Calculus 1 question about limits

Limit of 3X/tanX as x approaches 0.

I said it was undefined because I ended up using the sandwich theorem to find the limit of XsinX as x approached 0 and I got 0 and the limit of X3XcosX is 0 (resulting in a limit of 0/0). However I am usually wrong (lol) so I was wondering if I could be shown the correct way to do it WITHOUT using l'hospital's rule (if it even applies)...I have not learned that yet :). Thank you so much in advance.

L'Hôpital's rule does apply, but as usual you don't need it. All you need to know is that limits distribute over basically everything, the special limit
lim[x->0] sin(x)/x = lim[x->0] x/sin(x) = 1

and that tan(x) = sin(x)/cos(x).

3x / tan(x) = 3x / (sin(x) / cos(x)) = 3xcos(x) / sin(x) = (x/sin(x))(3cos(x))

Then limits distribute over multiplication. So
lim[x->0] 3x/tan(x) = (lim[x->0] x/sin(x))(lim[x->0] 3cos(x)) = 1 * 3cos(0) = 1 * 3 * 1 = 3

lim 3x/tanx
x->0

lim 3x/[sinx/cosx]
x->0

lim 3x/{[sin(x)/x]x/cosx]
x->0

lim sin(x)/x =1
x->0

lim 3x/{1*x/cosx]
x->0

lim 3cosx
x->0

as lim x->0 then

3cos(0) =3

the limit is 3

limit (3x / tanx) = 0 / 0
x–>0


3x / tanx

3x cosx / sinx <<———— cos(0) = 1

3x / sinx <<———— limit (x–>0) [ sin(x) / x ] = 1


limit (3x / sinx) = 3 ★
x–>0