Using the limit definition, find the direction derivative of $f(x,y,z)=\sin (xyz)$ at the point $P(1,1,0)$ in the direction of $v=⟨2,1,1⟩$.
$D_vf(a) := \lim_{t \to 0} \frac{f(a+tv) - f(a)}{t}$
Using this definition I get:
$$\lim_{t \to 0} \frac{f((1+2t,1+t,0+t)) - f(1,1,0)}{t} =\lim_{t \to 0} \frac{\sin{(1+2t)(1+t)(t)} \ -\ \sin (1\cdot1\cdot0)}{t}\\ =\lim_{t \to 0} \frac{\sin{2t^3 + 3t^2 +t} \ -\ \sin 0}{t} \overset{L'Hôp}{=}\lim_{t \to 0} \frac{\cos{(2t^3 + 3t^2 + t)}\ \cdot\ (6t^2 + 6t +1)}{1}= 1$$
Is this the correct approach or did I do something wrong?