How do you find the limit of #(sin(7 x)) / (tan(4 x))# as x approaches 0?

1 Answer
Mar 3, 2016

7/4

Explanation:

Let #f(x)=sin(7x)/tan(4x)#

#implies f(x)=sin(7x)/(sin(4x)/cos(4x))#

#implies f(x)=sin(7x)/sin(4x)*cos(4x)#

#implies f'(x)=lim_(x to 0){sin(7x)/sin(4x)*cos(4x)}#

#implies f'(x)=lim_(x to 0){(7*sin(7x)/(7x))/(4*sin(4x)/(4x))*cos(4x)}#

#implies f'(x)=7/4lim_(x to 0){(sin(7x)/(7x))/(sin(4x)/(4x))*cos(4x)}=7/4{lim_(x to 0)sin(7x)/(7x))/(lim_(x to 0)sin(4x)/(4x))*lim_(x to 0)cos(4x)=7/4*1/1*cos(4*0)=7/4*cos0=7/4*1=7/4#