#lim_(x->oo)3^x tan(k/3^x)#
Now, as #x->oo#, #3^x->oo#. So, setting #t=3^x#,
#=lim_(t->oo)t tan(k/t)#
#=lim_(t->oo)t/cot(k/t)#
The limit is of the form of #oo/oo#, so L'Hôpital's rule applies. Find, separately, the derivative with respect to #t# of the numerator and denominator.
Numerator: #d/dt(t)=1#
Denominator: #d/dt(cot(k/t))=d/dt(cos(k/t)/sin(k/t))=k/(t^2sin^2(k/t))#
Thus, the limit is equivalent to
#=lim_(t->oo)1/(k/(t^2sin^2(k/t)))#
#=lim_(t->oo)(t^2sin^2(k/t))/k#
#=lim_(t->oo)k(sin^2(k/t))/(k/t)^2#
As #t->oo#, #k/t->0#. Setting #s=k/t#,
#=k lim_(s->0)(sin^2(s))/s^2#
Since we know that #lim_(x->0)sin(x)/x=1# (you can confirm it with L'Hôpital's rule), we have the final answer
#=k#
