How to do this?

Quick method

Note that:

#lim_(t->0) sin t / t = 1#

#lim_(t->0) tan t / t = 1#

That is, both #sin t# and #tan t# behave like #t# for small values of #t#.

So:

#lim_(x->0) (sin 4x) / (tan 5x) = lim_(x->0) (4x)/(5x) = 4/5#

L'Hôpital's rule method

Note that:

#lim_(x->0) sin 4x = 0#

#lim_(x->0) tan 5x = 0#

So the requested limit is an indeterminate form #0/0# and hence L'Hôpital's rule applies.

#lim_(x->0) (sin 4x)/(tan 5x) = lim_(x->0) (d/(dx) sin 4x)/(d/(dx) tan 5x)#

#color(white)(lim_(x->0) (sin 4x)/(tan 5x)) = lim_(x->0) ((d/(dx) 4x)(cos 4x))/((d/(dx) 5x)(sec^2 5x))#

#color(white)(lim_(x->0) (sin 4x)/(tan 5x)) = lim_(x->0) (4 cos 4x)/(5 sec^2 5x)#

#color(white)(lim_(x->0) (sin 4x)/(tan 5x)) = (4 * 1)/(5 * 1)#

#color(white)(lim_(x->0) (sin 4x)/(tan 5x)) = 4/5#

#lim_(x->0)sin(4x)/tan(5x)#

Since #tan(theta)=sin(theta)/cos(theta)#, we have
#=lim_(x->0)(sin(4x)cos(5x))/sin(5x)#
#=lim_(x->0)sin(4x)/sin(5x)#

Now, arrange to
#=lim_(x->0)4/5*sin(4x)/(4x)*(5x)/sin(5x)#

If the factors are canceled out, this will be equal to the previous expression. The reason for this arrangement is to exploit the identity #lim_(x->0)sin(x)/x=1#.

Then, we have
#=lim_(x->0)4/5*lim_(x->0)sin(4x)/(4x)*lim_(x->0)(5x)/sin(5x)#

As #x->0#, #4x->0# and #5x->0#. Therefore, the above expression becomes
#=lim_(x->0)4/5*lim_(4x->0)sin(4x)/(4x)*lim_(5x->0)(5x)/sin(5x)#

Now we can use #lim_(x->0)sin(x)/x=lim_(x->0)x/sin(x)=1# to evaluate the individual limits:
#=4/5*1*1#
#=4/5#

We can verify this limit with a graph:
graph{sin(4x)/tan(5x) [-1,1,-1,1]}

As seen, as #x->0#, #sin(4x)/tan(5x)->4/5#.