could someone please answer this trig limit without l'hopital's rule?

I’ve got no idea about maths or why this popped up on my homepage so I’ll definitely need the hospital if I attempt to solve it
In my introductory calculus course, we learned that the limit as x approaches zero of sin(x)/x is 1. Do you know that? If so, you can do something like:

sin(4x)

= sin(2\*2x)

= 2sin(2x)cos(2x)

= 2( 2sin(x)cos(x) )( 1-2sin^(2)(x) )

= 4 ( sin(x)cos(x) )( 2cos^(2)(x)-1 )

So, the expression in our limit can be written as

(4sin(x)-sin(4x))/x^(3)

= ( 4sin(x)-4 (sin(x)cos(x))( 2cos^(2)(x)-1 ) )/x^(3)

= 4 \* sin(x)/x \* ( 1-cos(x)( 2cos^(2)(x)-1 ) )/x^(2)

= 4 \* sin(x)/x \* ( 1-2cos^(3)(x)+cos(x) )/x^(2)

= 4 \* sin(x)/x \* ( 2cos^(2)(x)+2cos(x)+1 )( 1-cos(x) )/x^(2)

= 4 \* sin(x)/x \* ( 2cos^(2)(x)+2cos(x)+1 )( 1-cos(x) )( 1+cos(x) )/( x^(2) (1+cos(x)) )

= 4 \* sin(x)/x \* ( 2cos^(2)(x)+2cos(x)+1 )( 1-cos^(2)(x) )/( x^(2) (1+cos(x)) )

= 4 \* sin(x)/x \* ( 2cos^(2)(x)+2cos(x)+1 )( sin^(2)(x) )/( x^(2) (1+cos(x)) )

= 4 \* sin^(3)(x)/x^(3) \* ( 2cos^(2)(x)+2cos(x)+1 )/(1+cos(x))

=4 \* (sin(x)/x)^(3) \* ( 2cos^(2)(x)+2cos(x)+1 )/(1+cos(x))

The limit of this as x approaches zero is 4 \* (1)^(3) \* (2+2+1)/(1+1)=10
by
I assume you're not supposed to use sin x = x - x\^3/6 + o(x\^3), as that makes the problem trivial.

In that case, use the identity

4 sin x - sin 4x = 4 sin\^3 x $1/(1 + cos x) + 2 cos x$

to show that the limit is 10.
My fist guess would be to use the double angle formula for sin twice.

sin(4x) = 4 cos^(3)(x) sin(x) - 4 cos(x) sin^(3)(x)

That might get you somewhere
How rigorous do you need to be? I mean, for a two-sided limit to exist, f(x) must move towards a fixed point, and the fixed point has to be the same coming from both directions. Plug in increasingly (very) small positive and negative x, you “clearly”  get arbitrarily close to 10 from both directions. Just trying to think what you can do without the tools of calculus or applicable trig identities.
Quotient and Chain Rule
Another answer using \*fairly simple\* trig equations:

You should know that sin 2x = 2 sin x cos x.

Using the fact, prove that lim x->0 (2 sin x - sin 2x) / (x\^3) = 1, and therefore lim x->0 (4 sin x - sin 4x) / (x\^3) = lim x->0 ((2 × (2 sin x - sin 2x) / (x\^3)) + 8 × (2 sin (2x) - sin (4x)) / ((2x)\^3)) = 2 + 8 = 10.
Use the maclaurin series for sin(x)
Can you use taylor expansions?