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

Question

Ezra Adams · Accepted Answer

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#

Adam Adkins · Answer

undefined To evaluate the limit of sin(3x)/x as x approaches 0, we can use the concept of L'Hôpital's Rule. By applying this rule, we differentiate both the numerator and denominator with respect to x. The derivative of sin(3x) is 3cos(3x), and the derivative of x is 1.

Taking the limit of the differentiated function, we have 3cos(3x)/1, which simplifies to 3cos(0) = 3.

Therefore, the limit of sin(3x)/x as x approaches 0 is equal to 3.