#lim_(x->0) sin(4x)/tan(5x) = 4/5#
In order to solve this problem, we will need to make use of L'hopital's Rule. L'hopital's Rule states that, for functions #f(x)# and #g(x)# differentiable on an open interval around a given point #c# with #g'(x)!=0# for any #x!=c#:
#lim_(x->c) f(x)/g(x) = lim_(x->c)(f'(x))/(g'(x))#
For this to apply, however, several conditions must be true. First, it must be the case that #lim_(x->c)f(x) = lim_(x->c)g(x) = 0 or +- oo#. Second, #lim_(x->c)(f'(x))/(g'(x))# must exist.
The question asks us to find #lim_(x->0) (sin(4x))/(tan(5x))#. These functions are differentiable for an interval around #x=0# (#sin x# is continuous and differentiable everywhere, while #tan x# is differentiable on the interval #((-npi)/2, (npi)/2)#. Since we have #tan(5x)# here, this means that instead of #-npi/2 < x < npi/2# being our interval, we have #-npi/2 < 5x < npi/2 -> -npi/10 < x < npi/10#.
Via use of the chain rule and the derivatives of the basic trig functions, using #f(x) = sin(4x)# and #g(x) = tan(5x)#, we have:
#f'(x) = d/dx sin(4x) = 4 cos(4x)# and #g'(x) = d/dx tan(5x) = 5 sec^2(5x)#, which in turn yields #(f'(x))/(g'(x)) = (4 cos (4x))/(5 sec^2 (5x))#
Note that we cannot simply convert to #cos^3# here because, although #sec u = 1/cos u#, we have different terms for #u# here (#4x# vs #5x#). However, we know that anything times 0 is simply equal to 0, and that #cos 0 = 1#. This gives us:
#lim_(x->0)(f'(x))/(g'(x)) = (f'(0))/(g'(0)) = (4 cos (4*0))/(5 * sec^2(0)) = (4*1)/(5*1) = 4/5#
