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#