarr0)(sinx(3cos^2x-sin^2x))/x$

$= lim_(xrarr0)((sinx/x)(3cos^2x-sin^2x))$

$= (1)(3(1)^2 - 0^2) = 3$

How do you evaluate the limit sin(3x)/xsin(3x)/x as x approaches 00?